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https://www.glassdoor.com.au/Interview/What-is-the-smallest-number-divisible-by-225-that-consists-of-all-1s-and-0s-QTN_116006.htm | 1,455,337,842,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701166141.55/warc/CC-MAIN-20160205193926-00314-ip-10-236-182-209.ec2.internal.warc.gz | 840,448,730 | 24,436 | Jane Street
## Interview Question
Interview New York, NY
# What is the smallest number divisible by 225 that consists
of all 1s and 0s?
7
11111111100
Interview Candidate on 12/10/2010
16
Two facts to note are: 1) Multiplies of 225 end in 00 25 50 75 2) A number is divisible by 9 if the digits sum to 9. Only those multiples ending in 00 could have only 1's and 0's. So, the smallest digit we can multiply 225 by to get 00 at the end is 4. That is, 225*4=900. Now, we want to find the smallest multiple of 900 that contains only 1's and 0's. Let us first focus on 9 and then tack on the 00 after. The smallest multple of 9 with only 1's and 0's is 111111111. This is a consquence of the fact that the digits must sum to 9. Now, we tack on the 00 at the end and obtain 11111111100.
Interview on 19/01/2011
0
^^^^^^WRONG 1111111110. Think of 225 as (5)^2(9). So this number must end in a 0 to be divisible by 5. Since every number that 5 divides ends in a 5 or 0, that number is also divisible by 5. Since a number divisible by 9 must have the digits sum to a number divisible by 9, then all we need is 9 1's and a 0 on the end for 1111111110.
Anonymous on 06/07/2012
4
1111111110 / 225 = 4938271.6
Anonymous is wrong on 14/07/2012
0
I fail to understand the above answers. Shouldn't it be 225 or 2250. 225/225 =1 and 2250/225 =10. Both these numbers contain just 1's and 0's :P
Anonymous on 11/10/2013
4
The above answers seem too complicated/not explained at all. First, notice that 225 = 25 * 9. (1) A number is divisible by 9 iff the sum of its digits is divisible by 9 => we must have 9 1's in our number. (2) A number is divisible by 25 iff the last two digits are divisible by 25 => our last two digits must be 0's. Putting (1) and (2) together, our number is 11111111100.
Anonymous on 09/09/2014 | 589 | 1,824 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2016-07 | latest | en | 0.911402 |
https://economics.stackexchange.com/questions/9559/how-to-relate-real-rate-of-return-on-capital-to-bond-interest-rate-lagrangian?noredirect=1 | 1,713,734,429,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817819.93/warc/CC-MAIN-20240421194551-20240421224551-00111.warc.gz | 197,383,983 | 40,236 | # How to relate real rate of return on capital to bond interest rate: Lagrangian
Suppose that household resource constrain equation is as follows: $$P_tC_t + Q_tB_t+ P_tI_t \leq W_tL_t+R_tK_t+B_{t-1}+D_t$$ where $P_t$ is price at time $t$, $Q_t$ is the price of one bond quantity, $B_t$ is bond quantity, $I_t$ is investment, $W_t$ is wage, $L_t$ is labor amount, $R_t$ is nominal capital rental rate, $K_t$ is capital at tie $t$, $D_t$ is dividends.
Taking it into Lagrangian at $t=0$: $E_0 \sum_{t=0}^{\infty}\beta^t U(C_t,L_t) - \lambda_t[P_tC_t + Q_tB_t+ P_tI_t - (W_tL_t+R_tK_t+B_{t-1}+D_t)]$
Taking partial derivative of lagrangian respect to $K_{t+1}$ where $K_{t+1} = (1-\delta)K_t + I_t$ seems to produce: $$\frac{\lambda_t}{\lambda_{t+1}} = \frac{R_{t+1}}{P_{t+1}}$$ (Dropped expectation sign, but should be there)
And taking partial derivative of the lagrangian respect to $B_t$:
$$\frac{\lambda_t}{\lambda_{t+1}} = \frac{P_t}{P_{t+1}}\frac{1}{Q_t}$$
Equating these two,
$$R_{t+1} = \frac{P_t}{Q_t}$$
Taking $-\log Q_t = i_t$
$$\hat{R_{t+1}} = \hat{P_t} - \hat{Q_t} = \hat{P_t}+i_t$$ where $\hat{X} = \log X$.
This does not seem to be a right formula to me, and I must have made some mistake. What did I do wrong here?
Ignoring the expectations operator, your lagrangean has two mistakes: first the constraint is per-period so it is also multiplied by the discount factor. Second, the way you have wrote the time indexes is inconsistent, as regards their interpretation for capital and bonds. If $K_t$ denotes "capital at the beginning of period $t$" (as it does), you should also write $B_{t}$ instead of $B_{t-1}$, to denote bonds held at the beginning of period $t$. This takes with it also the time index on $Q$.
Also I would advise to write your constraint in "normal form" (as they say in micro), namely with the constraint as "higher than or equal to zero".
Let's try this, up to a point.
$$\Lambda = \sum_{t=0}^{\infty}\beta^t \Big[U(C_t,L_t) + \lambda_t\big(W_tL_t+R_tK_t+B_{t}+D_t - (P_tC_t + Q_{t+1}B_{t+1}+ P_tI_t)\big)\Big]$$
$$= \sum_{t=0}^{\infty}\beta^t \Big[U(C_t,L_t)\\ + \lambda_t\big(W_tL_t+R_tK_t+B_{t}+D_t - P_tC_t - Q_{t+1}B_{t+1}- P_t\big[K_{t+1} - (1-\delta)K_t\big]\big)\Big]$$
Then
$$\frac{\partial \Lambda}{\partial K_{t+1}} = 0 \implies -\beta^{t}\lambda_t P_t + \beta^{t+1}\lambda_{t+1}[(1-\delta)P_{t+1} +R_{t+1}] =0$$
and
$$\frac{\partial \Lambda}{\partial B_{t+1}} = 0 \implies -\beta^{t}\lambda_t Q_{t+1} + \beta^{t+1}\lambda_{t+1} =0$$
Equating we get
$$\frac {[(1-\delta)P_{t+1} +R_{t+1}]}{P_t} = \frac 1{Q_{t+1}}$$
Now, $R_{t+1}$ is the nominal gross return to capital, so a $P_{t+1}$ is implicitly in there. If we set $r_{t+1}\equiv R_{t+1}/P_{t+1}$ the relation becomes
$$\frac {[(1+(r_{t+1} -\delta)]P_{t+1}}{P_t} = \frac 1{Q_{t+1}}$$
Taking logs, we get the Fisher equation written for period $t+1$.
• Just a minor comment: In equilibrium the return on the two assets $B_{t+1}$ and $K_{t+1}$ are identical as a result of the need for rate of change of marginal utilities $\lambda_{t}/\lambda_{t+1}$ to be equal across the two assets. Feb 17, 2021 at 2:31 | 1,163 | 3,131 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-18 | latest | en | 0.892638 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-10th-edition/chapter-2-differentiation-2-4-exercises-page-136/3 | 1,547,825,697,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583660175.18/warc/CC-MAIN-20190118151716-20190118173716-00626.warc.gz | 822,707,433 | 12,458 | ## Calculus 10th Edition
$u=x^3-7.$ $y=\sqrt{u}.$
Working from the inside out: Take $x$; cube it and then subtract $7$ from it $\leftarrow$ "inner function." Take $x^3-7$; raise it to the power of $\dfrac{1}{2}\leftarrow$ "outer function." $u=x^3-7.$ $y=\sqrt{u}.$ | 101 | 265 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2019-04 | latest | en | 0.486244 |
https://socratic.org/questions/how-do-you-factor-by-grouping-y-3-y-2y-2-2 | 1,601,102,502,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400234232.50/warc/CC-MAIN-20200926040104-20200926070104-00389.warc.gz | 632,088,595 | 5,724 | # How do you factor by grouping y^3+y+2y^2+2?
May 9, 2015
${y}^{3} + y + 2 {y}^{2} + 2$
$\left({y}^{3} + y\right) + \left(2 {y}^{2} + 2\right)$
Factor $y$ out of the first term.
$y \left({y}^{2} + 1\right)$
Factor $2$ out of the second term.
$2 \left({y}^{2} + 1\right)$
The common factor is $\left({y}^{2} + 1\right)$ .
$y \left({y}^{2} + 1\right) + 2 \left({y}^{2} + 1\right)$ =
$\left(y + 2\right) \left({y}^{2} + 1\right)$ | 210 | 436 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-40 | latest | en | 0.530506 |
null | null | null | null | null | null | Pecora in Perspective: Everything Old is New Again — the Pecora Commission Redux
pecora-ferdinand.thumbnail.jpgNancy Pelosi has called for the creation of a new Pecora Commission, and have been quite a few mentions of the Pecora Commission in the news lately. However, I have seen little that actually goes to first sources and explains exactly what the Pecora Commission was and what it did. I will attempt to fill that void.
I have been reading the final report and some of the hearing transcripts. What strikes me, first and foremost, is that they read as if the testimony was taken last week, instead of in the mid-1930’s. You don’t even have to change the names , very much. The same entities, for example Citigroup the known as National City Bank of New York and its stock trading arm National City Company, Inc. were advancing the “too big to fail” argument back then, too.
Executive compensation consisting of a “base pay” that represented a small percentage of total compensation plus various “bonus pools” were the norm for top executives back then as well. There were even off shore subsidiary tax avoidance schemes uncovered. And there were even pyramid scheme scandals.
Reading the “Report of the Committee on Banking and Currency” pursuant to Senate Resolution 84 of the 72nd Congress and Senate Resolutions 56 and 97 of the 73rd Congress is much like reading any newspaper today; the same schemes, the same scandals, the same excuses.
So, what is different? Well, back in the 1930s the concentration of wealth and power into the hands of a few Masters of the Universe was seen as dangerous and undemocratic. After all, the fewer the number of people making momentous decisions that would vastly affect the economy, the more likely that they will begin to act in tandem, so that mistakes in judgment by a few, and a few who mostly talked only to each other, could have disastrous consequences for the many.
Also, a lack of transparency in the markets created a climate where the temptation to fraud and self dealing was almost overwhelming, because the chances of being caught were so slight, so long as Ponzi scheme or fund continued to make payments to “investors”.
The report also decries the practice of making loans on margin to fuel speculation in stocks. Speculation is essentially a “bet” on whether a stock will go up or down, as opposed to investment which is a judgment about whether or not the investor thinks a particular company is being run well and profitably. Would you encourage a system whereby people take out loans to fund a gambling trip to Las Vegas or Atlantic City? I doubt it, yet investors, even small investors are encouraged to buy and sell “on margin” as if this were some kind of prudent practice. Today we call speculators “day traders,” but they are really just placing bets using the stock market as their betting forum.
Congress, in the response to the revelations that came out during testimony held from April 1932 to May 1934, passed a series of laws intended to remedy some of the worst abuses and to provide oversight and create some transparency in the markets. During this period, the Banking Act of 1933 (more commonly referred to as Glass-Steagall) was passed. It required the divorcement of commercial banks from their investment affiliates and created federal bank deposit insurance.
Congress also passed the Securities Act of 1933, which required issuers of securities to disclose material information about those securities to the public before offering those securities for sale. The Securities Exchange Act of 1934 which created the SEC. The Trust Indenture Act of 1940 placed additional registration requirements on a class of securities that included Bonds, debentures and notes offered for sale. The Investment Company Act of 1940 regulates mutual funds and similar entities. You can reach the full text of these laws via hotlinks at the SEC’s website.
This is the first part of a continuing series on the original Pecora Commission and its relevance today.
Comments are closed. | null | null | null | null | null | null | null | null | null |
http://freqnbytes.com/standard-error/calculate-standard-error-in-regression-analysis.php | 1,524,573,725,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946688.88/warc/CC-MAIN-20180424115900-20180424135900-00215.warc.gz | 122,141,765 | 5,488 | Home > Standard Error > Calculate Standard Error In Regression Analysis
# Calculate Standard Error In Regression Analysis
## Contents
Sign in to add this video to a playlist. Why I Like the Standard Error of the Regression (S) In many cases, I prefer the standard error of the regression over R-squared. The standard error statistics are estimates of the interval in which the population parameters may be found, and represent the degree of precision with which the sample statistic represents the population The only difference is that the denominator is N-2 rather than N. http://freqnbytes.com/standard-error/calculate-regression-standard-error.php
What is the Standard Error of the Regression (S)? And the uncertainty is denoted by the confidence level. Transcript The interactive transcript could not be loaded. What does it all mean - Duration: 10:07. http://blog.minitab.com/blog/adventures-in-statistics/regression-analysis-how-to-interpret-s-the-standard-error-of-the-regression
## The Standard Error Of The Estimate (for The Regression) Measures
It is only the context of your analysis that lets you infer that the "independent" variabes "cause" the variation in the "dependent" variable. The 9% value is the statistic called the coefficient of determination. http://blog.minitab.com/blog/adventures-in-statistics/multiple-regession-analysis-use-adjusted-r-squared-and-predicted-r-squared-to-include-the-correct-number-of-variables I bet your predicted R-squared is extremely low. Roman letters indicate that these are sample values.
In an example above, n=16 runners were selected at random from the 9,732 runners. Then subtract the result from the sample mean to obtain the lower limit of the interval. The Minitab Blog Data Analysis Quality Improvement Project Tools Minitab.com Regression Analysis Regression Analysis: How to Interpret S, the Standard Error of the Regression Jim Frost 23 January, 2014 How To Calculate Standard Error In Regression Model As a result, we need to use a distribution that takes into account that spread of possible σ's.
Model diagnostics When analyzing your regression output, first check the signs of the model coefficients: are they consistent with your hypotheses? How To Calculate Standard Error Of Regression Coefficient Somebody else out there is probably using the same data to prove that your dependent variable is "causing" one of your independent variables! National Center for Health Statistics typically does not report an estimated mean if its relative standard error exceeds 30%. (NCHS also typically requires at least 30 observations – if not more Sign in to report inappropriate content.
Standard error of regression slope is a term you're likely to come across in AP Statistics. Standard Error Linear Regression This is not supposed to be obvious. Since we are trying to estimate the slope of the true regression line, we use the regression coefficient for home size (i.e., the sample estimate of slope) as the sample statistic. Loading...
## How To Calculate Standard Error Of Regression Coefficient
Jim Name: Nicholas Azzopardi • Wednesday, July 2, 2014 Dear Mr.
A natural way to describe the variation of these sample means around the true population mean is the standard deviation of the distribution of the sample means. The Standard Error Of The Estimate (for The Regression) Measures Fitting so many terms to so few data points will artificially inflate the R-squared. How To Calculate Standard Error Of Regression In Excel Thanks for the beautiful and enlightening blog posts.
That's too many! this website Phil Chan 25,889 views 7:56 Understanding Standard Error - Duration: 5:01. Unlike R-squared, you can use the standard error of the regression to assess the precision of the predictions. If values of the measured quantity A are not statistically independent but have been obtained from known locations in parameter space x, an unbiased estimate of the true standard error of How To Calculate Standard Error Of Regression Slope
Does insert only db access offer any additional security How much should I adjust the CR of encounters to compensate for PCs having very little GP? Previously, we showed how to compute the margin of error, based on the critical value and standard error. In most cases, the effect size statistic can be obtained through an additional command. http://freqnbytes.com/standard-error/calculate-standard-error-regression.php more hot questions question feed about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Arts Culture / Recreation Science
The obtained P-level is very significant. Standard Error Multiple Regression The confidence level describes the uncertainty of a sampling method. Usually we do not care too much about the exact value of the intercept or whether it is significantly different from zero, unless we are really interested in what happens when
## The correlation coefficient is equal to the average product of the standardized values of the two variables: It is intuitively obvious that this statistic will be positive [negative] if X and
Specifically, the term standard error refers to a group of statistics that provide information about the dispersion of the values within a set. Available at: http://damidmlane.com/hyperstat/A103397.html. A medical research team tests a new drug to lower cholesterol. Confidence Interval Regression Analysis Also, the estimated height of the regression line for a given value of X has its own standard error, which is called the standard error of the mean at X.
However... 5. Get a weekly summary of the latest blog posts. Sign in to make your opinion count. see here Available at: http://www.scc.upenn.edu/čAllison4.html.
The confidence interval for the slope uses the same general approach. The range of the confidence interval is defined by the sample statistic + margin of error. Because these 16 runners are a sample from the population of 9,732 runners, 37.25 is the sample mean, and 10.23 is the sample standard deviation, s. The standard error is a measure of the variability of the sampling distribution.
The standard deviation of the age for the 16 runners is 10.23. The standard deviation of the age was 3.56 years. That statistic is the effect size of the association tested by the statistic. When calculating the margin of error for a regression slope, use a t score for the critical value, with degrees of freedom (DF) equal to n - 2.
v t e Statistics Outline Index Descriptive statistics Continuous data Center Mean arithmetic geometric harmonic Median Mode Dispersion Variance Standard deviation Coefficient of variation Percentile Range Interquartile range Shape Moments The standard error of the forecast gets smaller as the sample size is increased, but only up to a point. The confidence thresholds for t-statistics are higher for small sample sizes. How to compare models Testing the assumptions of linear regression Additional notes on regression analysis Stepwise and all-possible-regressions Excel file with simple regression formulas Excel file with regression formulas in matrix | 1,419 | 7,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-17 | latest | en | 0.813981 |
null | null | null | null | null | null | [box cover]
With her marriage coming to an end, Nadine Hightower (Kim Basinger) wants to get hold of some "artistic" photos she took on the hopes of getting into Playboy magazine. But when she goes to photographer Raymond Escobar's (Jerry Stiller) studio, the grumpy old coot winds up murdered with a knife in his back before he can give them to her. Desperate to get ahold of the pictures, Nadine enlists the help of her soon-to-be ex-husband, Vernon (Jeff Bridges), a dreamer who owns a run-down bar and can barely keep a roof over his head. However, when they return to the studio, the pair grab a file called "Nadine," which is not nudie picks but the State of Texas's secret blueprints for a new freeway. It's valuable information, worth untold riches for anybody savvy enough to scoop up some cheap land nearby, and that's what puts the Nadine and Vernon in the crosshairs of Buford Pope (Rip Torn), who would kill for that information, and already has. But with the prospect of a huge payday, the two team together while Buford sets his men out to find them. Nadine also has a secret she hasn't told Vernon, yet: She's pregnant. Writer-Director Robert Benton has had an odd career, working as an editor for Esquire, his big break came when he (with David Newman) wrote the screenplay for 1967's Bonnie and Clyde. This led to more scripts and a shot at directing in 1972 with the western Bad Company. Though he still works on screenplays (like 1978's Superman and most recently 2005's Ice Harvest) Benton won two Oscars for writing and directing 1979's Kramer vs. Kramer, and a third for his screenplay for 1984's Places in the Heart. Yet for such a successful career, his filmography is filled with long pauses (he's directed 10 films over the last 33 years), and he's jumped from genre to genre. 1987's Nadine is notable for being a screwball comedy, and — for a film that's dropped off the radar — it's a charming entertainment, notable for a very sharp third act. Basinger and Bridges have a great chemistry together (they later re-teamed as a broken up couple in 2004's Door in the Floor), and as the situation escalates, Benton stages some wonderful cat-and-mouse sequences with the right amount of danger and humor. It's a very modest picture, and noticeably brief (running a scant 82 minutes), but engaging as long as it plays. Sony/Columbia TriStar presents Nadine in a good anamorphic transfer (1.78:1) and Dolby 2.0 Stereo audio. Bonus trailers, keep-case.
Back to Main Page | null | null | null | null | null | null | null | null | null |
https://topwriters4me.com/category/mathemtics/ | 1,659,890,377,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570651.49/warc/CC-MAIN-20220807150925-20220807180925-00564.warc.gz | 529,104,022 | 12,917 | ## Solving Differential Equations Word Problems and Initial-Value Problems
1. Find a general solution of the differntial equation Then find a particular solution that satisfies the intiial condition . 2. A bacteria population is incresing according to the natural growth formula and numbers 100 at 12 noon and 156 at 1 p.m. Write a formula giving after hours. 3. Apply Euler’s method to the […]
## Ratio Calculation
Please see attachment. Selected Financial Ratios [L03, l04] The financial statements for Castile Products, Inc., an importer of consumer products, are given below: Castile Products, Inc. Balance Sheet December 31 Assets Current assets: Cash . Accounts receivable, net . Merchandise inventory . Prepaid expenses . Total current assets . Property and equipment, net . Total […]
## Domain and range of y=2x – 3
y=2x – 3 The solution gives the domain and range of y=2x – 3. To continue with the answer check on topwriters4me.com/
## Probability
A silver dollar is flipped twice, calculate the probability of each of the following occurring: A, a head on the first flip b, a tail on the second flip given that the first toss was head c, two tails d, tail on the first tail on the second e, tail on the first a head […]
## Mosquito growth rate
The population of a colony of mosquitoes obeys the law of uninhibited growth. If there are 1000 mosquitoes initially and there are 1500 after 1 day, what is the size of the colony after 2 days? How long is there until there are 30,000 mosquitoes? What is the size of the colony after 2 days? […]
## Green’s Theorem Used to Evaluate Integrals
Use Green’s Theorem to evaluate the line integral along the given positively oriented curve. Below, ‘S’ represents the integral sign. 1) S_c xy dx + y^5 dy C is the triangle with verticies (0,0), (2,0) and (2,1). 2) S_c (y + e^sqrt(x) )dx + (2x + cos(y^2) ) dy C is bounded by the region […]
## Probability : Addition and Multiplication Rules
It can be easy enough to get the addition rule and the multiplication rule confused. Tell me the difference between the two. Provide the notations and then tell me what type of problem I would use each one for. Addition Rule: Notation for Addition Rule: P(A or B) = P(event A occurs or event B […]
## Mathematical Statistics
Mathematical Statistics
## Equal Number Of Values
Equal Number Of Values | 567 | 2,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2022-33 | latest | en | 0.890005 |
http://mathhelpforum.com/advanced-statistics/22884-quick-distribution-help-please.html | 1,501,002,337,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425339.22/warc/CC-MAIN-20170725162520-20170725182520-00573.warc.gz | 211,088,645 | 10,553 | 1. ## quick distribution help please!
The figure below shows the standard normal distribution from statistics, which is given by the following equation.
Statistics books often contain tables such as the following (below right), which show the area under the curve from 0 to b for various values of b.
bArea1.34132.47723.49874.5000
Use the information given in the table and the symmetry of the curve about the y-axis to find the following.
(a) =
(b) =
some help...
2. Originally Posted by mathaction
The figure below shows the standard normal distribution from statistics, which is given by the following equation.
Statistics books often contain tables such as the following (below right), which show the area under the curve from 0 to b for various values of b.
bArea1.34132.47723.49874.5000
Use the information given in the table and the symmetry of the curve about the y-axis to find the following.
(a) =
(b) =
some help...
a) $\frac{1}{\sqrt{2 \pi}} \int_1^2 e^{-~\frac{x^2}{2}} ~dx$ $
=~\frac{1}{\sqrt{2 \pi}} \int_0^2 e^{-~\frac{x^2}{2}} ~dx-\frac{1}{\sqrt{2 \pi}} \int_0^1 e^{-~\frac{x^2}{2}} ~dx
$
That is the required area is written as the difference between two areas both of which are of the form given in the table.
RonL
3. b) Again write the area as the combination of two areas with $x=0$ as one end point:
$\frac{1}{\sqrt{2 \pi}} \int_{-3}^2 e^{-~\frac{x^2}{2}} ~dx$ $
=~\frac{1}{\sqrt{2 \pi}} \int_{-3}^0 e^{-~\frac{x^2}{2}} ~dx+\frac{1}{\sqrt{2 \pi}} \int_0^2 e^{-~\frac{x^2}{2}} ~dx$
Now use the symmetry about $x=0$ to change the first of the integrals abouve into the form given in the table:
$
\frac{1}{\sqrt{2 \pi}} \int_{-3}^0 e^{-~\frac{x^2}{2}} ~dx=
\frac{1}{\sqrt{2 \pi}} \int_{0}^3 e^{-~\frac{x^2}{2}} ~dx
$
So:
$\frac{1}{\sqrt{2 \pi}} \int_{-3}^2 e^{-~\frac{x^2}{2}} ~dx$ $
=~\frac{1}{\sqrt{2 \pi}} \int_{0}^3 e^{-~\frac{x^2}{2}} ~dx+\frac{1}{\sqrt{2 \pi}} \int_0^2 e^{-~\frac{x^2}{2}} ~dx$
RonL | 666 | 1,937 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-30 | longest | en | 0.831111 |
null | null | null | null | null | null | Some Demonstration Programs for Use in Teaching Elementary Probability and Statistics: Parts 3 and 4
Bruce E. Trumbo
California State University, Hayward
Copyright (c) 1995 by Bruce E. Trumbo, all rights reserved. This text may be freely shared among individuals, but it may not be republished in any medium without express written consent from the author and advance notification of the editor.
Restrictions. The programs accompanying this article are also copyrighted 1995 by Bruce E. Trumbo, all rights reserved. Permission to use the programs and any portion of this article for any nonprofit educational purpose is hereby granted, except that any use that involves gambling for money or for items or services of value is not permitted.
Although extensively tested, the computer programs should be regarded as developmental. They may contain errors, some of which may cause unpredictable results, including computer "crashes." Source code is not available. Please report errors to the author. NO WARRANTY OR REPRESENTATION OF FITNESS FOR ANY PURPOSE IS MADE OR IMPLIED. The user assumes all risks of any kind and waives the right to claim damages.
Requirements. Programs in this series are intended for use with IBM-PC compatible machines equipped with EGA graphics or better, although some of them may run successfully on other machines.
Download Programs to a Local File
Key Words: Bivariate normal distribution; Correlation; Expectation; Hypergeometric distribution; Gambling; Keno; Pedagogy; Simulation.
In this second paper of a series, two programs for EGA-equipped IBM-PC compatible machines are included with indications of their pedagogical uses in the teaching of elementary probability and statistics. Concepts illustrated include the coefficient of correlation, the expectation of a discrete distribution, the concept of a fair game, and the hypergeometric distribution. Three datasets useful for illustrating correlation are also documented and appended.
1. Introduction
1 As in the first paper in this series (Trumbo 1994), the emphasis here is on building student intuition and understanding of probability concepts through the use of simple computer programs in class. Both of the programs made available with this article are based on simulation. Of course, the results of simulations can be presented without having a computer, but the advantage of interactive computer use is that simulations can be repeated often enough that general principles can be discerned above the inevitable eccentricities of each individual simulation run.
2 The package of probability demonstration programs included with Trumbo (1994) consisted of the following: BRUN40.EXE (Microsoft utility), PROBDEMO.EXE (Entry program), PDMEN.EXE (Menu program), PDLLN.EXE (Part 1), and PDPPR.EXE (Part 2). The present article adds PDBNS.EXE (Part 3) and PDKEN.EXE (Part 4) to the list of programs released.
3 Each of the four demonstration programs released to date is called from a Main Menu activated by the command PROBDEMO. The menu program is written to detect the presence of these four programs (as well as others that may be released later), displaying only programs currently loaded into the same directory. The title page of each program tells how to start running it. For all except the program in Part 4, the F1 and F2 keys give technical details and additional options for more advanced users.
4 These programs were written for student use in computer labs, and most of the material provided here is intended to help instructors and students use them to the best advantage in that setting. However, if suitable projection equipment is available, the programs could also be used in large lecture sessions, perhaps with commentary partially based on the laboratory notes provided.
5 Readers are referred to Trumbo (1994) for further comments on the rationale, design, development, testing, advantages, limitations, and classroom use of the programs in this series.
6 Part 3 is intended to give students an intuitive grasp of the correlation coefficient, including an understanding of the meaning of various numerical values of correlation between -1 and +1. (Advanced students can benefit from establishing some of the distribution theory involved in the workings of the program.)
7 Part 4 allows students to play a simulation of the casino game Keno. It provides insight into the lure of a gambling game in which highly advertised large winnings are possible, but at which the player will, on average, lose more than a quarter of the money bet. The hypergeometric probability distribution is used to compute probabilities and expected winnings for this game---two crucial kinds of information not generally provided by casinos.
2. Part 3: Simulated Samples from a Bivariate Normal Distribution to Illustrate Various Correlations
2.1. Operation of the Program and Discussion of Its Purpose
8 This is a simulated drawing of 500 observations from a bivariate normal distribution. Both means are fixed at zero and both standard deviations are fixed at unity. The user may select any value of the population correlation \rho between -1 and 1.
9 A bivariate normal density function can be viewed as a mound-shaped surface above a plane. The shape of the mound varies as \rho varies, but it is always possible to find an ellipse in the plane above which exactly 95% of the probability (volume) of the mound lies. The shape of this ellipse can vary from circular (for \rho = 0) to "football" shaped (for \rho around +.7 or -.7) to "cigar" shaped (for \rho nearer to +1 or -1). In the present graphical display, the inclination of the principal axis of the ellipse changes from +45 degrees to -45 degrees depending on whether \rho is positive or negative. As the simulation begins, this ellipse appears in deep blue background on the plotting axes. Its shape provides a valuable visual link to the numerical value of \rho. When the simulation is complete this ellipse should contain about 95% of the points. Thus, about 5% of the 500 points (about 25) will lie outside the ellipse.
10 As the simulated points are sampled from the bivariate normal distribution they are plotted on the axes, and the updated sample correlation r is printed after every 5 observations. After all 500 points are plotted, the user is asked if he or she wants to do another simulation. The possible responses are K for "Keep the same value of \rho and run again," Y for "Yes, with the opportunity to select a different value of \rho," and N for "No, return to the title page."
11 Not surprisingly, most students who encounter the idea of correlation for the first time have no feel for what the various numbers in the allowable range from -1 to +1 mean. In particular, many expect that any correlation above .9 or below -.9 will correspond to almost perfect fit to a line. Exploring with this program helps to build intuition about sample correlations in several ways:
1. Students can experiment with several runs each of various values of the population correlation. The sample correlation in each case will be fairly near the selected value of \rho. Thus, they can form an impression of what a correlation of .5 or .9 or -.99 looks like.
2. The distinction between population and sample correlation is illustrated. The population correlation \rho is a parameter, which is chosen when one specifies a particular bivariate probability model. The sample correlation is a random quantity, slightly different with each run, resulting from the sampling process. The near agreement of the sample correlation with the population correlation shows that the sample correlation can be used as an estimate of the population correlation.
3. The ellipse in the background ("football" or "cigar" shape) gives students a vivid graphical image to associate with the population correlation.
12 The F1 key shows technical details about the generation of the bivariate normal observations. The Box-Muller method is used to convert pairs of independent uniform random variables, generated easily by the computer, into pairs of independent standard normal random variables, which are then transformed to yield a bivariate observation with the appropriate correlation. (See Zelen and Severo in Sec. 26.8 of Abramowitz and Stegun (1972) or Sec. 6.5.1 of Kennedy and Gentle (1980).) Plotting peculiarities for the special cases where \rho is equal to +1 or -1 are also discussed briefly. The F2 key shows additional options (monochrome operation, if needed for projection, and speed controls). Return to the title page from either the F1 or F2 screen by pressing ESC.
2.2. Before Going to the Lab
13 Ideally, the following should be done just after the concept of correlation or formula for the coefficient of correlation has been introduced in lecture:
14 Ask students what their reaction would be if you were to run your eye around the classroom and guess that the average height of the students present is about 75 inches (6'3"). Ask what a better guess would be.
15 Recall the Empirical Rule, which says that (for roughly normal data) 95% of the observations will fall within two standard deviations of the mean. Armed with this information, ask them what their reaction would be if you guessed the standard deviation of the weights in the class to be about two pounds. Ask what a better guess would be.
16 Ask for 10-15 student volunteers willing to disclose (truthfully) their heights and their weights. As each student provides this information, plot the point on an overhead projector or blackboard for all to see. Also input these numbers into a computer with projection display running a statistical package such as Minitab, or enter the numbers into a hand calculator. Prepare to reveal the correlation, but do not do so yet.
17 Alternative scenario: Gather the height and weight information at the end of the previous class and have an overhead projection slide and sample correlation figure ready. (I like to appear to live dangerously, but always have a back-up dataset from some previous class just in case not enough volunteers emerge or the computer crashes.)
18 Typically, of course, there is a positive association between heights and weights. Ask what the class thinks would be a good guess for the correlation represented by the scatterplot of heights and weights. If this discussion is taking place when the concept of correlation is very new to the class, you will mainly get blank stares and bad guesses. Reveal the correct answer.
19 Tell students that a major purpose of the computer demonstration to follow in lab is for them to be able to gain enough intuition that they will be able to make rough, but intelligent guesses as to the correlation represented by a scatterplot.
20 Optional additional demonstration: Show overhead projection slides with scatterplots of several bivariate datasets. (These should be scaled so that the height and width of each scatterplot are approximately equal, and so that each scatterplot uses most, but not all, of the height of the slide.) Ask students which slides represent a clear positive association, which represent clear negative association, and which represent no significant association at all. Comparing two datasets that exhibit positive association, point out the one with the larger association and stress that it will have a correlation nearer to +1 than the other.
21 Several datasets are documented in Appendix B and provided with this article. Other datasets might be obtained from the text you are using. The height and weight data collected in class might also be made available. For the larger datasets you will probably want to use Minitab or some other statistical package to produce the scatterplots.
22 In designing a scatterplot, attention needs to be given to the selection of the scales. The amount of unused space surrounding the data cloud can influence perception of the correlation (Cleveland et al. 1982). I think it is a mistake to introduce this complication too early while students are learning to use scatterplots to understand correlation. Thus the program has been designed so that the only changeable parameter is \rho. (Both population means are fixed at 0 and both population standard deviations are fixed at 1.) The parenthetical suggestion just above on scaling scatterplots for classroom demonstration is in this same spirit.
23 After students have had experience using scatterplots and begin to feel comfortable using them to interpret correlation, you may want to show them examples in which the same data are presented on several scatterplots with different scales. (See, for example, Moore 1995, page 112.) My preference for an elementary course is to do just enough of this so that students will be aware that proper scaling is important. Many computer packages for statistical analysis (including Minitab) give you the ability to manipulate scatterplot scales at will. At the suggestion of a referee I have added to my program the capability to double the scale of either or both axes; the blue ellipse (which, of course, is not a feature of ordinary scatterplots where scaling may be an issue) may also be suppressed. (From the Title Page of the program, press F2 for instructions on the use of the scaling and ellipse suppression options.)
24 There is considerable literature on the making and interpretation of scatterplots. Additional references, some dealing with issues beyond correlation, are Strahan and Hansen (1978), Shipp and Margolin (1982), Cleveland and McGill (1984), Raveh (1985), Huber (1987), Lewandowsky and Spence (1989), Meyer and Shinar (1992), and Spence and Garrison (1993).
2.3. Lab Instructions for Students
25 Type PROBDEMO at the DOS prompt and press ENTER at the "copyright" page to get to the "Main Menu," where you should select item 3. To adjust your monitor for the most effective use of this program, press ENTER to start the program and type in the value 0 for "rho". On this adjustment run, ignore the points and concentrate on the blue figure in the background.
(a) Adjust the brightness of your monitor so that this figure is clearly visible, but not really bright, and
(b) if your monitor has a vertical size adjustment, try to adjust it so that the figure is a perfect circle.
26 Notes: (1) The symbol \rho is not conveniently available on the PC screen, so the name of this symbol "rho" is spelled out. We use the symbol \rho in the rest of these lab notes. (2) In your work with this program, you may wish to change the speed at which it runs. Press \ (slow) or / (fast) at the title page or continuation prompt to do so; return to normal speed with |.
27 Type N to return to the title page. On the title page it is suggested that you try the following values of \rho: 0, -.2, .4, -.8, .9, -.99, .999, and 1. Do this. (Press Y after each run to do another run with a different value of \rho.) Then let a neighbor pick several of these values (perhaps changing the sign of some of them), covering the bottom line of the screen so that the values of \rho and r are not visible to you. Can you figure out which values of \rho your neighbor chose for each run? Pick several values of \rho for your neighbor to guess from scatterplots.
28 Some special features of the program may help in such guessing games: If the person guessing does not watch the simulation run, the person running the computer can press L when it is complete to hide the legend on the bottom line. If you are working on your own, return to the Title Page and press G. This will put the program into a mode in which it selects values of \rho for you, and hides the resulting numbers until the run is completed and you have had a chance to make your guess. Also, you can press X, at the Title Page or when asking for the next simulation run, to keep the ellipse from being plotted in the background; you will probably find that this makes guessing more challenging.
29 How accurate are your guesses expected to be? You are not trying to substitute your eyeball for a computer. You are only trying to get a rough idea what various values of r look like in practice. Correlations range from - 1 to +1. For values near +1 and -1 you should be able to guess r correct to the nearest 0.1 or even better; for values nearer to 0 you may be off by as much as 0.2 or even 0.3. If r is far enough from 0 you should have no trouble seeing whether it is positive or negative.
2.4. Questions
1. Do five runs of the program using any values of \rho you wish (for easiest viewing, it may be best to keep \rho between -.9 and +.9). The blue ellipse shown for each run is supposed to contain 95% of the 500 points. Thus, it is expected that roughly (.05)(500) = 25 points in each run will fall outside the ellipse. For each of the five runs do your best to count the points that fall outside. (For some of the ones near the boundary you may have to guess.) Do your results seem to confirm that about 5% of the points fall outside of the ellipse?
2. The value of the POPULATION correlation \rho that you choose is a constant for any one probability model; it is a fixed "population parameter." Notice, however, that the SAMPLE correlation r is a random variable. It will be different for each sample of 500 observations you select from the population. Why is r different from \rho?
(a) Choose \rho = .9, then do five runs with this same value of \rho (press K after all but the last run), writing down the five values of r that result.
(b) Repeat the steps in (a), but use \rho = .3.
(c) Are your values of r more variable in (a) or in (b)?
(Answer: Of course we cannot say for sure what happened in your particular simulation runs, but the theoretical variance of r [i.e., V(r)] is smaller when \rho = .9 than when \rho = .3. It would be very surprising if your sampling did not produce results in the same direction.)
3. For this exploration put the program into slow mode by pressing \ at the title page or the continuation prompt before starting each run. (You will see the \-symbol in the lower right-hand corner of the screen when you do.) In slow mode the first 20 points are plotted very slowly. For five runs with \rho = .9, try to note the value of r when n is about 10. (Values print only at n = 5, 10, 15, etc. Do not worry if you have to settle for 15 instead of 10 or if you botch a run completely and must try again.) For each of the five runs, note the values of r very early in the run (n = 10 or 15) and at the end of the run (n = 500). Anytime after you have written down the value of r early in each run, you can press | for medium or / for fast speed.
The sample correlation r is an estimator of the population correlation \rho. As with other widely used estimators in statistics, r tends to be a better estimator when it is based on a larger sample. Do your records from the five runs confirm this principle? (Answer: During a run, the values of r tend to fluctuate at first before beginning to settle down to something near \rho. Of course, it is possible for a run to begin with nearly the correct value of \rho and then to drift somewhat afield, but this is much less common.)
30 Note: Questions 4-6 use datasets provided in Appendix B. The instructions below are written for use with Minitab (and the Minitab worksheet versions of the data files), but they can be adapted for use with other statistical packages (and the ASCII versions of the data files). Especially if you are using Minitab, you may want to provide your students with a copy of Appendix C which shows explicitly how to proceed with Question 4.
4. In a study of the concentration of red blood cells in blood samples from newborn babies, two measures of red blood cells were used: The "hematocrit" is a measure of the percent by volume of blood that consists of red blood cells. The "hemoglobin" is a quantitative chemical analysis for the protein hemoglobin, the substance that gives these cells their characteristic red color. These are two ways to try to attach quantitative measures to the concept "concentration of red blood cells." They should be highly correlated. Anemic babies (ones with not enough red blood cells) should be low on both scales; polycythemic babies (ones with too many red cells) should be high on both scales.
In Minitab, retrieve the worksheet REDCELL.MTW and make a scatterplot of hematocrit (HCrit) against hemoglobin (Hgb). Try to find the center of gravity of the data cloud by estimating the sample means of each variable and finding the corresponding point on the plot. Try to imagine an ellipse centered there, of the right shape to match the data cloud, and of the right size to contain 95% of the data points---all but perhaps a couple of them. Try to make a rough intuitive guess as to the value of r. (Write down your guesses for the means and the correlation before you continue. It would be very surprising if your guesses were exactly correct, but you won't develop your ability to make educated guesses without practicing.)
Finally, use Minitab to find the means of the two variables and the correlation between them. How well did you locate the center of gravity? How good was your estimate of r?
5. The worksheet EUROPREC.MTW contains annual precipitation (in mm) for three European cities, Manchester, Paris, and Madrid, for the 100-year period 1870-1969. "Precipitation" is rainfall plus snowfall converted to equivalent amounts of rain. Weather patterns in Europe would lead to the supposition that rainfall for Manchester and Paris will show a significant correlation, whereas precipitation for Manchester and Madrid (distant cities in different climates) would not. Follow the same procedures (scatterplot, guessing, correlation, etc.) as in Question 4 twice: once for Manchester-Paris and once for Manchester-Madrid. What values of r did you guess? What are the actual computed values? Which pair of cities shows the highest correlation?
6. The worksheet RAINGRAD.MTW shows high school graduation rates (in percents) and typical annual precipitation (in inches) for the 50 United States plus the District of Columbia. Follow the same procedures as in Question 4 for plotting and guessing r. Can you think of a mechanism or rationale to explain this correlation? If so, write down your speculation. If not, write an explanation of how you think it was possible to find data with a value of r so far from 0. (Do this before you look at the answer.)
(Answer: Here is how the data were actually obtained: For such a small sample size (here n = 51) it does not take much looking around through an almanac to find two variables that happen to show a correlation quite different from 0. It would be very difficult to imagine a causative link between rainfall and high school graduation rates.)
2.5. Theoretical Problems for Advanced Students
31 Students near the end of a first calculus-prerequisite course in probability theory or students in a second probability course should be able to demonstrate the validity of the Box-Muller transformation and of the transformation used to obtain bivariate normal observations (as presented briefly on the page that shows when F1 is pressed from the title page of Part 3). These exercises are phrased formally below as Advanced Problems 1 and 2. Students can also be expected to derive the equation for the ellipse that contains 95% of the probability---Advanced Problems 3 and 4 below.
Advanced Problem 1. Let U and V be independent random variables, each distributed uniformly on the interval [0, 1). Show that the random variables W and X defined below are independent random variables, each distributed standard normal:
W = \sqrt{(-2 ln U)} \sin{(2 \pi V)},
X = \sqrt{(-2 ln U)} \cos{(2 \pi V)}.
Advanced Problem 2. If W and X are independent random variables, each distributed standard normal, and if Y is defined as below, then show that (X,Y) has a bivariate normal distribution with E(X) = E(Y) = 0, V(W) = V(Y) = 1, and correlation \rho:
Y = \rho X + \sqrt{(1 - {\rho}^2)} W.
Advanced Problem 3. With W and X defined as in Advanced Problem 2, show that the random vector (W,X) falls inside the circle
W^2 + X^2 = 5.99
with probability .95.
Advanced Problem 4. Use the result of Advanced Problem 3 and the definitions of X and Y in Advanced Problem 2 to find the equation of the ellipse that contains the point (X,Y) with probability .95.
2.6. Concluding Comments on Part 3
32 I usually use this program in class as soon as the idea of correlation has been introduced. Some elementary books introduce the sample correlation r without mentioning the population correlation \rho. In this case, I explain briefly in lecture that just as \mu is the population parameter corresponding to \bar{X} (quantifying centrality) and \sigma is the population parameter corresponding to s (dispersion), so---for bivariate data---\rho is the population parameter corresponding to r (association). Later, when the population correlation has been formally introduced, we take another look at the program, emphasizing the distinction between the population parameter and the statistic, which can be used as its estimate.
33 Finally, towards the end of a first calculus-based probability course or in a second course, I have found that students are motivated to look at the mathematics behind the program, as outlined in the advanced problems given above.
34 A demonstration somewhat parallel to this program, but not quite as graphically elegant or easy to use, can be done using Minitab (see Appendix A). This demonstration uses Minitab's procedure for generating standard normal random variables, so the procedure for obtaining standard normal variates from uniform ones is not explicitly displayed.
35 Note: Instructors interested in a comprehensive collection of simulation experiments in statistics using Minitab may wish to consider Keller (1994). Although this book has no experiments specifically dealing with correlation, and its primary emphasis is computational rather than graphical, many of the author's purposes are similar to the ones that prompted my programs. Spurrier et al. (1995) also contains laboratory material appropriate for elementary statistics and probability courses.
3. Part 4: The Casino Game "Keno"
3.1. Operation of the Game and Discussion of Its Purpose
36 Keno is a lottery game played in many gambling casinos. Typically, 80 balls numbered from 1 through 80 are agitated in an air stream by a machine so that 20 of them can be selected at random during play. Before play begins the gambler marks a ticket printed with the numbers from 1 through 80 in an effort to predict some of the numbers that will be among the 20 drawn. Many variations of the game are possible, but we concentrate here on the simplest, in which the gambler decides to "mark" a certain number of "spots" (i.e., predict a certain number of balls that will be selected); for us the number of predictions must be between one and nine.
37 After the 20 balls are selected by the machine, the casino notes how many "hits" (successful predictions) the gambler has accomplished. Each casino publishes lists of payoffs that depend on the amount bet, the number of spots marked, and the number of hits achieved. Relatively small proportions of hits receive no payoff. Relatively large proportions of hits (although extremely unlikely) can yield very large payoffs. The payoff schedules used in this program are ones recently advertised by Harrah's casinos in Reno and South Lake Tahoe, Nevada. The maximum possible payoff, for perfect tickets with large numbers of spots marked, is $50,000. Here is the payoff schedule for $2 bet on a game with only five spots marked, in which the maximum payoff is $1640.
Number Corresponding
of Hits Payoff
0 0
1 0
2 0
3 2
4 18
5 1640
38 One goal of the casino, evident from the way Keno games are promoted, is to get the gambler to focus on the largest possible payoff. The minuscule probabilities of such bonanzas are not advertised. Another goal seems to be to foster the illusion among gamblers that Keno is a game of strategy and skill in which there is some advantage in making wise guesses. It would be illegal to make this claim forthrightly. (But it is not illegal to encourage players to give very careful consideration to the numbers they pick, nor to act as if it is worthwhile to go through this thought process afresh for each new game.) Honestly played, Keno is a game of pure chance owing to the randomness of the selection of balls.
39 The probabilities of each number of hits can be computed using the hypergeometric probability distribution. The total number of outcomes from the drawing is the combinations of 80 things taken 20 at a time: C(80, 20) = 3.54 x 10^18. The number of ways to achieve exactly four hits is the number of ways to pick four hits out of five spots marked, C(5, 4) = 5, times the number of ways to pick 16 non-hits out of the 75 spots not marked, C(75, 16) = 8.55 x 10^15. The product is 4.28 x 10^16. Thus, the probability of getting exactly four hits is 4.28/354 = .01209.
40 Similar computations are used to fill in the rest of the probabilities in the table below for our example of a five-spot game, on which $2 is bet.
Hits Payoff Probability Product
0 0 .22718 0.00
1 0 .40569 0.00
2 0 .27046 0.00
3 2 .08394 0.17
4 18 .01209 0.22
5 1640 .00064 1.06
1.00000 1.44
41 The expected amount won E(W) is computed as the sum of the products shown in this table: $1.44. Thus, in our example, the expected return on a bet of $2.00 is about $1.44. It is typical of most Keno games that the gambler loses on average a little more than 25% of the amount bet. Based on the criterion of the expected percentage of the bet lost on each play, Keno is among the least favorable among the common "honest" games of pure chance legally available anywhere in the United States. Only the state lotteries with their payoffs of approximately 50% are worse, but with these there is the hope that the proceeds will be used more or less efficiently for some public good.
42 One can only speculate on the popularity of a game with such miserable odds. One reason is surely the possibility (if not the probability) of winning big for only a small bet. Many players will have a vivid image in their minds of how they would feel if they "win big" and what they might do with the money. They may feel a thrill as each number is drawn and posted. Another reason might be that the game requires a minimum of knowledge or concentration---or even sobriety---to play. In fact, "Keno runners" are available throughout the casinos, even in restaurants and cocktail lounges, to submit marked Keno tickets for play, and the results of each game appear on ubiquitous screens. A third reason might be that it is possible to imagine that one's failure on the game just finished was, nevertheless, "nearly" a success. "If only I had picked 15 instead of 25" (15 instead of 14, 15 instead of 16, 15 instead of 5, Aunt Sue's birthday which is the 15th instead of Uncle Dan's birthday which is the 27th, etc.) There are so many ways to imagine that one "almost" won that the probability of "almost" winning (if quantifiable at all) may be quite large indeed.
43 The program in Part 4 gives students the opportunity to try their luck at a simulated game of Keno. They begin with a "grant" of $20 in play money and can continue playing until it is lost. Then it is easy enough to start the program again with a fresh $20 stake. The illusion that it makes a difference what numbers are chosen is maintained to an extent by the ability to make changes until the ticket is marked just the way the player desires. However, unlike in the casino atmosphere, the probabilities of winning and the expected loss are clearly posted for each game.
44 From the title page the user has the option to start play at once or to read an introduction in which the game is explained and an indication (much less thorough than provided above in this paper) is given as to how probabilities are computed. Since several pages of explanation are provided in the introduction, there is no F1-page to provide technical details.
45 Since the look and feel of this program depends on the display of color text and since different computers with monochrome displays treat text so diversely, I could see no feasible way to write a program that would use a monochrome display with appealing and predictable results. Also the game seems to lose something if the speed is changed much either way. For these reasons there is no F2-page offering monochrome or speed options. For similar reasons, no Minitab analogue is offered.
3.2. Before Going to the Lab
46 Some preliminary words of caution are necessary. Somehow computers and recreational games seem to go together in the minds of students at all levels. Before introducing ANY demonstration into a statistics or probability lab, it is important to make sure that there is something tangible to be gained and that adequate thought is given to preventing confusion or abuse. In the case of a game, however, an extra degree of foresight may be necessary.
47 Before using this game in class it is important to understand exactly what you expect to accomplish by its presentation and to make sure students understand what this is. Written instructions on how to proceed and a requirement for a written report on what was accomplished are especially important here. It is also wise to make Keno the last item presented at a particular lab session so that there is a natural ending point to playing it.
48 One approach for introducing the Keno program into an elementary or intermediate probability course is to explain in lecture how Keno is played and to present the payoff table for some specific instance, such as the one given above for a $2 bet on a five-spot game. (Any relevant payoff table can be obtained by running the program.) Ask if anyone has ever played the game in a casino. Ask students if it looks like a game they think they could win. Ask how they would judge whether the game is "fair" and what they mean by fair.
49 If the idea of using expectation as a criterion for fairness does not emerge, propose a simpler lottery in which 1000 tickets have been sold and there is only one prize---$1500 for the single winner. Does $1.50 emerge as the "fair" price for a ticket in such a lottery?
50 Depending on how the discussion goes and on the interests of the instructor, there might be some room at this point for a brief and intuitive mention of the idea commonly called "utility" by game theorists. (A presentation of the formal ideas would not be difficult, but would take the discussion away from its main purpose.) Here are some points that might be profitably mentioned: Some people may consider $1.50 to be too small to be of any value, except for the momentary entertainment value it might have. After all, people put such sums of money into video games with NO chance of a monetary return. On the other hand, $1500 might seem to be a large enough sum to buy them some truly valued item. For such a person $1500 might be worth more than a thousand times $1.50---formally, a "non-linear utility function." If so, $1.50 might seem like a bargain price for the lottery ticket. What about such a person who gets carried away and participates in 1000 lotteries within some short span of time---the beginnings of gambling fever?
51 Next, whether or not the hypergeometric distribution is covered in the text for the course, one might show from combinatorial principles how to compute several of the probabilities that go with the payoff table. The additional probabilities can be supplied without computation and left as an exercise. Then the expected winnings can be computed, and the profitability of the game for the casino can be discussed.
52 It might also be worth showing an overhead projection slide of a Keno ticket, using the layout on the computer screen as a model. Show a game with five spots marked. Then show an overlay with 20 balls selected that produce only three hits but lots of "near" hits in the immediate vicinity of spots marked. Would students feel that they had "almost" won $18---or maybe even $1640---and be encouraged to try again?
53 Finally, one might discuss the public perception of how often gamblers win compared with the known, computable probabilities of winning. How likely is someone who does win big at Keno to let everyone know? How likely is someone who gambles all weekend and loses heavily to advertise his or her lack of success? Certainly, within the confines of the casino, games with no winners are quietly ignored and the next game is started at once. A game with a big winner is the subject of hoopla for hours if not days or weeks in every medium of exposure available to the casino.
54 (It might also be added here that hoopla governs the public perception of events other than gambling. The news media cover the most interesting events, and do not emphasize more common but uninteresting occurrences. What are the real chances of being devastated by an earthquake in California? Is it more dangerous to fly from Chicago to Atlanta than it is to drive? Statistical methods based on random sampling are necessary in the pursuit of truth precisely because our intuitive data gathering about the relative frequencies of even simple events is so easily biased.)
3.3. Lab Instructions For Students
55 Type PROBDEMO at the DOS prompt and then press ENTER at the "copyright" page to get to the "Main Menu," where you should select Item 4. Begin by reading the introduction. Then select a $2 bet on a five-spot game. Before you play the game notice the number of hits that is most likely. In the long run, over 40% of five-spot games will result in one hit (which yields no payoff).
56 Continue by playing several more five-spot games with $2 bets. Keep track of the number of hits you get on each game. (Also keep track of the number of games with no payoff in which you feel you "almost won" either an $18 or a $1640-payoff.) Compare notes with other students nearby. Does the 40% figure seem reasonable? (According to your definition of "almost winning," what percentage of the games fall into that category?)
3.4. Questions
57 Pedagogical Note: The Keno game works well at a variety of academic levels. Letters in parentheses indicate the level of difficulty: E for elementary, M for intermediate, and A for advanced.
58 Questions 1-4 refer to $2 five-spot games.
1. (E) If you play one game, what are the chances of losing your $2? (Answer: .22718 + .40569 + .27046 = .90333.)
2. (E) If you started with $20,000 and played 10,000 games, about how much money would you expect to have left? (Answer: $14,400.)
3. (M) In Problem 2, what is the probability that you will lose all of your money? (Answer: (.90333)^10,000 or very nearly zero.)
4. (M) Find the variance of the amount won in playing one game. (Answer: Using rounded numbers from the table in the program, we have E(W^2) = 4(.08394) + 324(.01209) + 2,689,600(.00064) = .33576 + 3.91716 + 1721.344 = 1725.59692, V(W) = 1725.60 - 1.44^2 = 1723.)
5. (M) (Continuation of 2 and 4) Find the standard deviation of the amount won in playing 10,000 independent games. Suppose that the Empirical Rule holds so that there is about a 95% chance that you will wind up with an amount of money that is within two standard deviations of the expected winnings. What interval of likely winnings does this give? (Answer: For the 10,000 games the variance is 10,000 times the answer to Question 4. The square root of this is $4151. The interval is $14,400 plus or minus twice $4151 or about $6100 to $22,700. This answer cannot be exact because we are using some rounded numbers from the computer screen for input.)
6. (E) Suppose you play a two-spot game and mark the spots 13 and 66. What is the probability that you will get exactly one hit? (Answer: From the program: .37975. Computed: C(2,1) C(78,19) / C(80,20).
7. (M) If you bet $2 on the game in Question 6, what payoff for two hits would make this a fair game? (Answer: The probability of two hits is .06013. Since fewer hits will not pay, we need the product of this probability and the payoff to be equal to the $2 bet; $33.26 comes very close.)
8. (A) (Continuation of 6) Further suppose in the two-spot game in Question 6 that you consider the numbers 3, 12, 14, and 23 to be "near" to 13, and the numbers 56, 65, 67, and 76 to be "near" to 66. What is the probability that you will get exactly one hit, but feel that one or more other numbers drawn "nearly" gave you a second hit? (Answer: First, we find the probability of exactly one hit and EXACTLY ONE near hit. In the numerator select which of the two marked numbers is the hit, select one of four allowable numbers for the near hit, and then select the 18 other numbers: thus, the probability of exactly one hit and exactly one near hit is:
C(2,1) C(4,1) C(74,18) / C(80,20) = .1644.
Probabilities of one hit along with exactly two, three, and four near hits are computed similarly. The total probability of one hit and one or more near hits is .2585. Over a quarter of the time you will think you "nearly" won.)
3.5. Concluding Comments on Part 4
59 I have used this program most often in elementary probability courses as soon as the concept of expectation for discrete distributions has been introduced. I have used it successfully even when the text does not include an explanation of the hypergeometric distribution. I either offer my own brief treatment keyed specifically to Keno, or just say that it is possible to compute the probabilities, provide them without proof, and then focus on how they are used to find the expected winnings. (In the latter case, several students usually insist on private outside-of-class explanations of how to find the probabilities.)
60 Keno can be used at a much lower academic level than the other programs in this package. At California State University at Hayward we have a variety of periodic events in which high school students and junior high school students come to campus (usually on a Saturday) for a day of classes, demonstrations, and exhibits. Some of these events are targeted at disadvantaged students or gifted students, and sometimes any students in the area and their parents are invited. The Keno program has become a standard attraction for these events. I admit that attempts to accompany it with some appreciation of the probability principles involved meet with varying degrees of success depending on the audience. The sessions always have waiting lines, even with 20 or more available computer stations, and they are never boring. I see no harm in this sort of carnival atmosphere if one goes into it with eyes open and worthy ulterior motives in mind.
4. Conclusion
61 The programs presented here can be used in an introductory probability course to build student interest and understanding of the concepts of randomness, expected value, and correlation. The programs' most effective use is in an interactive laboratory setting, with (1) adequate introduction and guidance to focus attention on the concepts to be learned, (2) an opportunity for students to compare and discuss results, and (3) a structure for students to provide written answers to relevant questions and to summarize what they have learned.
Preparation of this article and development of the programs provided with it were partially supported by NSF Grant USE 91-50433 and by California State University, Hayward. Newton Wai, a statistics graduate student at Cal State Hayward, has read drafts of this paper and made helpful suggestions. The author also wishes to thank the editor and the referees for their careful and helpful comments.
Computer programs were compiled using Microsoft QuickBasic (Version 4.0). The utility BRUN40.EXE, which must be present to run the programs, is property of Microsoft Corporation and is used with permission.
Appendix A
MINITAB Alternative to Part 3
The attached programs should run on any IBM-PC compatible machine with a display that is EGA or better. In a Windows environment it is usually possible to run DOS programs; many recent Macintosh machines are also able to emulate or run DOS. However, for those users who need or prefer to use Minitab, macros are included that capture most of the spirit of Part 3. (Values of r based on only part of a simulation, such as those used in Question 3 of Part 3, are not available in the Minitab macros.) The stored programs BIVCOR.MTB, BIVCOR1.MTB, and BIVCOR2.MTB are intended to be used together and should work in most pre-Windows releases of Minitab. The stored programs WBIVCOR.MTB, WBIVCOR1.MTB, and WBIVCOR2.MTB will run on Minitab for Windows.
Technical notes:
(1) The three non-Windows Minitab stored programs were written using Release 7 for DOS. BIVCOR2.MTB uses the GPLOT command with a LINE subcommand, neither of which is supported by Windows versions. WBIVCOR2.MTB uses the new Windows PLOT * command and the new LINE subcommand that goes with it. All of the Minitab stored programs are annotated with comments (following #-symbols) that explain the key steps.
(2) To operate these programs make sure Minitab addresses the path that contains them (use the CD command, "change drive/disk," if necessary), and at the MTB > prompt type the appropriate one of the following commands:
MTB > exec 'bivcor'
MTB > exec 'wbivcor'
The appropriate command must be repeated for each run.
Appendix B
Documentation of Data Sets in Part 3
Three datasets are presented; each is provided in three formats. The format with the extension .MTW is a worksheet in the format produced by Minitab Release 7 for DOS and readable by many other DOS and Windows versions of Minitab. The extension .MTP indicates a Minitab worksheet saved using the PORTABLE subcommand and retrievable by all versions of Minitab. The extension .dat.txt indicates DOS ASCII format with unlabeled variables, which appear in columns specified below for each dataset. These can be used with almost any statistical computer software.
Data Set 1. Data are taken from Herzog and Felton (1994) based on blood samples from 43 newborn babies at a Northern California community hospital. Variables are labeled HCrit (hematocrit in percent) and Hgb (hemoglobin in grams per deciliter). Data files are named REDCELL.MTB, etc. In the ASCII file, hematocrit appears in columns 1-4, hemoglobin in columns 8-11. Both variables carry one digit after the decimal point.
A printout of a Minitab session in response to Question 4 of Part 3 is shown in Appendix C.
Data Set 2. Data are taken from Mitchell (1975). Variables are Year (1870-1969), Manchester (Manchstr), Paris, Madrid. The columns for cities give annual precipitation in millimeters. Data files are named EUROPREC.MTW, etc. In the ASCII file, Year is in columns 1-4, Manchester in columns 6-9, Paris in columns 11-15, and Madrid in columns 17-20. All variables are integer-valued. I thank Jason Stover, a graduate student in statistics at California State University, Hayward, for calling these data to my attention.
Data Set 3. Data are taken from The World Almanac and Book of Facts (1994). Variables are State (two letter abbreviation), typical annual rainfall (data in inches, credited to National Climatic Data Center, U.S. Department of Commerce), and percentage graduation rates from public high schools (credited to National Center for Educational Statistics, U.S. Department of Education). Files are named RAINGRAD.MTB, etc. In the ASCII file, State appears in columns 1-2, Rainfall in columns 8-9, Graduation Rate in columns 16-17. State is an alphanumeric variable and the other two are integer variables.
All three of these datasets are taken from Trumbo (forthcoming), a compendium of classroom materials based the exploration of real datasets.
Appendix C
Sample MINITAB Run for a Data Set Used in Part 3
This Appendix presents a printout for Question 4 of Part 3 made using Minitab Release 7 for DOS. A character graphics scatterplot was used here so that results could be printed using ASCII characters. A higher quality plot could be obtained using the Minitab command GPLOT (or the new PLOT command or menu selection in Minitab for Windows).
Technical notes: (1) To be readable, this output should be printed in a format with line widths of at least 76 characters and in a mono-spaced font (such as Courier). (2) It is assumed that the Minitab worksheet is located in the path C:\JSE.
MTB > retr 'c:\jse\redcell'
[output confirming successful retrieval suppressed]
MTB > plot c1 c2
HCrit -
- * ** *
60+ 3 * *
- * *
- * 2 *
- * *
- 2222
50+ 2* 3 *
- * * 2
- *
- *
- 2
40+ * * *
14.0 16.0 18.0 20.0 22.0
MTB > desc c1 c2
N MEAN MEDIAN TRMEAN STDEV SEMEAN
HCrit 43 52.43 52.10 52.44 6.62 1.01
Hgb 43 17.826 17.500 17.810 2.292 0.349
MIN MAX Q1 Q3
HCrit 39.40 64.70 48.80 57.00
Hgb 13.400 22.700 16.600 19.800
MTB > corr c1 c2
Correlation of HCrit and Hgb = 0.990
Analogous Minitab sessions should be performed in response to Questions 5 and 6 in Part 3.
Cleveland, W. S., Diaconis, P., and McGill, R. (1982), "Variables on Scatterplots Look More Highly Correlated When the Scales Are Increased," Science, 216, 1138-1141.
Cleveland, W. S., and McGill, R. (1984), "The Many Faces of a Scatterplot," Journal of the American Statistical Association, 79, 807-822.
Herzog, B., and Felton, B. (1994), "Hemoglobin Screening for Normal Newborns," Journal of Perinatology, 14(4), 285-289.
Huber, P. J. (1987), "Experiences With Three-Dimensional Scatterplots," Journal of the American Statistical Association, 82, 448-453.
Keller, G. (1994), Statistics Laboratory Manual: Experiments Using Minitab, Belmont, CA: Duxbury Press.
Kennedy, W. J., Jr., and Gentle, J. E. (1980), Statistical Computing, New York and Basel: Marcel Dekker, Inc.
Lewandowsky, S., and Spence, I. (1989), "Discriminating Strata in Scatterplots," Journal of the American Statistical Association, 84, 682-688.
Meyer, H., and Shinar, D. (1992), "Estimating Correlations From Scatterplots," Human Factors, 34, 335-349.
Mitchell, B. R. (1975), European Historical Statistics, New York: Columbia University Press.
Moore, D. S. (1995), The Basic Practice of Statistics, New York: W. H. Freeman and Company.
Raveh, A. (1985), "On Quick Estimates of Pearson's r From Scatter Diagrams [Letter]," The American Statistician, 39, 239-240.
Shipp, C. E., and Margolin, C. G. (1982), "Graphical Display of Scatter Data Using the Standard Deviation Ellipse," in Proceedings of SAS Users Group International Conference, 7, pp. 171-175.
Spence, I., and Garrison, R. F. (1993), "A Remarkable Scatterplot," The American Statistician, 47, 12-19.
Spurrier, J. D., Edwards, D., and Thombs, L. A. (1995), Elementary Statistics Laboratory Manual, Belmont, CA: Duxbury Press.
Strahan, R. F., and Hansen, C. J. (1978), "Underestimating Correlation From Scatterplots," Applied Psychological Measurement, 2, 543-594.
Trumbo, B. E. (1994), "Some Demonstration Programs for Use in Teaching Elementary Probability: Parts 1 and 2," Journal of Statistics Education, v.2, n.2.
Trumbo, B. E. (forthcoming), Exploring Real Data With Minitab (provisional title), Belmont, CA: Duxbury Press.
The World Almanac and Book of Facts (1994), Mahwah, NJ: World Almanac Books.
Zelen, M., and Severo, N. C. (1964), "Probability Functions," in Handbook of Mathematical Functions (1974 ed.), eds. M. Abramowitz and I. A. Stegun, U.S. Department of Commerce, National Bureau of Standards, Applied Mathematics Series #55 (Tenth Printing with corrections), Washington: U.S. Government Printing Office, pp. 927-995.
Bruce E. Trumbo
Department of Statistics
California State University, Hayward
Hayward, CA 94542
Download Programs to a Local File
To unpack the files BRUN40.EXE, PROBDEMO.EXE, PDMEN.EXE, PDLLN.EXE, PDPPR.EXE, PDBNS.EXE, and PDKEN.EXE, type prob02 at the DOS prompt.
Download Minitab Programs and Datasets to a Local File
Return to Table of Contents | Return to the JSE Home Page | null | null | null | null | null | null | null | null | null |
https://abrobotics.tripod.com/ControlLaws/clpid.htm | 1,725,744,053,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650920.0/warc/CC-MAIN-20240907193650-20240907223650-00248.warc.gz | 68,112,907 | 7,341 | # PID Equations
The PID equations are the core of most control systems. They consist of three different ways of computing a control command from the error between an input reference value and the actual value that the system is achieving.
The Proportional equation creates a correction signal which is proportional to the error.
The Derivative equation creates a correction which is proportional to the rate of change of the error signal.
The Integral equation creates a correction signal which is proportional to the error signal multiplied by the time the error exists.
Intuitive PID
Before going into these equations further, let's try to get a feel for what these equations really do. I'm going to use the example of steering a car since it is something we all think we understand.
Imagine you are driving down the road trying to steer so as to stay in the center of your lane. We'll assume your cars steering works perfectly and the steering wheel is straight up when the car is going straight and it takes no force on the wheel for you to hold it straight up. So when the car is going down the center of the lane, it should take "zero" input from you to keep going straight.
So, let's say you reach down to tune the radio, and when you look back the car has drifted two feet to the right of center. Your first reaction will be to turn the steering wheel a bit to the left to turn the car back toward the center. You'll probably turn the wheel farther than if the car had drifted over only one foot; and not as far as if it had drifted over 4 feet. This is the proportional part. All it means is that the farther you are away from your reference, the more you will turn the wheel to make your correction.
Now, having been distracted for another second, you look up to see that the car has turned toward the left such that it is moving rapidly toward the center of the lane. You realize that if this goes on for much longer, the car will go over onto the left side of the lane. So, you turn the steering wheel to the right to reduce the rate at which the car is moving left. If the car is moving left VERY rapidly, you move the wheel right a lot. If it is moving left slowly, you move the wheel right just a bit. This is the Derivative part. It seems very similar to the proportional gain except that it reacts to the speed at which you are moving sideways rather than the distance that the car is displaced sideways.
So, the proportional term says to turn toward the center, and the derivative term says to turn away from the center (as you approach it, or toward the center if you are moving farther away). If both of these terms are used, there will be a balance such that a large offset will cause a large wheel turn toward the center which will be balanced out by a large rate toward the center causing a large wheel turn to away from the center.. The two equations will tend to cancel each other out; but the amount that they differ will direct the car to the left or right in such a way as to push the equations toward balance. Hence, as the car approaches the center, the rate will become smaller and smaller until the car is in the center with zero rate. This is pretty much the way you steer your car when things are running smoothly.
What's the integral term good for then? Integral terms are good for when there are biases in the system. Maybe your car has one low tire, or a bent suspension, and is pulling to the right. Or for our simple case, let's just say the steering wheel is installed so that the car goes straight forward when the wheel is turned left 20 degrees. So, if you try to steer your car straight ahead by placing the steering wheel at zero (straight up), the car will turn right. You'll soon figure out that you have to hold the wheel 20 degrees left, but let's speculate on how you come to that conclusion. You'll probably start holding the wheel straight up (zero degrees) and observe that the car is moving right. You'll probably turn the car a bit to the left and notice that it is still turning right, so you'll add a bit more to the left. You'll keep adding more turn to the left until you see that it isn't turning right anymore. At this point you'll have figured out that you have to add 20 degrees left to all your steering calculations and you'll just do that from now on. You may even rotate your hands on the wheel so that your hands are steering straight when the wheel is 20 degrees left. But, the thing to note is that you kept adding more correction over time until the problem went away. This is exactly what an integrator does.
So let's look a little more at each of these terms: (and, no, they are not in the P - I - D order, but it makes more sense to do Proportional then Derivative then Integral) | 1,049 | 4,804 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-38 | latest | en | 0.947234 |
null | null | null | null | null | null | ‘Color’ is the new ‘black and white’ for EU trademarks
This article explains a recent rule change and provides recommendations to optimize logo and other color trademark protection in the EU
Previously, a logo or design trademark that was registered as community trade marks (CTM) in the European Union (EU) in black and white protected versions of the logo in any color, but this is no longer true. Under a new CTM rule effective July 15, 2014, logo trademarks filed in black and white or grey will not necessarily protect the same mark used predominantly in color. This new approach, one of several changes resulting from the Office for Harmonization in the Internal Market (OHIM) Convergence Programme, should inform both filing and enforcement strategies. This article explains the rule change and provides recommendations to optimize logo and other color trademark protection in the EU.
Why the change?
The goal of the OHIM Convergence Programme is to harmonize the differing practices of national trademark offices (TMOs) in the EU. Previously, in general, a trademark registered in black and white protected use in any color. Moving forward, however, the approach is “what you see is what you get,” so a mark registered in black and white only covers use of the logo in black and white. This new practice is intended to reduce uncertainty for trademark owners since all participating TMOs should apply the same rules. Nevertheless, this quest for harmonization may create more uncertainty, at least in the short term, since implementation and interpretation of this new approach may vary among national courts. Moreover, a few EU countries, namely Sweden, Denmark, Finland, France and Italy, are not currently participating because their national laws conflict with this rule; they will only adopt this rule if and when their laws are amended.
Why is this important?
There are two situations, one defensive and one offensive, in which a black and white logo registration is now of questionable value. First, an action to cancel a logo registration is likely to be successful — even if the logo is used extensively — if the use is only in color. (If the logo is used in black and white, this could avoid cancellation, even if the predominant use of the logo is in color.)
Second, in an infringement action, a trademark owner will find it difficult to prove that use of its logo is senior to, and has priority over, an infringer’s mark, if the logo is registered in black and white but used only in color. Moreover, an infringement action often triggers a cancellation counterclaim which puts the entire registration at risk. Finally, if the logo is registered in black and white, an infringer’s use of the same color as the logo owner’s actual use will have little weight in proving likelihood of confusion.
How is the rule applied?
The new rule states that a trademark in black and white is not identical to the same mark in color, unless the differences in color are insignificant. An “insignificant difference” is defined as a difference that a reasonably observant consumer will perceive only upon side-by-side examination. The examples below, provided by OHIM, illustrate that the “insignificant differences” test is interpreted very narrowly.
Perkins Coie 1
The above comparisons show that a slight difference in grey scale or a minimal use of color, e.g., the touch of yellow on the girl’s dress, would be deemed insignificant differences. Therefore, in these cases, a black and white registration would not be cancelled based on these uses.
The examples below illustrate significant differences, highlighting the restrictive scope of the new rule.
Perkins Coie 2
To be clear, each of the “Used Marks” illustrates a significant difference, including Number 4, which use minimal color, and Number 7, which has no color at all. Therefore, each of the above “Used Marks” would not be covered by the black and white registration. In other words, the Registered Mark could be cancelled for non-use if only a “Used Mark” were displayed.
Finally, word marks registered in black and white but used in color may fall outside of the rule because the word, rather than the color, would be viewed as the distinctive element of the mark. Nevertheless, if the color has a distinctive character, e.g., a word mark with each letter in a different color, then the word mark may not be protected by a black and white registration.
In light of these changes, brand owners should review their CTM registrations and consider the following recommendations for both logos used in color and word marks for which color is a distinctive element:
1. Applications to register any new CTM logo should be filed in color if the logo is used only, or predominantly, in color.
2. An existing black and white logo registration should be supplemented with a new filing in color. This is particularly true if the black and white registration is more than five years old, and therefore susceptible to cancellation for non-use.
3. If a word mark is integrated into a logo, a separate trademark registration for the word alone should be considered.
4. If color is a distinctive element of a word mark, the word mark should be filed in color.
As a practical matter, it could be more cost-effective to file new color applications than to renew black and white registrations. Therefore, once a new color application matures to registration, the black and white registrations could be allowed to lapse. The new color registration has additional benefits: It cannot be challenged for non-use during the first five years, and it provides an opportunity to update, and perhaps expand, the scope of the goods and services covered by the registration.
In short, EU CTM trademarks should be registered in color if they are predominantly used in color, color provides distinctive character to the mark, or there is a significant difference between the color mark and the registered mark in a side-by-side comparison. Filing a new color mark may be a better strategy, and more economical, than renewing a black and white registration. Finally, for optimal protection, we recommend separate registrations for any word mark, even if it is integrated into a logo; the color version of a word mark if color is a distinctive element; and any logo used in color.
The author thanks Lydia Z. Ansari, a 2014 summer associate at Perkins Coie LLP and a law student at the University Washington School of Law, who significantly contributed to this article.
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Heidi L. Sachs
Heidi L. Sachs, a partner at Perkins Coie LLP in Seattle, focuses her practice on domestic and international trademark law, as well as copyright...
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EE103 Section 7
EE103 Section 7 - EE103 Winter 2009 Lecture Notes(SEJ...
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EE103 Winter 2009 Lecture Notes (SEJ) Section 7 SECTION 7: INTRODUCTION TO OPTIMIZATION: NONLINEAR LEAST SQUARES, THE NEWTON AND GAUSS-NEWTON METHODS ....................... 129 Newton’s Method for Systems of Nonlinear Equations ......................................... 129 Newton’s Algorithm .............................................................................................. 130 Introduction to Optimization and Nonlinear Least Squares ................................ 133 Philosophy of Unconstrained Nonlinear Optimization ......................................... 134 The Newton Method for Optimization .................................................................... 136 Numerical Example of Steepest Decent vs. Newton’s Method ......................... 137 Newton’s Method Applied to Nonlinear Least-squares .................................... 138 A Modification: The Gauss-Newton Method ..................................................... 139 Tutorial Codes (i.e., not elegantly written) . ...................................................... 141
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EE103 Winter 2009 Lecture Notes (SEJ) Section 7 129 SECTION 7: INTRODUCTION TO OPTIMIZATION: NONLINEAR LEAST SQUARES, THE NEWTON AND GAUSS-NEWTON METHODS This Section introduces the student to Newton’s method for solving a system of nonlinear equations, where the number of variables is equal to the number of equations. Subsequently, we consider the case where there are more nonlinear equations than there are variables. As in the linear case ( , Ax b m n = > ), this leads us to consider the nonlinear least squares problem. The latter problem often arises when considering the estimation of parameters for known functional forms. Newton’s Method for Systems of Nonlinear Equations We are interested in a numerical method for computing a solution of the system of nonlinear equations ( ) () 112 212 12 ,,, 0 , , , 0 0 n n nn fxx x x x = = = # Let T n x xx x = and let T n f ff f = . We have () () () () ,, , T n f xf x f x f x = and we abbreviate the above system, using vector notation, to ( ) 0 fx = an n x n system of nonlinear equations. Given a point , T kk k k n x x = , we use the 1 st order approximation for each of the functions; that is () ( ) ( )( ) () () () ( ) 11 1 22 2 k k k n f x f x x x f x f x x x f x f x x x ≈+ # or in vector-matrix notation:
EE103 Winter 2009 Lecture Notes (SEJ) Section 7 130 () () ( ) () () () () ( ) 1 2 k k kk k n k f fx f xf x x x fx fx J x x x Δ ⎡⎤ ⎢⎥ ≈+ ⎣⎦ # As in the single variable case, we solve the (system of) linear equations ( ) ( )( ) 0 k f + −= or, assuming ( ) k f Jx is nonsingular, we write ( notation, not computation! ) ( ) ( ) 1 1 1 k f k k f xx J x x xJ x f x + =− ( ) ( ) ( ) 1 Nf gx x J x f x ∴= However, we do not solve for 1 k x + by inverting ( ) k f . Rather, letting ( ) k yx x , we solve the system of equations (e.g., by LU methods) ( ) ( ) f Jxy fx Let k y be the solution. Therefore, 1 k y x xy + = + Newton's algorithm for a system of nonlinear equations may be succinctly stated. Newton’s Algorithm Select an initial vector x DoWhile (some appropriate stopping condition has not occurred) Compute the Jacobian, f Let y solve the system of linear equations f (assuming f is nonsingular) x ←+ End DoWhile
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EE103 Winter 2009 Lecture Notes (SEJ) Section 7 131 Example (“Using Matlab as a calculator”): Consider the circle and ellipse 22 12 10 xx + −= 0.5 0 += We wish to find an intersection point. Starting with (1, 1.5) t x = , we’ll use Matlab as a calculator to find a solution and we’ll stop when the addition to the current value of x is less than approximately 3 10 .
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EE103 Section 7 - EE103 Winter 2009 Lecture Notes(SEJ...
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Ask a homework question - tutors are online | 1,081 | 4,336 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-09 | latest | en | 0.62905 |
https://mathoverflow.net/questions/70272/puzzle-in-martin-gardner-book/70316 | 1,579,952,076,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251672440.80/warc/CC-MAIN-20200125101544-20200125130544-00173.warc.gz | 558,685,902 | 24,327 | # Puzzle in Martin Gardner book [closed]
What is the official name of this problem? Martin Gardner gives introduction in his book "Math circus". The problem belongs to 1D random walk. What can be read to gain deep insight into this problem? Or other useful resources.
We can complicate matters by allowing transition probabilities to vary from 1/2 and by allowing steps longer than one unit. Consider the following curious paradox first called to my attention (in betting terms) by Enn Norak, a Canadian mathematician. A walker starts 100 steps to the right of 0 on a line that has no barriers. Instead of a coin a packet of 10 playing cards-five red and five black-is used as a randomizer. The cards are shuffled and spread face down and any card is selected. After its color is noted it is discarded. If it is red, the walker steps to the right: if black, he steps to the left. This continues until all 10 cards have been taken. (The transition probability varies with each step. It is 1/2 only when there is an equal mixture of red and black cards before the draw.) The walk differs also from walks discussed above in that before each card is noted the walker chooses the length (which need not be integral) of his next step. Assume that the walker adopts the following halving strategy in choosing step lengths. After each card is noted he takes a step (left or right) equal to exactly half of his distance from 0. His first step is 100/2 = 50 units. If the card is red, he goes to the 150 mark. His next step will then be 150/2 = 75. If the first card drawn is black, he goes left to the 50 mark, and so his next step will be 50/2 = 25. He continues in this manner until the tenth card is noted. Will he then be to the right or to the left of the 100 mark where he began the walk? The answer is that he is sure to be to the left. This may not be very surprising, but it is surely astonishing that, regardless of the order in which the cards are drawn, he will end the walk at exactly the same spot.
• So... on black cards his distance from 0 is multiplied by $\frac 12$, and on red cards it's multiplied by $\frac 32$. There are 5 of each, so after 10 steps his distance from 0 has been multiplied by $\frac {3^5}{2^{10}}$, regardless of order. There doesn't seem to be much more to it than that, or am I missing something? If you take logs, it turns back into a regular random walk, and you know how many total steps are taken each way. – Vaughn Climenhaga Jul 13 '11 at 23:05
• The name I'd give it is "multiplication is commutative". – Robert Israel Jul 13 '11 at 23:42
• The question seems to presuppose that such puzzles have official names. Rather few have names at all, and I know of no way for a name to become "official". – Andreas Blass Jul 13 '11 at 23:51
• I don't see much of a future in Robert Israel's becoming a puzzle writer. (If he shouts spoilers out in theatres. I don't think I'll go to the movies with him either.) Gerhard "Multiplication Can Also Be Associative" Paseman, 2011.07.13 – Gerhard Paseman Jul 14 '11 at 1:02
• I don't think Gerhard is objecting to the presence of Robert's comment here but to the idea of using "multiplication is commutative" as the name of the puzzle. Standard practice is that the name of a puzzle should not give away the answer. – Andreas Blass Jul 14 '11 at 1:22
Sure, multiplication is commutative, but there is more to it than that. While being reasonably easy, this puzzle suggests variations in ways that the equation $x\cdot y = y\cdot x$ doesn't.
In his wonderful paper Games People Don't Play, http://www.teorver.ru/newkatalog/1193689162.pdf, Peter Winkler describes essentially the same game as Next card color betting'' (a bit of googling also turns up http://www.maxdama.com/?p=137 and http://www.dartblog.com/data/2008/08/007950.php). But there the player, Victor, wants to end up as far to the right as possible (increasing his bankroll). It turns out that there is a strategy that guarantees him to end up with $2^{10}/\binom{10}{5} = 256/63$ times his initial bankroll, or about 406 steps to the right of the origin.
This is a game that children can understand, but if we pursue the analysis, it doesn't stop until we have developed, besides insights into hedging strategies, a good deal of nontrivial mathematics including information theory (the amount of information Victor has about the red-black sequence dictates exactly how much money he will ideally make by betting), the Wallis product formula (showing that his final bankroll is asymptotically $\sqrt{\pi n}$ for a deck of $n$ red and $n$ black cards), and even the central limit theorem. | 1,148 | 4,628 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2020-05 | latest | en | 0.966842 |
https://jsteinhardt.wordpress.com/tag/theory-practice/ | 1,638,634,368,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362999.66/warc/CC-MAIN-20211204154554-20211204184554-00120.warc.gz | 390,123,175 | 28,581 | ## Nobody Understands Probability
The goal of this post is to give an overview of Bayesian statistics as well as to correct errors about probability that even mathematically sophisticated people commonly make. Hopefully by the end of this post I will convince you that you don’t actually understand probability theory as well as you think, and that probability itself is something worth thinking about.
I will try to make this post somewhat shorter than the previous posts. As a result, this will be only the first of several posts on probability. Even though this post will be shorter, I will summarize its organization below:
• Bayes’ theorem: the fundamentals of conditional probability
• modeling your sources: how not to calculate conditional probabilities; the difference between “you are given X” and “you are given that you are given X”
• how to build models: examples using toy problems
• re-evaluating a standard statistical test
I bolded the section on models because I think it is very important, so I hope that bolding it will make you more likely to read it.
Also, I should note that when I say that nobody understands probability, I don’t mean it in the sense that most people are bad at combinatorics. Indeed, I expect that most of the readers of this blog are quite proficient at combinatorics, and that many of them even have sophisticated mathematical definitions of probability. Rather I would say that actually using probability theory in practice is non-trivial. This is partially because there are some subtleties (or at least, I have found myself tripped up by certain points, and did not realize this until much later). It is also because whenever you use probability theory in practice, you end up employing various heuristics, and it’s not clear which ones are the “right” ones.
If you disagree with me, and think that everything about probability is trivial, then I challenge you to come up with a probability-theoretic explanation of the following phenomenon:
Suppose that you have never played a sport before, and you play soccer, and enjoy it. Now suppose instead that you have never played a sport before, and play soccer, and hate it. In the first case, you will think yourself more likely to enjoy other sports in the future, relative to in the second case. Why is this?
Or if you disagree with the premises of the above scenario, simply “If X and Y belong to the same category C, why is it that in certain cases we think it more likely that Y will have attribute A upon observing that X has attribute A?”
Bayes’ Theorem
Bayes’ theorem is a fundamental result about conditional probability. It says the following:
$p(A \mid B) = \frac{p(B \mid A)p(A)}{p(B)}$
Here $A$ and $B$ are two events, and $p(A \mid B)$ means the probability of $A$ conditioned on $B$. In other words, if we already know that $B$ occurred, what is the probability of $A$? The above theorem is quite easy to prove, using the fact that $p(A \cap B) = p(A \mid B)p(B)$, and thus also equals $p(B \mid A)p(A)$, so that $p(A \mid B)p(B) = p(B \mid A)p(A)$, which implies Bayes’ theorem. So, why is it useful, and how do we use it?
One example is the following famous problem: A doctor has a test for a disease that is 99% accurate. In other words, it has a 1% chance of telling you that you have a disease even if you don’t, and it has a 1% chance of telling you that you don’t have a disease even if you do. Now suppose that the disease that this tests for is extremely rare, and only affects 1 in 1,000,000 people. If the doctor performs the test on you, and it comes up positive, how likely are you to have the disease?
The answer is close to $10^{-4}$, since it is roughly $10^4$ times as likely for the test to come up positive due to an error in the test relative to you actually having the disease. To actually compute this with Bayes’ rule, you can say
p(Disease | Test is positive) = p(Test is positive | Disease)p(Disease)/p(Test is positive),
which comes out to $\frac{0.99 \cdot 10^{-6}}{0.01 \cdot (1-10^{-6}) + 0.99 \cdot 10^{-6}},$ which is quite close to $10^{-4}$.
In general, we can use Bayes’ law to test hypotheses:
p(Hypothesis | Data) = p(Data | Hypothesis) p(Hypothesis) / p(Data)
Let’s consider each of these terms separately:
• p(Hypothesis | Data) — the weight we assign to a given hypothesis being correct under our observed data
• p(Data | Hypothesis) — the likelihood of seeing the data we saw under our hypothesis; note that this should be quite easy to compute. If it isn’t, then we haven’t yet fully specified our hypothesis.
• p(Hypothesis) — the prior weight we give to our hypothesis. This is subjective, but should intuitively be informed by the consideration that “simpler hypotheses are better”.
• p(Data) — how likely we are to see the data in the first place. This is quite hard to compute, as it involves considering all possible hypotheses, how likely each of those hypotheses is to be correct, and how likely the data is to occur under each hypothesis.
So, we have an expression for p(Hypothesis | Data), one of which is easy to compute, the other of which can be chosen subjectively, and the last of which is hard to compute. How do we get around the fact that p(Data) is hard to compute? Note that p(Data) is independent of which hypothesis we are testing, so Bayes’ theorem actually gives us a very good way for comparing the relative merits of two hypotheses:
p(Hypothesis 1 | Data) / p(Hypothesis 2 | Data) = [p(Data | Hypothesis 1) / p(Data | Hypothesis 2)] * p(Hypothesis 1) / p(Hypothesis 2)
Let’s consider the following toy example. There is a stream of digits going past us, too fast for us to tell what the numbers are. But we are allowed to push a button that will stop the stream and allow us to see a single number (whichever one is currently in front of us). We push this button three times, and see the numbers 3, 5, and 3. How many different numbers would we estimate are in the stream?
For simplicity, we will make the (somewhat unnatural) assumption that each number between 0 and 9 is selected to be in the stream with probability 0.5, and that each digit in the stream is chosen uniformly from the set of selected numbers. It is worth noting now that making this assumption, rather than some other assumption, will change our final answer.
Now under this assumption, what is the probability, say, of there being exactly 2 numbers (3 and 5) in the stream? By Bayes’ theorem, we have
$p(\{3,5\} \mid (3,5,3)) \propto p((3,5,3) \mid \{3,5\}) p(\{3,5\}) = \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^{10}.$
Here $\propto$ means “is proportional to”.
What about the probability of there being 3 numbers (3, 5, and some other number)? For any given other number, this would be
$p(\{3,5,x\} \mid (3,5,3)) \propto p((3,5,3) \mid \{3,5,x\}) p(\{3,5,x\}) = \left(\frac{1}{3}\right)^3 \left(\frac{1}{2}\right)^{10}.$
However, there are 8 possibilities for $x$ above, all of which correspond to disjoint scenarios, so the probability of there being 3 numbers is proportional to $9 \left(\frac{1}{3}\right)^3 \left(\frac{1}{2}\right)^{10}$. If we compare this to the probability of there being 2 numbers, we get
p(2 numbers | (3,5,3)) / p(3 numbers | (3,5,3)) = 3/8.
Even though we have only seen two numbers in our first three samples, we still think it is more likely that there are 3 numbers than 2, just because the prior likelihood of there being 3 numbers is so much higher. However, suppose that we made 6 draws, and they were 3,5,3,3,3,5. Then we would get
p(2 numbers | (3,5,3,3,5)) / p(3 numbers | (3,5,3,3,5)) = (1/2)^6 / [9 * (1/3)^6] = 81/64.
Now we find it more likely that there are only 2 numbers. This is what tends to happen in general with Bayes’ rule — over time, more restrictive hypotheses become exponentially more likely than less restrictive hypotheses, provided that they correctly explain the data. Put another way, hypotheses that concentrate probability density towards the actual observed events will do best in the long run. This is a nice feature of Bayes’ rule because it means that, even if the prior you choose is not perfect, you can still arrive at the “correct” hypothesis through enough observations (provided that the hypothesis is among the set of hypotheses you consider).
I will use Bayes’ rule extensively through the rest of this post and the next few posts, so you should make sure that you understand it. If something is unclear, post a comment and I will try to explain in more detail.
An important distinction that I think most people don’t think about is the difference between experiments you perform, and experiments you observe. To illustrate what I mean by this, I would point to the difference between biology and particle physics — where scientists set out to test a hypothesis by creating an experiment specifically designed to do so — and astrophysics and economics, where many “experiments” come from seeking out existing phenomena that can help evaluate a hypothesis.
To illustrate why one might need to be careful in the latter case, consider empirical estimates of average long-term GDP growth rate. How would one do this? Since it would be inefficient to wait around for the next 10 years and record the data of all currently existing countries, instead we go back and look at countries that kept records allowing us to compute GDP. But in this case we are only sampling from countries that kept such records, which implies a stable government as well as a reasonable degree of economics expertise within that government. So such a study almost certainly overestimates the actual average growth rate.
Or as another example, we can argue that a scientist is more likely to try to publish a paper if it doesn’t agree with prevalent theories than if it does, so looking merely at the proportion of papers that lend support to or take support away from a theory (even if weighted by the convincingness of each paper) is probably not a good way to determine the validity of a theory.
So why are we safer in the case that we forcibly gather our own data? By gathering our own data, we understand much better (although still not perfectly) the way in which it was constructed, and so there is less room for confounding parameters. In general, we would like it to be the case that the likelihood of observing something that we want to observe does not depend on anything else that we care about — or at the very least, we would like it to depend in a well-defined way.
Let’s consider an example. Suppose that a man comes up to you and says “I have two children. At least one of them is a boy.” What is the probability that they are both boys?
The standard way to solve this is as follows: Assuming that male and female children are equally likely, there is a $\frac{1}{4}$ chance of two girls or two boys, and a $\frac{1}{2}$ chance of having one girl and one boy. Now by Bayes’ theorem,
p(Two boys | At least one boy) = p(At least one boy | Two boys) * p(Two boys) / p(At least one boy) = 1 * (1/4) / (1/2+1/4) = 1/3.
So the answer should be 1/3 (if you did math contests in high school, this problem should look quite familiar). However, the answer is not, in fact, 1/3. Why is this? We were given that the man had at least one boy, and we just computed the probability that the man had at two boys given that he had at least one boy using Bayes’ theorem. So what’s up? Is Bayes’ theorem wrong?
No, the answer comes from an unfortunate namespace collision in the word “given”. The man “gave” us the information that he has at least one male child. By this we mean that he asserted the statement “I have at least one male child.” Now our issue is when we confuse this with being “given” that the man has at least one male child, in the sense that we should restrict to the set of universes in which the man has at least one male child. This is a very different statement than the previous one. For instance, it rules out universes where the man has two girls, but is lying to us.
Even if we decide to ignore the possibility that the man is lying, we should note that most universes where the man has at least one son don’t even involve him informing us of this fact, and so it may be the case that proportionally more universes where the man has two boys involve him telling us “I have at least one male child”, relative to the proportion of such universes where the man has one boy and one girl. In this case the probability that he has two boys would end up being greater than 1/3.
The most accurate way to parse this scenario would be to say that we are given (restricted to the set of possible universes) that we are given (the man told us that) that the man has at least one male child. The correct way to apply Bayes’ rule in this case is
p(X has two boys | X says he has >= 1 boy) = p(X says he has >= 1 boy | X has two boys) * p(X has two boys) / p(X says he has >=1 boy)
If we further assume that the man is not lying, and that male and female children are equally likely and uncorrelated, then we get
p(X has two boys | X says he has >= 1 boy) = [p(X says he has >= 1 boy | X has two boys) * 1/4]/[p(X says he has >= 1 boy | X has two boys) * 1/4 + p(X says he has >= 1 boy | X has one boy) * 1/2]
So if we think that the man is $\alpha$ times more likely to tell us that he has at least one boy when he has two boys, then
p(X has two boys | X says he has >= 1 boy) = $\frac{\alpha}{\alpha+2}$.
Now this means that if we want to claim that the probability that the man has two boys is $\frac{1}{3}$, what we are really claiming is that he is equally likely to inform us that he has at least one boy, in all situations where it is true, independent of the actual gender distribution of his children.
I would argue that this is quite unlikely, as if he has a boy and a girl, then he could equally well have told us that he has at least one girl, whereas he couldn’t tell us that if he has only boys. So I would personally put $\alpha$ closer to 2, which yields an answer of $\frac{1}{2}$. On the other hand, situations where someone walks up to me and tells me strange facts about the gender distribution of their children are, well, strange. So I would also have to take into account the likely psychology of such a person, which would end up changing my estimate of $\alpha$.
The whole point here is that, because we were an observer receiving information, rather than an experimenter acquiring information, there are all sorts of confounding factors that are difficult to estimate, making it difficult to get a good probability estimate (more on that later). That doesn’t mean that we should give up and blindly guess $\frac{1}{3}$, though — it might feel like doing so gets away without making unwarranted assumptions, but it in fact implicitly makes the assumption that $\alpha = 1$, which as discussed above is almost certainly unwarranted.
What it does mean, though, is that, as scientists, we should try to avoid situations like the one above where there are lots of confounding factors between what we care about and our observations. In particular, we should avoid uncertainties in the source of our information by collecting the information ourselves.
I should note that, even when we construct our own experiments, we should still model the source of our information. But doing so is often much easier. In fact, if we wanted to be particularly pedantic, we really need to restrict to the set of universes in which our personal consciousness receives a particular set of stimuli, but that set of stimuli has almost perfect correlation with photons hitting our eyes, which has almost perfect correlation with the set of objects in front of us, so going to such lengths is rarely necessary — we can usually stop at the level of our personal observations, as long as we remember where they come from.
How to Build Models
Now that I’ve told you that you need to model your information sources, you perhaps care about how to do said modeling. Actually, constructing probabilistic models is an extremely important skill, so even if you ignore the rest of this post, I recommend paying attention to this section.
This section will have the following examples:
• determining if a coin is fair
• finding clusters
Determining if a Coin is Fair
Suppose that you have occasion to observe a coin being flipped (or better yet, you flip it yourself). You do this several times and observe a particular sequence of heads and tails. If you see all heads or all tails, you will probably think the coin is unfair. If you see roughly half heads and half tails, you will probably think the coin is fair. But how do we quantify such a calculation? And what if there are noticeably many more heads than tails, but not so many as to make the coin obviously unfair?
We’ll solve this problem by building up a model in parts. First, there is the thing we care about, namely whether the coin is fair or unfair. So we will construct a random variable X that can take the values Fair and Unfair. Then p(X = Fair) is the probability we assign to a generic coin being fair, and p(X = Unfair) is the probability we assign to a generic coin being unfair.
Now supposing the coin is fair, what do we expect? We expect each flip of the coin to be independent, and have a $\frac{1}{2}$ probability of coming up heads. So if we let F1, F2, …, Fn be the flips of the coin, then p(Fi = Heads | X = Fair) = 0.5.
What if the coin is unfair? Let’s go ahead and blindly assume that the flips will still be independent, and furthermore that each possible weight of the coin is equally likely (this is unrealistic, as weights near 0 or 1 are much more likely than weights near 0.5). Then we have to have an extra variable $\theta$, the probability that the unfair coin comes up heads. So we have p(Unfair coin weight = $\theta$) = 1. Note that this is a probability density, not an actual probability (as opposed to p(Fi = Heads | X = Fair), which was a probability).
Continuing, if F1, F2, …, Fn are the flips of the coin, then p(Fi = Heads | X = Fair, Weight = $\theta$) = $\theta$.
Now we’ve set up a model for this problem. How do we actually calculate a posterior probability of the coin being fair for a given sequence of heads and tails? (A posterior probability is just the technical term for the conditional probability of a hypothesis given a set of data; this is to distinguish it from the prior probability of the hypothesis before seeing any data.)
Well, we’ll still just use Bayes’ rule:
p(Fair | F1, …, Fn) $\propto$ p(F1, …, Fn | Fair) p(Fair) = $\left(\frac{1}{2}\right)^n$ p(Fair)
p(Unfair | F1, …, Fn) $\propto$ p(F1, …, Fn | Unfair) p(Unfair) = $\int_{0}^{1} \theta^{H}(1-\theta)^{T} d\theta$ p(Unfair)
Here H is the number of heads and T is the number of tails. In this case we can fortunately actually compute the integral in question and see that it is equal to $\frac{H!T!}{(H+T+1)!}$. So we get that
p(Fair | F1, …, Fn) / p(Unfair | F1, …, Fn) = p(Fair)/p(Unfair) * $\frac{(H+T+1)!}{2^n H!T!}$.
It is often useful to draw a diagram of our model to help keep track of it:
Now suppose that we, being specialists in determining if coins are fair, have been called in to study a large collection of coins. We get to one of the coins in the collection, flip it several times, and observe the following sequence of heads and tails:
HHHHTTTHHTTT
Since there are an equal number of heads and tails, our previous analysis will certainly conclude that the coin is fair, but its behavior does seem rather suspicious. In particular, different flips don’t look like they are really independent, so perhaps our previous model is wrong. Maybe the right model is one where the next coin value is usually the same as the previous coin value, but flips with some probability. Now we have a new value of X, which we’ll call Weird, and a parameter $\phi$ (basically the same as $\theta$) that tells us how likely a weird coin is to have a given probability of switching. We’ll again give $\phi$ a uniform distribution over [0,1], so p(Switching probability of weird coin = $\phi$) = 1.
To predict the actual coin flips, we get p(F1 = Heads | X = Weird, Switching probability = $\phi$) = 1, p(F(i+1) = Heads | Fi = Heads, X = Weird, Switching probability = $\phi$) = $1-\phi$, and p(F(i+1) = Heads | Fi = Tails, X = Weird, Switching probability = $\phi$) = $\phi$. We can represent this all with the following graphical model:
Now we are ready to evaluate whether the coin we saw was a Weird coin or not.
p(X = Weird | HHHHTTTHHTTT) $\propto$ p(HHHHTTTHHTTT | X = Weird) p(X = Weird) = $\int_{0}^{1} \frac{1}{2}(1-\phi)^8 \phi^3 d\phi$ p(X = Weird)
Evaluating that integral gives $\frac{8!3!}{2 \cdot 12!} = \frac{1}{3960}$. So p(X = Weird | Data) = p(X = Weird) / 3960, compared to p(X = Fair | Data), which is p(X = Fair) / 4096. In other words, positing a Weird coin only explains the data slightly better than positing a Fair coin, and since the vast majority of coins we encounter are fair, it is quite likely that this one is, as well.
Note: I’d like to draw your attention to a particular subtlety here. Note that I referred to, for instance, “Probability that an unfair coin weight is $\theta$“, as opposed to “Probability that a coin weight is $\theta$ given that it is unfair”. This really is an important distinction, because the distribution over $\theta$ really is the probability distribution over the weights of a generic unfair coin, and this distribution doesn’t change based on whether our current coin happens to be fair or unfair. Of course, we can still condition on our coin being fair or unfair, but that won’t change the probability distribution over $\theta$ one bit.
Finding Clusters
Now let’s suppose that we have a bunch of points (for simplicity, we’ll say in two-dimensional Euclidean space). We would like to group the points into a collection of clusters. Let’s also go ahead and assume that we know in advance that there are $k$ clusters. How do we actually find those clusters?
We’ll make the further heuristic assumption that clusters tend to arise from a “true” version of the cluster, and some Gaussian deviation from that true version. So in other words, if we let there be k means for our clusters, $\mu_1, \mu_2, \ldots, \mu_k$, and multivariate Gaussians about their means with covariance matrices $\Sigma_1, \Sigma_2, \ldots, \Sigma_k$, and finally assume that the probability that a point belongs to cluster i is $\rho_i$, then the probability of a set of points $\vec{x_1}, \vec{x_2}, \ldots, \vec{x_n}$ is
$W_{\mu,\Sigma,\rho}(\vec{x}) := \prod_{i=1}^n \sum_{j=1}^k \frac{\rho_j}{2\pi \det(\Sigma_j)} e^{-\frac{1}{2}(\vec{x_i}-\mu_j)^T \Sigma_j^{-1} (\vec{x_i}-\mu_j)}$
From this, once we pick probability distributions over the $\Sigma$, $\mu$, and $\rho$, we can calculate the posterior probability of a given set of clusters as
p($\Sigma$, $\mu$, $\rho$ | $\vec{x}$) $\propto$ p($\Sigma$) p($\mu$) p($\rho$) $W_{\mu,\Sigma,\rho}(\vec{x})$
This corresponds to the following graphical model:
Note that once we have a set of clusters, we can also determine the probability that a given point belongs to each cluster:
p($\vec{x}$ belongs to cluster $(\Sigma, \mu, \rho)$) $\propto$ $\frac{\rho}{2\pi \det(\Sigma)} e^{-\frac{1}{2}(\vec{x}-\mu)^T \Sigma^{-1} (\vec{x}-\mu)}$.
You might notice, though, that in this case it is much less straightforward to actually find clusters with high posterior probability (as opposed to in the previous case, where it was quite easy to distinguish between Fair, Unfair, and Weird, and furthermore to figure out the most likely values of $\theta$ and $\phi$). One reason why is that, in the previous case, we really only needed to make one-dimensional searches over $\theta$ and $\phi$ to figure out what the most likely values were. In this case, we need to search over all of the $\Sigma_i$, $\mu_i$, and $\rho_i$ simultaneously, which gives us, essentially, a $3k-1$-dimensional search problem, which becomes exponentially hard quite quickly.
This brings us to an important point, which is that, even if we write down a model, searching over that model can be difficult. So in addition to the model, I will go over a good algorithm for finding the clusters from this model, known as the EM algorithm. For the version of the EM algorithm described below, I will assume that we have uniform priors over $\Sigma_i$, $\mu_i$, and $\rho_i$ (in the last case, we have to do this by picking a set of un-normalized $\rho_i$ uniformly over $\mathbb{R}^k$ and then normalizing). We’ll ignore the problem that it is not clear how to define a uniform distribution over a non-compact space.
The way the EM algorithm works is that we start by initializing $\Sigma_i,$ $\mu_i$, and $\rho_i$ arbitrarily. Then, given these values, we compute the probability that each point belongs to each cluster. Once we have these probabilities, we re-compute the maximum-likelihood values of the $\mu_i$ (as the expected mean of each cluster given how likely each point is to belong to it). Then we find the maximum-likelihood values of the $\Sigma_i$ (as the expected covariance relative to the means we just found). Finally, we find the maximum-likelihood values of the $\rho_i$ (as the expected portion of points that belong to each cluster). We then repeat this until converging on an answer.
For a visualization of how the EM algorithm actually works, and a more detailed description of the two steps, I recommend taking a look at Josh Tenenbaum’s lecture notes starting at slide 38.
The Mind Projection Fallacy
This is perhaps a nitpicky point, but I have found that keeping it in mind has led me to better understanding what I am doing, or at least to ask interesting questions.
The point here is that people often intuitively think of probabilities as a fact about the world, when in reality probabilities are a fact about our model of the world. For instance, one might say that the probability of a child being male versus female is 0.5. And perhaps this is a good thing to say in a generic case. But we also have a much better model of gender, and we know that it is based on X and Y chromosomes. If we could look at a newly conceived ball of cells in a mother’s womb, and read off the chromosomes, then we could say with near certainty whether the child would end up being male or female.
You could also argue that I can empirically measure the probability that a person is male or female, by counting up all the people ever, and looking at the proportion of males and females. But this runs into two issues — first of all, the portion of males will be slightly off of 0.5. So how do we justify just randomly rounding off to 0.5? Or do we not?
Second of all, you can do this all you want, but it doesn’t give me any reason why I should take this information, and use it to form a conjecture about how likely the next person I meet is to be male or female. Once we do that, we are taking into account my model of the world.
Statistics
This final section seeks to look at a result from classical statistics and re-interpret it in a Bayesian framework.
In particular, I’d like to consider the following strategy for rejecting a hypothesis. In abstract terms, it says that, if we have a random variable Data’ that consists of re-drawing our data assuming that our hypothesis is correct, then
p(Hypothesis) < p(p(Data’ | Hypothesis) <= p(Data | Hypothesis))
In other words, suppose that the probability of drawing data less likely (under our hypothesis) than the data we actually saw is less than $\alpha$. Then the likelihood of our hypothesis is at most $\alpha$.
Or actually, this is not quite true. But it is true that there is an algorithm that will only reject correct hypotheses with probability $\alpha$, and this algorithm is to reject a hypothesis when p(p(Data’ | Hypothesis) <= p(Data | Hypothesis)) < $\alpha$. I will leave the proof of this to you, as it is quite easy.
To illustrate this example, let’s suppose (as in a previous section) that we have a coin and would like to determine whether it is fair. In the above method, we would flip it many times, and record the number H of heads. If there is less than an $\alpha$ chance of coming up with a less likely number of heads than H, then we can reject the hypothesis that the coin is fair with confidence $1-\alpha$. For instance, if there are 80 total flips, and H = 25, then we would calculatae
$\alpha = \frac{1}{2^{80}} \left(\sum_{k=0}^{25} \binom{80}{k} + \sum_{k=55}^{80} \binom{80}{k} \right)$.
So this seems like a pretty good test, especially if we choose $\alpha$ to be extremely small (e.g., $10^{-100}$ or so). The mere fact that we reject good hypotheses with probability less than $\alpha$ is not helpful. What we really want is to also reject bad hypotheses with a reasonably large probability. I think you can get around this by repeating the same experiment many times, though.
Of course, Bayesian statistics also can’t ever say that a hypothesis is good, but when given two hypotheses it will always say which one is better. On the other hand, Bayesian statistics has the downside that it is extremely aggressive at making inferences. It will always output an answer, even if it really doesn’t have enough data to arrive at that answer confidently. | 7,177 | 29,399 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 106, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2021-49 | latest | en | 0.968317 |
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Zeckendorf Representation
Tool to apply / check the Zeckendorf theorem stipulating that any integer can be written in the form of sum of non consecutive Fibonacci numbers also called Zeckendorf representation.
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# Zeckendorf Representation
## Zeckendorf Representation Calculator
Representation Zeckendorf (classic positive Fibonacci) NegaFibonacci (Negative Fibonacci indices)
### What is the Zeckendorf theorem? (Definition)
Every natural integer $n \in \mathbb {N}$ has a unique representation in the form of a sum of non-consecutive Fibonacci numbers. Its formula is written: $$n = \sum_{i=0}^{k} \alpha_i F_{i}$$ with $F_i$ the ith Fibonacci number, $\alpha_i$ is a binary number $0$ or $1$ (a way to indicate that the number of Fibonacci is in the sum, or it is not) and $\alpha_i \times \alpha_{i + 1} = 0$ (a way to prevent 2 numbers consecutive Fibonacci).
This proprety is used in Fibonacci coding (a binary representation of any integer based on the values of $\alpha_i$ in the formula above)
### How to calculate a Zeckendorf representation?
Enter a value of a number $N$ and dCode will do the calculation automatically.
Example: 10000 is the sum of $6765 + 2584 + 610 + 34 + 5 + 2$, respectively the 20th, 18th, 15th, 9th, 5th and 3rd Fibonacci numbers
Algorithmically, dCode uses Binet's formula to obtain Fibonacci numbers close to a given number and subtracts them recursively until finding the Zeckendorf representation.
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# A set has 16 numbers. If a number is three times the average
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A set has 16 numbers. If a number is three times the average [#permalink] 09 Nov 2006, 01:28
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A set has 16 numbers. If a number is three times the average of the other 15 numbers, the number is what fraction of the sum of the total numbers?
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Let the sum of the 15 numbers be x
Let the number be A
Then, A = x/15
Sum of total numbers = x+A
Fraction = A/x+A = (x/15)/x+(x/15) = 1/16
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ywilfred wrote:
Let the sum of the 15 numbers be x
Let the number be A
Then, A = x/15
Sum of total numbers = x+A
Fraction = A/x+A = (x/15)/x+(x/15) = 1/16
Ywilfred, A=3x/15 so their sum is 18x/15
The fraction is therefore 1/6
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hosam wrote:
ywilfred wrote:
Let the sum of the 15 numbers be x
Let the number be A
Then, A = x/15
Sum of total numbers = x+A
Fraction = A/x+A = (x/15)/x+(x/15) = 1/16
Ywilfred, A=3x/15 so their sum is 18x/15
The fraction is therefore 1/6
oops... hahaha.. yeah, thanks for pointing that out. the method is still the same though..
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Let the average of 15 numbers is x
Average of 16th number is 3x
the proportion of 16th number = 3x/(3x+15x) = 1/6
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the way is difficult, but perfect trust makes it easy.
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null | null | null | null | null | null | Posted on January 28th, 2011
Dr. Daya Hewapathirane
It is often overlooked but is necessary to highlight in a most prominent manner that Buddhism, the foundation of our nation’s culture prevented many discriminatory and criminal practices and cultural norms of Hindu Indians from being adopted in our country. We are indebted to and revere the founders and guardians of our nation “”…” our Sinhala royalty and nobility and our esteemed Mahasangha for protecting our nation from these extreme forms of Hindu Indian prejudices, injustice and violence. In spite of being exposed to Hindu India from historic times and in spite of our country being under the control of Hindu Dravidians for varying periods of time in the past, Buddhist principles and cultural norms which form the basis of our nations culture protected our nation and disallowed our people from observing unhealthy and undesirable Hindu cultural norms widely practiced in India. These immoral practices of Indian Hindus, being completely contrary to Buddhist teachings and social norms were not acceptable to our people. In fact, they shunned these gruesome practices of Hindus. This is a clear case of our people not being prone to blindly accepting whatever cultural norms that came to us from India. Compatibility with Buddhist principles and adaptability to suit Buddhist norms were fundamental considerations in the incorporation of elements of other cultures into to our own. This has resulted in the development in Sri Lanka of a national culture that is unique to us, a culture that strongly reflects Buddhist ethics and Buddhist principles. Our nation’s unique Buddhist social structure is indigenous to our nation, it is home grown, and therefore gives our nation a distinct identity as a highly refined, virtuous, ethical, gracious and honourable nation.
Cultural norms of our nation are founded on values cherished by our Sinhala Buddhist civilization. These values inculcated in us by the Buddha Dhamma have been upheld by our people for over 2300 years. A fundamental value that Buddhism imbibes in people is that life is precious, that all sentient beings desire to live, fear death, harm and suffering and that one should act in such a way that would guarantee the safety of life of all, along with their possessions and belongings. The Buddha’s teaching emphasizes boundless compassion to all living beings. The first precept of a Buddhist is that “I shall commit to myself that I shall not kill”. Buddhism does not teach or give a license to kill to suit one’s needs or because it is in-keeping with religious beliefs. The Buddha admonished his followers not to blindly hang on to beliefs, upholding them as the sole truth and denouncing the rest as false, for such an attitude is a primary source of conflict. Buddhism teaches that the act of harm or kill will come behind you and you will have to face the same repercussions either in this birth or the next birth, and when it so happens you will have no place to hide. What is taught in Buddhism is not only to avoid taking any kind of life, but also not to praise and promote violence, brutality and hostility. Non-violence and non-cruelty are the fundamental principles in Buddhism.
The Buddha expounded that the phenomenal world that we inhabit, is engulfed in the “fires” of greed, hatred and delusion raging fiercely in the hearts of people and they are the basic cause of suffering in human existence. It is the struggle associated with the impulses of these three mental poisons that causes so much pain, distress and destruction in human society. This is clearly evident when we try to understand the Hindu attitude towards children and women as enumerated in the foregoing paragraphs titled “ƒ”¹…”Killing infant females’ and “ƒ”¹…”Dowry Related Killings’ in India.
It is a common and widespread practice among the Hindus of India to kill infant females and abort female fetuses because of the obsession among these Hindu Indians for sons. This has been one of the most horrendous traditional practices and in fact, a criminal social norm of Hindu India from early times. The dowry system as practiced in India as an indispensable custom in Hindu marriages is the reason for this appalling community practice. The Indian Hindu dowry system requires the family to pay out a huge amount of money when a female child is married. Therefore, for a poor family, the birth of a girl can signal the beginning of financial ruin and hardship. Much of this utterly discriminatory criminal anti-female bias is to do with cultural beliefs and social norms of Hindu Indians. What is noteworthy is the fact that this ghastly custom is by no means, exclusive to poor Hindus, but prevails irrespective of the social and economic status and educational exposure of Hindu Indians, wherever they may live in the sub continent. By avoiding or preventing a girl being born, a family can avoid paying a large dowry on the marriage of their daughter. Girls with older sisters are often subject to the highest risk of mortality.
In Hindu India, sons are the parent’s only source of security in old age. This is particularly so where women have little economic independence or cannot inherit property. Son preference is also strong because in India women have fewer opportunities to earn income and invest household resources in female children. In Hindu tradition, sons are needed by Hindus for the cremation of deceased parents. Only sons can light the funeral pyre. It is a strong belief among Hindus that sons help in the salvation of the souls of dead parents by performing the ritual called “pind daan”. In the Hindu tradition, only sons can pray for and release the souls of dead parents, and only males can perform birth, death and marriage rituals. In contrast, Buddhist funeral ceremony is a very simple rite that can be performed by the widow, the daughter, or anyone else. According to Buddhist thought, future happiness does not depend on funeral rites, but on an individual’s actions while living.
India is a country with the most pervasive preference for sons and one of the highest levels of child mortality for girls in the world. Child mortality for girls exceeds that of boys by 43%. According to a recent report of the United Nations Children’s Fund, up to 50 million females are missing from India’s population as a result of systematic gender discrimination. In other words, about 50 million girls and women are killed or disappear in India every year owing to this criminal socio-cultural norm among Hindus. In India, there are less than 93 women for every 100 men in the population. The accepted reason for such disparity is the popular practice of female infanticide in India. In the year 2000, 68 of 1000 girls reported to have been born in India did not live beyond their first birthday and 86 of 1000 births did not live beyond age five.
According to the UNESCO, the problem is getting worse as scientific methods of detecting the sex of a baby and of performing abortions is improving in India. Diagnostic teams with ultrasound scanners detecting the sex of a child are commonly available all over India. They advertise with catch lines such as “spend 600 rupees now and save 50,000 rupees later”. These methods are becoming increasingly available in rural areas of India, and the trend towards the abortion of female fetuses is on the increase. Tamil Nadu, Karnataka, Rajasthan, Uttar Pradesh, Bihar, Punjab, and Gujarat are the provinces with the strongest preference for sons. This was revealed by the National Family Health Survey of India conducted in 1992-93. The first two provinces “”…” Tamil Nadu and Karnataka are the foremost Dravidian provinces of India with a combined population that exceeds 120 million. Tamil Nadu is the homeland of the Tamils wherever in the world they live, and culturally linked closely with the Tamils of Sri Lanka. It is well known that caste is a primary component of Tamil culture in Sri Lanka. Hindu religious practices among Tamils have reinforced their caste system. In fact, people of so-called low castes are not allowed in Hindu Shrines patronized by the so called high caste Hindu Tamils. The savage Tamil LTTE terrorist movement clearly reflected caste differences and rivalries. These terrorists were generally drawn from the Tamil fisherman caste which is a so-called low caste among the Tamils.
It is ironical, in fact hypocritical, that alongside the differential treatment and for that matter discrimination in the most extreme forms against females, the Hindus hold several Goddesses in high esteem and special kovils are built to venerate them. According to the Hindu religion, the Supreme Being has both masculine and feminine traits and the female part is as important as the male part. Among prominent goddesses in Hinduism are Durga symbolizing moral order and righteousness, Lakshmi symbolizing wealth and fortune “”…” material and spiritual, Saraswathi symbolizing arts, music, knowledge and wisdom and considered as the consort of Brahma the creator of the universe in the Hindu religion, Kali who is the violent and ferocious of all goddesses and is referred to as goddess of death and Parvathi who the consort of Shiva, the trinity god in Hinduism. How can Hindu males venerate female goddesses while indulging in or subscribing to gruesome criminal actions against females in their day to day lives?
For whatever reasons some Sinhala Buddhists also venerate a few of these Hindu goddesses but what is important to note is that there is no hypocrisy as far as the attitude of these Buddhists are concerned, because unlike their Hindu counterparts, Sinhala Buddhists do not subscribe to the differential treatment and discrimination against women in the real world they live in. Also, it is noteworthy that Buddhists hold motherhood in high esteem irrespective of the gender composition of her family. Mothers may have special attention for her sons but not at the expense of her daughters. In Buddhism, motherhood is considered as a position of high responsibility and respectability. Highlighting this special position of the mother, the Buddha raised the status of women in society. The Buddha said that a person has none else as worthy of honour and respect as one’s own mother. In Buddhism, the woman as the mother is always mentioned first when referring to the parental pair in the compound form “mata-pita”. The father is considered as playing a secondary role in bringing up children. In his discourses the Buddha highlights the close intimacy in the relationship between mother and her offspring, showing that none else can provide that love and protection of a mother to her offspring. When a mother fails in her duty as a mother, neglects or ill-treats her children, the impact of such action on the children is highly injurious.
There are deep rooted prejudices against women in India. Cultural practices such as the payment of dowry tend to subordinate women in Indian society. Although the dowry was legally prohibited in 1961, it continues to be highly institutionalized and grooms often demand a dowry consisting of a large sum of money, farm animals, furniture, and electronics. The practice of dowry abuse is rising in India. When the dowry amount is not considered sufficient or is not forthcoming, the bride is often harassed, abused and made miserable. This abuse can escalate to the point where the husband or his family burns the bride, often by pouring kerosene on her and lighting it and killing her. Such killings are reported as accidents or suicides by the family.
In Delhi, a woman is burned to death almost every twelve hours. The number of dowry murders shows an increasing trend. A 1997 government report stated that at least 5000 women die each year because of dowry deaths, and at least a dozen die each day in “kitchen fires” which are intentional. These official records are under-reported to a great extent. According to the 1997 report dowry deaths are showing an increasing trend. Convictions being rare, and judges who are usually men are often uninterested and susceptible to bribery. The lack of official registration of this crime is well evident in Delhi, where 90% of cases of women burnt were recorded as accidents, 05% as suicide and only the remaining 05% percent were shown as murder. The maternal mortality in India is the second highest in the world.
The low status of women in India is confirmed by the Human Development Report (1995) of the United Nations Development Programme where the status of women in India is placed in the bottom one fourth (1/4) of all 195 countries in the world. There are approximately 10 million prostitutes in India (Human Rights Watch, Robert I. Freidman, “India’s Shame: Sexual Slavery”¦,” The Nation, 8 April 1996). There are 300,000-500,000 children in prostitution in India (Rahul Bedi, “Bid To Protect Children As Sex Tourism Spreads”, London’s Daily Telegraph, 23 August, 1997). The “ƒ”¹…”devadasi’ tradition prevalent in many parts of India, prominently in Dravidian Karnataka province, continues to legitimize child prostitution. Tamil Nadu, Karnataka, Andhra Pradesh and Maharashtra are considered “high supply zones” for women in prostitution (Central Welfare Board, Meena Menon, and “The Unknown Faces”). In the Districts bordering Karnataka and Maharashtra known as the “devadasi belt”, women are in prostitution either because their husbands deserted them, or they are trafficked through coercion and deception. A devadasi is a woman married to a god and many of them are dedicated into prostitution for a goddess named Yellamma. Tamil Nadu Hindu leaders such as Jayalalitha will be doing a better service by her Tamil people if her attention is focused on uplifting the low status of women in Tamil Nadu where prostitution is rampant.
In 1997, there were reports of Indian armed forces arresting, torturing and molesting women and girls in Kashmir. Every day the local newspapers report such incidences. (KASHNet, Human Rights Information Network, 14 August 1997). Women and girls have been systematically brutalized and raped by Indian forces in house to house searches in Kashmir between October 1996 and December 1997. (“Rape and Molestation: A Weapon of War in Kashmir,” The Institute of Kashmir Studies,” 1998). Here, it is relevant to note that, withdrawing their earlier allegation against our armed forces, the USA State Department declared later that there have been no reports that rape and sexual abuse were used as tools of war in Sri Lanka by our Military forces unlike in other conflict areas around the world. It is noteworthy that our Military forces are basically composed of Sinhala Buddhists and are guided by Buddhist values.
The Buddha’s liberal attitude toward women had a great impact on the behavior of both men and women in Buddhist societies. The Buddhist doctrine of salvation through an individual’s own efforts presupposes the spiritual equality of all beings, male and female. This assertion of women’s spiritual equality, explicitly enunciated in the texts has had a significant impact on social structures and how women are viewed in the world. Women and men alike are able to attain the Buddhist goal by following the prescribed path; no external assistance in the form of a priestly intermediary or veneration of a husband is necessary as in the case of Hinduism. The Buddha condemned the caste structure dominated by Brahmins and denounced excessive ritual and sacrifice.
Understanding the culture of our nation involves knowing the basic values developed and promoted in Buddhism. They have exerted the greatest impact on all aspects of life in our motherland. Whether one calls Buddhism a religion or a philosophy, it is a way of life for mankind so that all beings who are here and everywhere live in comfort and security, without any threat on anybody’s account, to their continuance and survival- “Sukhino va khemino hontu sabbe satta bhavantu sukhitatta”.
As an integral part of the Buddhist spiritual path, compassion is a state of mind that is non-violent, non-harming and non-aggressive. It is a mental attitude based on the wish for others to be free of their suffering and is associated with a sense of commitment, responsibility and respect towards the other. Genuine compassion is based on the rationale that all human beings have an innate desire to be happy and overcome suffering. They have the natural right to fulfill this fundamental aspiration. On the basis of the recognition of this equality and commonality, one develops a sense of affinity and closeness with others. With this foundation one can feel compassion regardless of whether one views the other person as a male or female, infant, child or adult, Hindu, Muslim Buddhist or Christian of whatever caste. It is based on the other’s fundamental rights rather than your own mental projection.
Gaining peace and tranquility of mind is what Buddhists seek because it is such a mind that enables the development of wisdom and insight and thereby the experience of true happiness. It is the cultivation of one’s mind that is central to Buddhism and not the observance of dogmas and beliefs. Inner transformation by one’s own efforts is what Buddhists strive to attain. Inner discipline is the basis of Buddhist spiritual life and this involves combating negative states of mind such as anger, hatred, greed, and jealousy and the cultivation of positive states of mind such as compassion, kindness, tolerance and caring.
“Greed” indicates uncontrolled desire for, and attachment to material and other forms of worldly comforts, for wealth, dowry, power, fame and sons instead of daughters. The impulse of “hatred” involves resentment, rage and envy that are triggered when our egocentric desires are not fulfilled. These escalate into various forms of destruction and violence like in the case of aborting female fetuses, getting rid of female infants and discriminatory treatment of daughters as opposed to sons in Hindu families. “Delusion” refers to willful ignorance of reality or the ignorance of the true nature of life and the world. The wisdom that illuminates and reveals the true nature of life is referred to in Buddhism as “enlightenment.” Delusion clouds and obscures the light by which one might see things in their true nature. It makes one believe in something that contradicts reality.
Buddhist values are geared at developing a social ethic which, would contribute to co-existence, mutual understanding, co-operation and total harmony. It is to achieve this goal that Buddhism very strongly upholds that mankind is of one species, and hence everyone should be charitable and liberal towards the others, be pleasant in speech to them, do whatever is beneficial to them and above all, be impartial and treat all equally. To strengthen impartiality, people are advised not to succumb to biases and prejudices not to give in to hatred, fear, confusion, but to rise above them and do what is righteous. This concept of righteousness, which is designated by the term “ƒ”¹…”Dhamma’ in fact, provides the firm foundation for the whole of Buddhist culture. The general admonition is to do what is righteous or Dhamma and avoid what is unrighteous or adhamma and what is righteous is what is beneficial to one and others, as well.
On this basis all that is beneficial to oneself and others is considered meritorious or Punna and wholesome or Kusala and their opposites as demeritorious or Paapa and unwholesome or akusala. As Stanza No. 183 of the Dhammapada states: Not to do any evil, to cultivate good, to purify one’s mind “”…” this is the Teaching of the Buddhas. It is on this basic teaching that Buddhist values are developed, lives are moulded and social relations are cultivated. This explains why Buddhist culture attempts to nurture in the people a feeling for others, to mutually share with others moments of happiness and joy, to show respect to elders, to care for parents to attend on the sick and destitute, to honour and respect those who are deserving, to treat guests and visitors with friendliness and affection. These values brought and taught by Buddhism have sunk deep into the ethos of our people. The Panchaseela or five ethical precepts for lay persons provide additional content to the Buddhist conception of social institutions and conceptions of the good. These precepts enjoin refraining from killing, stealing, lying, sexual misconduct and intoxication. Put together, and viewed in a social context, they together constitute advice against violence and actions likely to sow discord, and advice towards openness and integrity. These more general values can inform the development of social institutions.
Sri Lanka’s Constitution guarantees equal rights without discrimination on grounds of sex and provides for affirmative action to ensure equal rights. In 1993, a “WOMEN’S CHARTER” became law, providing greater policy coherence on women’s issues and this Charter has been accepted as the cornerstone of all policy decisions on women by successive governments. A democratic society founded on Buddhist principles enshrined in its social order and institutions has a greater prospect for success, greater prospects for providing its citizenry with good lives and in fact greater claim to moral legitimacy. Generally in Sri Lanka, especially among the Sinhala people, there is no sex preference for males in child birth. In fact, having a daughter as the first child is considered as a blessing and a good omen for the new family. Also, dowry is not a primary consideration in most marriages in the Sinhala community. In some arranged marriages the girls’ party may declare prior to decision on marriage whatever that is offered as dowry for their child. Often it is left to parents to offer what they could in whatever form, for their daughter, as a form of material assistance for the new couple to establish themselves comfortably. In fact, in most arranged marriages the socio-economic standing of both parties, including property ownership, financial standing, education and professional status are primary considerations.
In contrast to the situation in India and other South Asian countries in general, in Sri Lanka, increased economic opportunities for women have resulted in parents regarding their daughters as economic assets rather than as liabilities. Increased opportunities of education for female children have led to an increase in their income-earning potential and thereby raise their economic value to their parents. Compared with other Southern and South Asian countries, the status of women in Sri Lanka is found to be more advanced. The many social welfare programmes of post-independence decades have helped to create favourable conditions for women, promoting greater participation of women in the development process. These include (a) rapid expansion of literacy and educational attainment of women, (b) improved life expectancy and decline in fertility and (c) wider participation of women in formal and informal economic activities.
Female literacy is quite high at 87% in Sri Lanka for several years. Female literacy in urban areas is 91%, while the rural rate is 78%. School attendance is equally high for both sexes at 84 per cent. In 1987, nearly 60 per cent of married women in the age group 15-49 years had an education beyond the primary level. The percentage of women entering universities in Sri Lanka has increased to 54.3%. About 50% of our school going population consists of girls. According to the New Internationalist, Issue 240 – February 1993, in developing countries of Asia and Africa, fewer girls than boys go to school and they spend fewer years there. Out of 100 million children not in primary school, two-thirds are girls.
The maternal mortality ratio is 32 per 100,000, the lowest in South Asia, whilst the child mortality rate under the age of 5 has declined to 10.2% per thousand live births. The average life expectancy of women in Sri Lanka is 76 years which surpasses the male life expectancy rate.
The economic participation of women in the modern sector has shown a marked increase in recent years and has helped to improve their social mobility. In manufacturing industries and export oriented modern industry about half of the total employed are women. Increased employment opportunities for women have given them a high degree of economic independence and personal freedom. With the opening up of foreign employment opportunities many women in our society play a constructive role in improving the economic conditions of their families by working abroad. Today 48% of our overseas workers comprise women. In the higher echelons of the public service and in the professional categories in the private sector women are increasingly playing a key role. In urban areas, many women are higher wage earners than their male spouses.
In Buddhism, unlike in Hinduism and Christianity, marriage is not a sacrament. It is a purely secular contract. In the Sigaalovaada Sutta the Buddha gives advice of a very practical nature to a young layman on how spouses should treat one another. The marital union is approached in a spirit of warm fellowship. The marital relationship is a reciprocal one with mutual rights and obligations, which is a momentous departure from Hindu ideas of marriage. The significant point is that the Buddha’s injunctions are applicable to both parties.
In ancient India a widow was expected to lead a life of strict celibacy and severe austerity upon the demise of her husband, for she was thought to be bound to him beyond death. With the death of husband she loses her social and religious status. In Buddhism, by contrast, death is considered a natural and inevitable end for all beings. As a result, a woman suffers no moral degradation on account of widowhood, nor is her social status altered in any way. In Sri Lankan society, a widow does not have to proclaim her widowhood in any tangible way, such as relinquishing her ornaments, shaving her head, or practicing self-mortification. The remarriage of widows is a common practice with no stigma attached. The disgusting sati ritual is unknown in Sri Lanka or any other Buddhist society. In India, in the year 1990, more than 50 widows were burnt alive when their husbands’ bodies were cremated in a ritual known as “sati,” based on the belief that a Hindu woman has no existence independent of her husband. (Sonali Verma, “Indian women still awaiting independence,” Human Rights Information Network: Indi News Network Digest, Volume 2, Issue1648, 16 August 1997).
What is important to note is that within the family, Sri Lankan women are less vulnerable to discrimination and oppression than their counterparts in India and other South Asian countries. The extreme situations of male dominance such as dowry deaths and widow immolation which are common in Hindu India are unheard of in Sri Lanka. So is the traditional practice of child marriages which is widely prevalent in India and totally non-existent in Sri Lanka. About 40% of the world’s child marriages occur in India. According to UNICEF’s “State of the World’s Children “”…” 2009″ report, 47% of India’s women aged 20to 24 were already married before their legal age of 18, with 56% in rural areas.
These positive comments on the status of women in Sri Lanka do not imply that everything is fine with Sri Lankan women. There are many serious issues and pervasive problems that remain unresolved, such as sexual assault, rape and spousal abuse associated with alcohol abuse, but the fact remains that on a comparative basis, the status of women in Sri Lanka is far better than that of women in India and other countries of South Asia. In general, women in Sri Lanka are relatively free from extreme forms of discrimination and harassment that are characteristic of Indian women and generally of women in other major Asian cultures. The social freedom enjoyed by women in Buddhist societies such as that of Sri Lanka, has evoked comment from many Western observers. A complete lack of segregation of the sexes has distinguished Buddhist societies from those of the Middle East, the Far East, and the Indian subcontinent, where segregation has often lead to the seclusion and confinement of women behind walls and veils.
Dr. Daya Hewapathirane
January 28, 2011
1. Lorenzo Says:
Thank you Dr Daya for stating facts as they are.
If SL was not a Buddhist country, it would have ended up in the lap of these barbarians.
Before Buddhism came to the country it was most probably a Hindu country. The king went hunting deer! Obviously the Hindu to Buddhist transformation was a move upwards.
Most LTTE cadres/leaders and Tamil racist political leaders were/are Hindus. They showed absolutely no morality whatsoever. The manner in which Hindu individuals killed young Buddhist monks, slayed hundreds in sacred Anuradhapura, the way they bombed the Sri Dalda Maligawa, destroyed hundreds of Buddhist temples, bombed the Kathankudi mosque and massacred babies in threatened villages show that they are the worse humans on the planet. There is no record of even Hindu political leaders condemning these attacks.
However, Hindus in SL have come a long way for the last 100 years from a very uncivilised Tamil Nadu tradition. One example is the Thesawalamei law. A woman cannot transfer her very own property without the written consent of her husband! The status of the woman was lower than an individual! In contrast even at that time Sinhala women held equal status to men. Charming Ceylon (1931) video clip makes this fact very clear.
2. jimmy Says:
I am sad for what happened to Innocent Budhist Monks , sinhala villagers who were massacred innocent lives lost by Brutal LTTErs. They do not represent all the Tamils
I know and am very sure my sinhala brothers feel the same way for what happened to innocent tamils during 77 , 83 riots . It does not represent the Sinhalese
I agree totally Tamils should abandon pipe dream of Eelam and be patriotic towards Srilanka . We have to thinklike Srilankans
It is the greatest responsibilty for SInhalese to make sure their Tamil brothers are protected in every way possible
Zero tolerence for any hate crimes or violence. There should be impartial investigation . Human rights should be upheld
Sinhaleese should forget the past and treat tamils as Srilankans and take care of them .
Equal opportunity in Education, jobs , Prmotions for All Srilankans.
I am sure there will be a better future for All of us and We can easily beat Singapore
3. Lorenzo Says:
What you say is contradictory.
“There should be impartial investigation.
Sinhaleese should forget the past and treat tamils as Srilankans and take care of them.”
So the Sinhalese should forget the past but Tamils can dwell in the past.
Forget what is bad for you and remember what is good for you.
That is not going to work. Investigations are all about the PAST and should not be attempted if we are going to move forward.
4. jimmy Says:
Yo have mistaken what I said sorry
I am also talking about future
What I meant is If any crimes takes place against any race from now onwards need full impartial investigation because we all are Srilankans
5. jimmy Says:
want to mention this newspaper is a great one to read which give an opportunnity for Sinahallese and Tamils to understand each other
I lived with Sinhalese friends , Great people who will not even kill a fly I am sure people in Lanka could work for Peace . Always think, dream love peace
I have Srilankan flag as background in my laptop . I agree with some writers in the forum Tamils did not have patriotic feelings . I am sure it will change when people love and respect each other I will pray for that. Majority should do more on this . It is like an elder brother should look after his little brother
I find I am addicted to this paper , I have decided to read it once a week
From this paper I could understand what sinhalese are looking from Tamils is nothing other than love for the Country
I love you all Peace Peace peace
6. M.S.MUdali Says:
Buddha was a Hindu and he was a Hindu until his death. Any disputes?
Sinhalese are tight lipped while Catholic Morons rampaged the country and allied with LTTE. Catholics target HINDUS and BUDDHISTS. But who is first? Hindus who are Tamils in Sri lanka is the easiest target. They tried and failed to carve out an enclave for Catholic rule.
How can anyone respect the Buddhist monks whne they talk politics? Because Monks showed that they are capable of murdering Prime Minister who brought respect to the Sangha in every way.
What is the status of the women of Sri Lanka now? Coming with nails from Muslim countries or throwing their kids in Kalu ganga or CATERING HIV infested whites in tourism industry is the best status?
How many children are forced in SEX trade in Sri lanka? Or exporting women to Muslims as their sex slaves is the best way?
just vomitting venom against Hindus is another face of racism. May be part of the Christian propaganda!
What is this author asking? To follow Somarama and Buddha Rakita?
How do these Buddhists justify Muslims and their 5 wives or their JIHAD( means killing non-muslims)? Muslims are allowed to have their “OWN” schools and courts. How? Where are the Buddhist principles in the case of Muslims?
7. Samson Says:
IPKF Son Mudali,
Truth hurts isn’t it?
There are no sex slaves in Muslim countries. Stop talking rubbish. Sex slaves were in Vanni. There is no racism here. Its all facts. What’s wrong with Muslims having 5 wives as long as it is within Islam law?
You have been writing racist slur against Catholics until now. Now your racism is directed at Muslims and Buddhists. So you hate everyone other than your race and religion which makes you a hardcore racist.
There is no Jihad in Sri Lanka. Only terrorists in Sri Lanka are Tamil Tigers. Now they too are also history. Dead pet dogs have better regard than Tamil Tiger terrorists.
8. Nanda Says:
Mudali is a Hindu extremist. He is a Tamil Tiger.
He had declared Jihad in Lanka Web. That is all.
He does not know Buddha did not “die”. He has insulted Buddha by calling him hindu. Hindus those days and even now sacrifice animals by mass killing. Buddha opposed this stupid cruel act. How can he be Hindu then ?
9. Lorenzo Says:
Got it.
Sure; fully agree. Futuristic.
I thought you are one of those war crimes losers like GTF, TAG. My apologies for the misunderstanding.
10. M.S.MUdali Says:
Buddha was not a Hindu? Then who was he? Catholic or Muslim?
Buddha was a Hindu reformer and tried to correct manythings in the Hindu religion. Only Hindus followed Buddhas reforms. Buddha never asked to build temples for him! Did he?
Nanda tells Buddha did not die! What a joke? Is he telling he became alive after his death like Christ? Then whose teeth in Dalada Maligava?
Buddha lived as a Hindu. That is why Hindu rituals are part of the Buddhism. Sri Lankan Sinhala Buddhism is known as PARANGI BUDDHISM. They simply tells lies about Buddha. Not only Buddha but many others too tried to correct Hindu religion. That is why Hindus venerate Buddha as BODHI MADHAVA.
All the Hindus never support animal sacrifice but few. Can we tell all the Buddhist monks are criminals and like Somarama?
In Sri Lanka Buddhas teachings not followed but Buddha became a political symbol. Buddha never wanted politics in hislife but Sri lankan monks have political party. That is the real insult for Buddha!
Buddha never asked his followers to bark TAMIL or Bengali but the so called Parangi Buddhists of Sri Lanka are barking all racists slurs because they sell Buddha for US dollars.
Islamic Law?
Immoral hindu practices must be banned but immoral Islamic practics to be coninued. Ohh What a logic? Does Buddhism tell to follow Islamic practices? Now the Muslim VOTE BANK is the issue and not Buddhism!
11. De Costa Says:
Parangiya again.
Don’t jump like a Hanuman.
Religion or race is a lable only. What do you know about Hinduism ? Teach us. What is Moksha ?
How do you explain the Dalits born from Brahma’s feet and fat Brhamins behave like animals born from mouth of Brhama?
Did Buddha taught this nonsense ? If so we can accept Buddha as a Hindu.
I am parangi but never heard about Parangi Buddhism. Some parangis are catholic and some are Buddhist. So what ?
Anyone can do anything to insult Buddha or Jesus or Mohamad. But they are out or hanuma’s comprehesion.
Why Samson gave you a name “IPKF son” ? -very suitable but you cannot be that young.
12. M.S.MUdali Says:
Parangiyas hate Buddha so much. That is why COSTAS appeared in Sri Lanka.
Buddha’s reforms are followed by Hindus and not by Catholics or Muslims.
Did Buddha ask you to eat meat? Did Buddha ask you to hate TAMIL people?
Samson and Costas are not Buddhists but Catholic racists and using Buddha for their hate propaganda. But Catholic Church publicly advocate for criminal deeds in Sri Lanka. Costa better talk about Catholic Church and not about Buddha because it is not your problem!
13. Nanda Says:
Mudali the Tiger,
Your true colours showing !
Buddhas are not born and do not die.
Hindus killed Mahatma. They are among the worst racist in the world.
Mahatma wanted to give equal opportunities to Datlits and Muslims so Hindus killed him. You are one of them. According to Buddha you and most of your fellow Tigers will definitely be born in hell.
14. Samson Says:
IPKF Son Mudali,
IPKFs were mostly Hindus. But how cruel were they to Jaffna Hindus, especially Hindu ladies.
The problem is certainly not in the Hindu religion but in the believers. Some are very primitive to worship the penis and the vagina. Hindu gods never told them to do so.
Siva sena is a Hindu extremist group with contacts to BJP. They burn down churches, mosques and Buddhist temples.
Recently a Tamil Hindu mob attacked a Buddhist temple in South India.
15. M.S.MUdali Says:
You are a son of a tourist! That is why you bark for Catholics. IPKF hunted LTTE and not the ordinary Tamils.
Buddhist temple in South India? what is that? Read the news better then bark.
BJP never touch a Buddhist temple but Muslims and Catholics burnt Buddhist temples.
Catholic Churches are supporting LTTE and deserved to be burned down in Sri Lanka too!
16. Samson Says:
IPKF Son Mudali,
You don’t have any originality. Just copy what others say. So very typical of you losers.
South Indian Hindu mobs attacked a Buddhist temple in India. You were among them. Weren’t you?
Anyway the best example of Hindu immoral practices is you who are a son of Hindu IPKF soldiers born to a Tamil Hindu woman in Jaffna. A Catholic priest looked after you. And you know the rest! That is why you hate them so much. You should claim damages from Vatican as many people like you did already. I have no hate or love for the Catholics.
17. jimmy Says:
can we have some decency please
This is the way Our people behaved for the last 60 years
we have to learn to respect each other .
18. cassandra Says:
Yes, Jimmy. Well said.
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# How many inches in 75 feet?
Wiki User
2009-12-04 03:16:13
12 inches = 1 foot
75 inches = (75 / 12) = 6.25 feet
Wiki User
2009-09-13 18:18:22
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# Compactness in Metric Spaces
## Math 331, Handout #2
We have proved the Heine-Borel Theorem for closed bounded intervals in R: If [a, b] is a closed
bounded interval, then every open cover of [a, b] has a finite subcover. This property can be
extended to certain other subsets of R and to certain subsets of metric spaces in general. The
sets in question are said to be “compact.” We can take the finite subcover property to be the
definition of compactness, although we will see that there are other equivalent properties that also
characterize compactness. Note that compactness can be defined entirely in terms of open sets,
without mentioning the distance measure. This means that compactness is really a topological
property rather than a metric property.
Definition 1. Let (M, d) be a metric space. A subset, K, of M is said to be compact if and only
if every open cover of K (by open sets in M ) has a finite subcover. If M itself has this property,
then we say that M is a compact metric space.
We start with the fact that in any metric space, a compact subset is closed and bounded.
Bounded here means that the subset “does not extend to infinity,” that is, that it is contained in
some open ball around some point. But we can give an alternative definition, which says that a set
is bounded if there is a limit on how far apart two points in the set can be. We can also define the
“diameter” of such a subset:
Definition 2. Let (M, d) be a metric space. A subset X of M is said to be bounded if and only if
the set {d(x, y) | x, y ∈ X} is bounded above. For a non-empty bounded set X, we define diam(X),
the diameter of X, to be the least upper bound of the set {d(x, y) | x, y ∈ X} (which exists by the
least upper bound property of R, since the set is bounded above).
Take careful note of how the finite subcover property is used in the following proof; the technique
is common in proofs about compactness.
Theorem 1. Let (M, d) be a metric space, and let K be a compact subset of M . Then K is a
closed subset of M , and K is bounded.
Proof. Let K be a compact subset of a metric space (M, d). To prove that K is closed, we show that
the complement, G = M rK, is open. Let z ∈ G. We need to find > 0 such that B (z) ⊆ G. Now,
for any x ∈ K, let x = d(x, z). Since z 6∈ K, x > 0. The collection of open sets {Bx (x) | x ∈ K}
is an open cover of K (since any x ∈ K is covered by Bx (x)). Since K is compact, there is a finite
subcover of this cover; that is, there is a finite set x1 , x2 , . . . , xn such that the corresponding open
balls already cover K. Let = 21 min(x1 , x2 , . . . , xn ). The claim is that B (z) ⊆ G. To show this,
let y ∈ B (z). We want to show y ∈ G, that is, y 6∈ K. Consider xi , where 1 ≤ i ≤ n. By the
triangle inequality, d(xi , y) + d(y, z) ≥ d(xi , z) = xi ≥ 2. So, d(xi , y) ≥ 2 − d(y, z) > 2 − = .
(The last inequality follows because d(y, z) < .) Then, since d(xi , y) > , y is not in the open ball
of radius xi about xi . Since the open balls Bxi (xi ) cover K, we have that y 6∈ K.
1
To prove that K is bounded, let x be some element of K, and consider the collection of
open balls of integral radius, {Bi (x) | i = 1, 2, . . . }. Since every element of K has some fi-
nite distance from x, this collection is an open cover of K, Since K is compact, it has a fi-
nite subcover {Bi1 (x), Bi2 (x), . . . , Bin (x)}, where we can assume i1 < i2 < · · · < in . But since
Bi1 (x) ⊆ Bi2 (x) ⊆ · · · ⊆ Bin (x), this means that Bin by itself already covers K. Then, for
y, z ∈ K, y and z are in Bxn (x), and d(y, z) ≤ d(y, x) + d(x, z) ≤ in + in . It follows that 2in is an
upper bound for {d(y, z) | y, z ∈ K}. So, K is bounded.
It is not true in general that every closed, bounded set is compact. However, that is true in
Rn , with the usual metric, and this fact gives a complete characterization of compact subsets of
Rn . We will not prove this at this time, but Exercise 5 proves it for the case n = 1.
The Bolzano-Weirstrass Theorem for closed, bounded intervals in R says that any infinite subset
of such an interval has an accumulation point in the interval. A similar result holds for a compact
subset of a metric space.
Theorem 2. Let (M, d) be a metric space and let K be a compact subset of M . Then any infinite
subset of K has an accumulation point in K.
Proof. We prove the contrapositive. Let X ⊆ K. Assume that X has no accumulation point in K.
We must show that X is finite. Let z ∈ K. Since z is not an accumulation point, there is an z > 0
such that X ∩ (Bz (z) r {z}) = ∅. Thus, X ∩ Bz (z) either is empty or is {z}. The set of open balls
{Bz (z) | z ∈ K} is an open cover of K. Since K is compact, there is a finite cover. That is, there
are finitely many points z1 , z2 , . . . , zn such that the corresponding open balls already cover K. But
each of the n open balls in that subcover contains at most one point of X, and it follows that X
has n or fewer points. So we have proved that X is finite.
In fact, a set being compact is actually equivalent to the property that every infinite subset of
that set has an accumulation point in the set. We have proved one direction of this equivalence.
We will not prove the other direction at this time.
Exercises
Exercise 1. Find an example of a closed and bounded set C in some metric space such that C is
not compact. (Hint: Consider the metric space (X, ρ) from exercise 1 in the first handout.)
Exercise 2. A subset X of a metric space is said to be totally bounded if for every > 0, X can
be covered by a finite collection of open balls of radius . Show that every compact set is totally
bounded.
Exercise 3. Let (M, d) be a metric space, let X be a non-empty subset of M , and let z be an
element of M . Define d(z, X) = glb{d(x, z) | x ∈ X}. Note that if z ∈ X, then d(z, X) = 0. Find
an example where z 6∈ X but d(z, X) = 0. Now suppose that K is a compact subset of M . Show
that d(z, K) = 0 if and only if z ∈ K.
Exercise 4. Let (M, d) be a metric space. Let K be a compact subset of M , and let C be a closed
subset of M . Then K ∩ C is compact. (Hint: The set M r C is an open set.)
Exercise 5. Show that any bounded, closed subset of R is compact. (Hint: Use the previous
exercise and the fact that any bounded, closed interval is compact.)
Exercise 6. Let (M, d) be a metric space, and let X be a subset of M . Show that X is bounded
(in the sense that it has a finite diameter) if and only if for any z ∈ M , there is a number N such
that X ⊆ BN (z). | 1,934 | 6,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2020-16 | latest | en | 0.937053 |
https://pressbooks.nscc.ca/algebratrigonometryopenstax/chapter/the-ellipse/ | 1,695,619,224,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506686.80/warc/CC-MAIN-20230925051501-20230925081501-00200.warc.gz | 522,557,002 | 36,125 | # 80 The Ellipse
### Learning Objectives
In this section, you will:
• Write equations of ellipses in standard form.
• Graph ellipses centered at the origin.
• Graph ellipses not centered at the origin.
• Solve applied problems involving ellipses.
Can you imagine standing at one end of a large room and still being able to hear a whisper from a person standing at the other end? The National Statuary Hall in Washington, D.C., shown in (Figure), is such a room.[1] It is an oval-shaped room called a whispering chamber because the shape makes it possible for sound to travel along the walls. In this section, we will investigate the shape of this room and its real-world applications, including how far apart two people in Statuary Hall can stand and still hear each other whisper.
### Writing Equations of Ellipses in Standard Form
A conic section, or conic, is a shape resulting from intersecting a right circular cone with a plane. The angle at which the plane intersects the cone determines the shape, as shown in (Figure).
Conic sections can also be described by a set of points in the coordinate plane. Later in this chapter, we will see that the graph of any quadratic equation in two variables is a conic section. The signs of the equations and the coefficients of the variable terms determine the shape. This section focuses on the four variations of the standard form of the equation for the ellipse. An ellipse is the set of all points$\,\left(x,y\right)\,$in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci).
We can draw an ellipse using a piece of cardboard, two thumbtacks, a pencil, and string. Place the thumbtacks in the cardboard to form the foci of the ellipse. Cut a piece of string longer than the distance between the two thumbtacks (the length of the string represents the constant in the definition). Tack each end of the string to the cardboard, and trace a curve with a pencil held taut against the string. The result is an ellipse. See (Figure).
Every ellipse has two axes of symmetry. The longer axis is called the major axis, and the shorter axis is called the minor axis. Each endpoint of the major axis is the vertex of the ellipse (plural: vertices), and each endpoint of the minor axis is a co-vertex of the ellipse. The center of an ellipse is the midpoint of both the major and minor axes. The axes are perpendicular at the center. The foci always lie on the major axis, and the sum of the distances from the foci to any point on the ellipse (the constant sum) is greater than the distance between the foci. See (Figure).
In this section, we restrict ellipses to those that are positioned vertically or horizontally in the coordinate plane. That is, the axes will either lie on or be parallel to the x– and y-axes. Later in the chapter, we will see ellipses that are rotated in the coordinate plane.
To work with horizontal and vertical ellipses in the coordinate plane, we consider two cases: those that are centered at the origin and those that are centered at a point other than the origin. First we will learn to derive the equations of ellipses, and then we will learn how to write the equations of ellipses in standard form. Later we will use what we learn to draw the graphs.
#### Deriving the Equation of an Ellipse Centered at the Origin
To derive the equation of an ellipse centered at the origin, we begin with the foci$\,\left(-c,0\right)\,$and$\,\left(c,0\right).\,$The ellipse is the set of all points$\,\left(x,y\right)\,$such that the sum of the distances from$\,\left(x,y\right)\,$to the foci is constant, as shown in (Figure).
If$\,\left(a,0\right)\,$is a vertex of the ellipse, the distance from$\,\left(-c,0\right)\,$to$\,\left(a,0\right)\,$is$\,a-\left(-c\right)=a+c.\,$The distance from$\,\left(c,0\right)\,$to$\,\left(a,0\right)\,$is$\,a-c$. The sum of the distances from the foci to the vertex is
$\left(a+c\right)+\left(a-c\right)=2a$
If$\,\left(x,y\right)\,$is a point on the ellipse, then we can define the following variables:
$\begin{array}{l}{d}_{1}=\text{the distance from }\left(-c,0\right)\text{ to }\left(x,y\right)\hfill \\ {d}_{2}=\text{the distance from }\left(c,0\right)\text{ to }\left(x,y\right)\hfill \end{array}$
By the definition of an ellipse,$\,{d}_{1}+{d}_{2}\,$is constant for any point$\,\left(x,y\right)\,$on the ellipse. We know that the sum of these distances is$\,2a\,$for the vertex$\,\left(a,0\right).\,$It follows that$\,{d}_{1}+{d}_{2}=2a\,$for any point on the ellipse. We will begin the derivation by applying the distance formula. The rest of the derivation is algebraic.
$\begin{array}{ll}\text{ }{d}_{1}+{d}_{2}=\sqrt{{\left(x-\left(-c\right)\right)}^{2}+{\left(y-0\right)}^{2}}+\sqrt{{\left(x-c\right)}^{2}+{\left(y-0\right)}^{2}}=2a\hfill & \text{Distance formula}\hfill \\ \sqrt{{\left(x+c\right)}^{2}+{y}^{2}}+\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}=2a\hfill & \text{Simplify expressions}\text{.}\hfill \\ \text{ }\sqrt{{\left(x+c\right)}^{2}+{y}^{2}}=2a-\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\hfill & \text{Move radical to opposite side}\text{.}\hfill \\ \text{ }{\left(x+c\right)}^{2}+{y}^{2}={\left[2a-\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\right]}^{2}\hfill & \text{Square both sides}\text{.}\hfill \\ \text{ }{x}^{2}+2cx+{c}^{2}+{y}^{2}=4{a}^{2}-4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}+{\left(x-c\right)}^{2}+{y}^{2}\hfill & \text{Expand the squares}\text{.}\hfill \\ \text{ }{x}^{2}+2cx+{c}^{2}+{y}^{2}=4{a}^{2}-4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}+{x}^{2}-2cx+{c}^{2}+{y}^{2}\hfill & \text{Expand remaining squares}\text{.}\hfill \\ \text{ }2cx=4{a}^{2}-4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}-2cx\hfill & \text{Combine like terms}\text{.}\hfill \\ \text{ }4cx-4{a}^{2}=-4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\hfill & \text{Isolate the radical}\text{.}\hfill \\ \text{ }cx-{a}^{2}=-a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\hfill & \text{Divide by 4}\text{.}\hfill \\ \text{ }{\left[cx-{a}^{2}\right]}^{2}={a}^{2}{\left[\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\right]}^{2}\hfill & \text{Square both sides}\text{.}\hfill \\ \text{ }{c}^{2}{x}^{2}-2{a}^{2}cx+{a}^{4}={a}^{2}\left({x}^{2}-2cx+{c}^{2}+{y}^{2}\right)\hfill & \text{Expand the squares}\text{.}\hfill \\ \text{ }{c}^{2}{x}^{2}-2{a}^{2}cx+{a}^{4}={a}^{2}{x}^{2}-2{a}^{2}cx+{a}^{2}{c}^{2}+{a}^{2}{y}^{2}\hfill & \text{Distribute }{a}^{2}.\hfill \\ \text{ }{a}^{2}{x}^{2}-{c}^{2}{x}^{2}+{a}^{2}{y}^{2}={a}^{4}-{a}^{2}{c}^{2}\hfill & \text{Rewrite}\text{.}\hfill \\ \text{ }{x}^{2}\left({a}^{2}-{c}^{2}\right)+{a}^{2}{y}^{2}={a}^{2}\left({a}^{2}-{c}^{2}\right)\hfill & \text{Factor common terms}\text{.}\hfill \\ \text{ }{x}^{2}{b}^{2}+{a}^{2}{y}^{2}={a}^{2}{b}^{2}\hfill & \text{Set }{b}^{2}={a}^{2}-{c}^{2}.\hfill \\ \text{ }\frac{{x}^{2}{b}^{2}}{{a}^{2}{b}^{2}}+\frac{{a}^{2}{y}^{2}}{{a}^{2}{b}^{2}}=\frac{{a}^{2}{b}^{2}}{{a}^{2}{b}^{2}}\hfill & \text{Divide both sides by }{a}^{2}{b}^{2}.\hfill \\ \text{ }\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\hfill & \text{Simplify}\text{.}\hfill \end{array}$
Thus, the standard equation of an ellipse is$\,\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1.$This equation defines an ellipse centered at the origin. If$\,a>b,$the ellipse is stretched further in the horizontal direction, and if$\,b>a,$ the ellipse is stretched further in the vertical direction.
#### Writing Equations of Ellipses Centered at the Origin in Standard Form
Standard forms of equations tell us about key features of graphs. Take a moment to recall some of the standard forms of equations we’ve worked with in the past: linear, quadratic, cubic, exponential, logarithmic, and so on. By learning to interpret standard forms of equations, we are bridging the relationship between algebraic and geometric representations of mathematical phenomena.
The key features of the ellipse are its center, vertices, co-vertices, foci, and lengths and positions of the major and minor axes. Just as with other equations, we can identify all of these features just by looking at the standard form of the equation. There are four variations of the standard form of the ellipse. These variations are categorized first by the location of the center (the origin or not the origin), and then by the position (horizontal or vertical). Each is presented along with a description of how the parts of the equation relate to the graph. Interpreting these parts allows us to form a mental picture of the ellipse.
### Standard Forms of the Equation of an Ellipse with Center (0,0)
The standard form of the equation of an ellipse with center$\,\left(0,0\right)\,$and major axis on the x-axis is
$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$
where
• $a>b$
• the length of the major axis is$\,2a$
• the coordinates of the vertices are$\,\left(±a,0\right)$
• the length of the minor axis is$\,2b$
• the coordinates of the co-vertices are$\,\left(0,±b\right)$
• the coordinates of the foci are$\,\left(±c,0\right)$, where$\,{c}^{2}={a}^{2}-{b}^{2}.\,$See (Figure)a
The standard form of the equation of an ellipse with center$\,\left(0,0\right)\,$and major axis on the y-axis is
$\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1$
where
• $a>b$
• the length of the major axis is$\,2a$
• the coordinates of the vertices are$\,\left(0,±a\right)$
• the length of the minor axis is$\,2b$
• the coordinates of the co-vertices are$\,\left(±b,0\right)$
• the coordinates of the foci are$\,\left(0,±c\right)$, where$\,{c}^{2}={a}^{2}-{b}^{2}.\,$See (Figure)b
Note that the vertices, co-vertices, and foci are related by the equation$\,{c}^{2}={a}^{2}-{b}^{2}.\,$When we are given the coordinates of the foci and vertices of an ellipse, we can use this relationship to find the equation of the ellipse in standard form.
### How To
Given the vertices and foci of an ellipse centered at the origin, write its equation in standard form.
1. Determine whether the major axis lies on the x– or y-axis.
1. If the given coordinates of the vertices and foci have the form$\,\left(±a,0\right)\,$and$\,\left(±c,0\right)\,$respectively, then the major axis is the x-axis. Use the standard form$\,\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1.$
2. If the given coordinates of the vertices and foci have the form$\,\left(0,±a\right)\,$and$\,\left(±c,0\right),$respectively, then the major axis is the y-axis. Use the standard form$\,\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1.$
2. Use the equation$\,{c}^{2}={a}^{2}-{b}^{2},\,$along with the given coordinates of the vertices and foci, to solve for$\,{b}^{2}.$
3. Substitute the values for$\,{a}^{2}\,$and$\,{b}^{2}\,$into the standard form of the equation determined in Step 1.
### Writing the Equation of an Ellipse Centered at the Origin in Standard Form
What is the standard form equation of the ellipse that has vertices$\,\left(±8,0\right)\,$and foci$\,\left(±5,0\right)?\,$
The foci are on the x-axis, so the major axis is the x-axis. Thus, the equation will have the form
$\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$
The vertices are$\,\left(±8,0\right),$so$\,a=8\,$and$\,{a}^{2}=64.$
The foci are$\,\left(±5,0\right),$so$\,c=5\,$and$\,{c}^{2}=25.$
We know that the vertices and foci are related by the equation$\,{c}^{2}={a}^{2}-{b}^{2}.\,$Solving for$\,{b}^{2},$ we have:
$\begin{array}{ll}{c}^{2}={a}^{2}-{b}^{2}\hfill & \hfill \\ 25=64-{b}^{2}\begin{array}{cccc}& & & \end{array}\hfill & \text{Substitute for }{c}^{2}\text{ and }{a}^{2}.\hfill \\ {b}^{2}=39\hfill & \text{Solve for }{b}^{2}.\hfill \end{array}$
Now we need only substitute$\,{a}^{2}=64\,$and$\,{b}^{2}=39\,$into the standard form of the equation. The equation of the ellipse is$\,\frac{{x}^{2}}{64}+\frac{{y}^{2}}{39}=1.$[/hidden-answer]
### Try It
What is the standard form equation of the ellipse that has vertices$\,\left(0,±4\right)\,$and foci$\,\left(0,±\sqrt{15}\right)?$
${x}^{2}+\frac{{y}^{2}}{16}=1$
Can we write the equation of an ellipse centered at the origin given coordinates of just one focus and vertex?
Yes. Ellipses are symmetrical, so the coordinates of the vertices of an ellipse centered around the origin will always have the form$\,\left(±a,0\right)\,$or$\,\left(0,\,±a\right).\,$Similarly, the coordinates of the foci will always have the form$\,\left(±c,0\right)\,$or$\,\left(0,\,±c\right).\,$Knowing this, we can use$\,a\,$and$\,c\,$from the given points, along with the equation$\,{c}^{2}={a}^{2}-{b}^{2},$to find$\,{b}^{2}.$
#### Writing Equations of Ellipses Not Centered at the Origin
Like the graphs of other equations, the graph of an ellipse can be translated. If an ellipse is translated$\,h\,$units horizontally and$\,k\,$units vertically, the center of the ellipse will be$\,\left(h,k\right).\,$This translation results in the standard form of the equation we saw previously, with$\,x\,$replaced by$\,\left(x-h\right)\,$and y replaced by$\,\left(y-k\right).$
### Standard Forms of the Equation of an Ellipse with Center (h, k)
The standard form of the equation of an ellipse with center$\,\left(h,\text{ }k\right)\,$and major axis parallel to the x-axis is
$\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$
where
• $a>b$
• the length of the major axis is$\,2a$
• the coordinates of the vertices are$\,\left(h±a,k\right)$
• the length of the minor axis is$\,2b$
• the coordinates of the co-vertices are$\,\left(h,k±b\right)$
• the coordinates of the foci are$\,\left(h±c,k\right),$where$\,{c}^{2}={a}^{2}-{b}^{2}.\,$See (Figure)a
The standard form of the equation of an ellipse with center$\,\left(h,k\right)\,$and major axis parallel to the y-axis is
$\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1$
where
• $a>b$
• the length of the major axis is$\,2a$
• the coordinates of the vertices are$\,\left(h,k±a\right)$
• the length of the minor axis is$\,2b$
• the coordinates of the co-vertices are$\,\left(h±b,k\right)$
• the coordinates of the foci are$\,\left(h,k±c\right),\,$where$\,{c}^{2}={a}^{2}-{b}^{2}.\,$See (Figure)b
Just as with ellipses centered at the origin, ellipses that are centered at a point$\,\left(h,k\right)\,$have vertices, co-vertices, and foci that are related by the equation$\,{c}^{2}={a}^{2}-{b}^{2}.\,$We can use this relationship along with the midpoint and distance formulas to find the equation of the ellipse in standard form when the vertices and foci are given.
### How To
Given the vertices and foci of an ellipse not centered at the origin, write its equation in standard form.
1. Determine whether the major axis is parallel to the x– or y-axis.
1. If the y-coordinates of the given vertices and foci are the same, then the major axis is parallel to the x-axis. Use the standard form$\,\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1.$
2. If the x-coordinates of the given vertices and foci are the same, then the major axis is parallel to the y-axis. Use the standard form$\,\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1.$
2. Identify the center of the ellipse$\,\left(h,k\right)\,$using the midpoint formula and the given coordinates for the vertices.
3. Find$\,{a}^{2}\,$by solving for the length of the major axis,$\,2a,$ which is the distance between the given vertices.
4. Find$\,{c}^{2}\,$using$\,h\,$and$\,k,$ found in Step 2, along with the given coordinates for the foci.
5. Solve for$\,{b}^{2}\,$using the equation$\,{c}^{2}={a}^{2}-{b}^{2}.$
6. Substitute the values for$\,h,k,{a}^{2},$ and$\,{b}^{2}\,$into the standard form of the equation determined in Step 1.
### Writing the Equation of an Ellipse Centered at a Point Other Than the Origin
What is the standard form equation of the ellipse that has vertices$\,\left(-2,-8\right)\,$and$\,\left(-2,\text{2}\right)$
and foci$\,\left(-2,-7\right)\,$and$\,\left(-2,\text{1}\right)?$
The x-coordinates of the vertices and foci are the same, so the major axis is parallel to the y-axis. Thus, the equation of the ellipse will have the form
$\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1$
First, we identify the center,$\,\left(h,k\right).\,$The center is halfway between the vertices,$\,\left(-2,-8\right)\,$and$\,\left(-2,\text{2}\right).\,$Applying the midpoint formula, we have:
$\begin{array}{l}\left(h,k\right)=\left(\frac{-2+\left(-2\right)}{2},\frac{-8+2}{2}\right)\hfill \\ \text{ }=\left(-2,-3\right)\hfill \end{array}$
Next, we find$\,{a}^{2}.\,$The length of the major axis,$\,2a,$ is bounded by the vertices. We solve for$\,a\,$by finding the distance between the y-coordinates of the vertices.
$\begin{array}{c}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2a=2-\left(-8\right)\\ 2a=10\\ a=5\end{array}$
So$\,{a}^{2}=25.$
Now we find$\,{c}^{2}.\,$The foci are given by$\,\left(h,k±c\right).\,$So,$\,\left(h,k-c\right)=\left(-2,-7\right)\,$and$\,\left(h,k+c\right)=\left(-2,\text{1}\right).\,$We substitute$\,k=-3\,$using either of these points to solve for$\,c.$
$\begin{array}{c}\,\,\,\,k+c=1\\ -3+c=1\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,c=4\end{array}$
So$\,{c}^{2}=16.$
Next, we solve for$\,{b}^{2}\,$using the equation$\,{c}^{2}={a}^{2}-{b}^{2}.$
$\begin{array}{c}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{c}^{2}={a}^{2}-{b}^{2}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,16=25-{b}^{2}\\ {b}^{2}=9\end{array}$
Finally, we substitute the values found for$\,h,k,{a}^{2},$ and$\,{b}^{2}\,$into the standard form equation for an ellipse:
$\,\frac{{\left(x+2\right)}^{2}}{9}+\frac{{\left(y+3\right)}^{2}}{25}=1$[/hidden-answer]
### Try It
What is the standard form equation of the ellipse that has vertices$\,\left(-3,3\right)\,$and$\,\left(5,3\right)\,$and foci$\,\left(1-2\sqrt{3},3\right)\,$and$\,\left(1+2\sqrt{3},3\right)?$
$\frac{{\left(x-1\right)}^{2}}{16}+\frac{{\left(y-3\right)}^{2}}{4}=1$
### Graphing Ellipses Centered at the Origin
Just as we can write the equation for an ellipse given its graph, we can graph an ellipse given its equation. To graph ellipses centered at the origin, we use the standard form$\,\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,\text{ }a>b\,$for horizontal ellipses and$\,\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1,\text{ }a>b\,$for vertical ellipses.
### How To
Given the standard form of an equation for an ellipse centered at$\,\left(0,0\right),$ sketch the graph.
1. Use the standard forms of the equations of an ellipse to determine the major axis, vertices, co-vertices, and foci.
1. If the equation is in the form$\,\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,\,$where$\,a>b,\,$then
• the major axis is the x-axis
• the coordinates of the vertices are$\,\left(±a,0\right)$
• the coordinates of the co-vertices are$\,\left(0,±b\right)$
• the coordinates of the foci are$\,\left(±c,0\right)$
2. If the equation is in the form$\,\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1,$where$\,a>b,\,$then
• the major axis is the y-axis
• the coordinates of the vertices are$\,\left(0,±a\right)$
• the coordinates of the co-vertices are$\,\left(±b,0\right)$
• the coordinates of the foci are$\,\left(0,±c\right)$
2. Solve for$\,c\,$using the equation$\,{c}^{2}={a}^{2}-{b}^{2}.$
3. Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.
### Graphing an Ellipse Centered at the Origin
Graph the ellipse given by the equation,$\,\frac{{x}^{2}}{9}+\frac{{y}^{2}}{25}=1.\,$Identify and label the center, vertices, co-vertices, and foci.
First, we determine the position of the major axis. Because$\,25>9,$the major axis is on the y-axis. Therefore, the equation is in the form$\,\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1,$where$\,{b}^{2}=9\,$and$\,{a}^{2}=25.\,$It follows that:
• the center of the ellipse is$\,\left(0,0\right)$
• the coordinates of the vertices are$\,\left(0,±a\right)=\left(0,±\sqrt{25}\right)=\left(0,±5\right)$
• the coordinates of the co-vertices are$\,\left(±b,0\right)=\left(±\sqrt{9},0\right)=\left(±3,0\right)$
• the coordinates of the foci are$\,\left(0,±c\right),\,$where$\,{c}^{2}={a}^{2}-{b}^{2}\,$Solving for$\,c,$ we have:
$\begin{array}{l}c=±\sqrt{{a}^{2}-{b}^{2}}\hfill \\ \,\,\,=±\sqrt{25-9}\hfill \\ \,\,\,=±\sqrt{16}\hfill \\ \,\,\,=±4\hfill \end{array}$
Therefore, the coordinates of the foci are$\,\left(0,±4\right).$
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. See (Figure).
### Try It
Graph the ellipse given by the equation$\,\frac{{x}^{2}}{36}+\frac{{y}^{2}}{4}=1.\,$Identify and label the center, vertices, co-vertices, and foci.
center:$\,\left(0,0\right);\,$vertices:$\,\left(±6,0\right);\,$co-vertices:$\,\left(0,±2\right);\,$foci:$\,\left(±4\sqrt{2},0\right)$
### Graphing an Ellipse Centered at the Origin from an Equation Not in Standard Form
Graph the ellipse given by the equation$\,4{x}^{2}+25{y}^{2}=100.\,$Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.
First, use algebra to rewrite the equation in standard form.
$\begin{array}{l} 4{x}^{2}+25{y}^{2}=100\hfill \\ \text{ }\frac{4{x}^{2}}{100}+\frac{25{y}^{2}}{100}=\frac{100}{100}\hfill \\ \text{ }\frac{{x}^{2}}{25}+\frac{{y}^{2}}{4}=1\hfill \end{array}$
Next, we determine the position of the major axis. Because$\,25>4,\,$the major axis is on the x-axis. Therefore, the equation is in the form$\,\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,\,$where$\,{a}^{2}=25\,$and$\,{b}^{2}=4.\,$It follows that:
• the center of the ellipse is$\,\left(0,0\right)$
• the coordinates of the vertices are$\,\left(±a,0\right)=\left(±\sqrt{25},0\right)=\left(±5,0\right)$
• the coordinates of the co-vertices are$\,\left(0,±b\right)=\left(0,±\sqrt{4}\right)=\left(0,±2\right)$
• the coordinates of the foci are$\,\left(±c,0\right),\,$where$\,{c}^{2}={a}^{2}-{b}^{2}.\,$Solving for$\,c,\,$we have:
$\begin{array}{l}c=±\sqrt{{a}^{2}-{b}^{2}}\hfill \\ \,\,\,=±\sqrt{25-4}\hfill \\ \,\,\,=±\sqrt{21}\hfill \end{array}$
Therefore the coordinates of the foci are$\,\left(±\sqrt{21},0\right).$
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.
### Try It
Graph the ellipse given by the equation$\,49{x}^{2}+16{y}^{2}=784.\,$Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.
Standard form:$\,\frac{{x}^{2}}{16}+\frac{{y}^{2}}{49}=1;\,$center:$\,\left(0,0\right);\,$vertices:$\,\left(0,±7\right);\,$co-vertices:$\,\left(±4,0\right);\,$foci:$\,\left(0,±\sqrt{33}\right)$
### Graphing Ellipses Not Centered at the Origin
When an ellipse is not centered at the origin, we can still use the standard forms to find the key features of the graph. When the ellipse is centered at some point,$\,\left(h,k\right),$we use the standard forms$\,\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1,\text{ }a>b\,$for horizontal ellipses and$\,\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1,\text{ }a>b\,$for vertical ellipses. From these standard equations, we can easily determine the center, vertices, co-vertices, foci, and positions of the major and minor axes.
### How To
Given the standard form of an equation for an ellipse centered at$\,\left(h,k\right),$ sketch the graph.
1. Use the standard forms of the equations of an ellipse to determine the center, position of the major axis, vertices, co-vertices, and foci.
1. If the equation is in the form$\,\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1,\,$where$\,a>b,\,$then
• the center is$\,\left(h,k\right)$
• the major axis is parallel to the x-axis
• the coordinates of the vertices are$\,\left(h±a,k\right)$
• the coordinates of the co-vertices are$\,\left(h,k±b\right)$
• the coordinates of the foci are$\,\left(h±c,k\right)$
2. If the equation is in the form$\,\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1,\,$where$\,a>b,\,$then
• the center is$\,\left(h,k\right)$
• the major axis is parallel to the y-axis
• the coordinates of the vertices are$\,\left(h,k±a\right)$
• the coordinates of the co-vertices are$\,\left(h±b,k\right)$
• the coordinates of the foci are$\,\left(h,k±c\right)$
2. Solve for$\,c\,$using the equation$\,{c}^{2}={a}^{2}-{b}^{2}.$
3. Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse.
### Graphing an Ellipse Centered at (h, k)
Graph the ellipse given by the equation,$\,\frac{{\left(x+2\right)}^{2}}{4}+\frac{{\left(y-5\right)}^{2}}{9}=1.\,$Identify and label the center, vertices, co-vertices, and foci.
First, we determine the position of the major axis. Because$\,9>4,$ the major axis is parallel to the y-axis. Therefore, the equation is in the form$\,\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1,\,$where$\,{b}^{2}=4\,$and$\,{a}^{2}=9.\,$It follows that:
• the center of the ellipse is$\,\left(h,k\right)=\left(-2,\text{5}\right)$
• the coordinates of the vertices are$\,\left(h,k±a\right)=\left(-2,5±\sqrt{9}\right)=\left(-2,5±3\right),$ or$\,\left(-2,\text{2}\right)\,$and$\,\left(-2,\text{8}\right)$
• the coordinates of the co-vertices are$\,\left(h±b,k\right)=\left(-2±\sqrt{4},5\right)=\left(-2±2,5\right),$ or$\,\left(-4,5\right)\,$and$\,\left(0,\text{5}\right)$
• the coordinates of the foci are$\,\left(h,k±c\right),\,$where$\,{c}^{2}={a}^{2}-{b}^{2}.\,$Solving for$\,c,$we have:
$\begin{array}{l}\begin{array}{l}\\ c=±\sqrt{{a}^{2}-{b}^{2}}\end{array}\hfill \\ \,\,\,=±\sqrt{9-4}\hfill \\ \,\,\,=±\sqrt{5}\hfill \end{array}$
Therefore, the coordinates of the foci are$\,\left(-2,\text{5}-\sqrt{5}\right)\,$and$\,\left(-2,\text{5+}\sqrt{5}\right).$
Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.
### Try It
Graph the ellipse given by the equation$\,\frac{{\left(x-4\right)}^{2}}{36}+\frac{{\left(y-2\right)}^{2}}{20}=1.\,$Identify and label the center, vertices, co-vertices, and foci.
Center:$\,\left(4,2\right);\,$vertices:$\,\left(-2,2\right)\,$and$\,\left(10,2\right);\,$co-vertices:$\,\left(4,2-2\sqrt{5}\right)\,$and$\,\left(4,2+2\sqrt{5}\right);\,$foci:$\,\left(0,2\right)\,$and$\,\left(8,2\right)$
### How To
Given the general form of an equation for an ellipse centered at (h, k), express the equation in standard form.
1. Recognize that an ellipse described by an equation in the form$\,a{x}^{2}+b{y}^{2}+cx+dy+e=0\,$is in general form.
2. Rearrange the equation by grouping terms that contain the same variable. Move the constant term to the opposite side of the equation.
3. Factor out the coefficients of the$\,{x}^{2}\,$and$\,{y}^{2}\,$terms in preparation for completing the square.
4. Complete the square for each variable to rewrite the equation in the form of the sum of multiples of two binomials squared set equal to a constant,$\,{m}_{1}{\left(x-h\right)}^{2}+{m}_{2}{\left(y-k\right)}^{2}={m}_{3},$ where$\,{m}_{1},{m}_{2},$ and$\,{m}_{3}\,$are constants.
5. Divide both sides of the equation by the constant term to express the equation in standard form.
### Graphing an Ellipse Centered at (h, k) by First Writing It in Standard Form
Graph the ellipse given by the equation$\,4{x}^{2}+9{y}^{2}-40x+36y+100=0.\,$Identify and label the center, vertices, co-vertices, and foci.
We must begin by rewriting the equation in standard form.
$4{x}^{2}+9{y}^{2}-40x+36y+100=0$
Group terms that contain the same variable, and move the constant to the opposite side of the equation.
$\left(4{x}^{2}-40x\right)+\left(9{y}^{2}+36y\right)=-100$
Factor out the coefficients of the squared terms.
$4\left({x}^{2}-10x\right)+9\left({y}^{2}+4y\right)=-100$
Complete the square twice. Remember to balance the equation by adding the same constants to each side.
$4\left({x}^{2}-10x+25\right)+9\left({y}^{2}+4y+4\right)=-100+100+36$
Rewrite as perfect squares.
$4{\left(x-5\right)}^{2}+9{\left(y+2\right)}^{2}=36$
Divide both sides by the constant term to place the equation in standard form.
$\frac{{\left(x-5\right)}^{2}}{9}+\frac{{\left(y+2\right)}^{2}}{4}=1$
Now that the equation is in standard form, we can determine the position of the major axis. Because$\,9>4,\,$ the major axis is parallel to the x-axis. Therefore, the equation is in the form$\,\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1,\,$where$\,{a}^{2}=9\,$and$\,{b}^{2}=4.\,$It follows that:
• the center of the ellipse is$\,\left(h,k\right)=\left(5,-2\right)$
• the coordinates of the vertices are$\,\left(h±a,k\right)=\left(5±\sqrt{9},-2\right)=\left(5±3,-2\right),\,$or$\,\left(2,-2\right)\,$and$\,\left(8,-2\right)$
• the coordinates of the co-vertices are$\,\left(h,k±b\right)=\left(\text{5},-2±\sqrt{4}\right)=\left(\text{5},-2±2\right),\,$or$\,\left(5,-4\right)\,$and$\,\left(5,\text{0}\right)$
• the coordinates of the foci are$\,\left(h±c,k\right),\,$where$\,{c}^{2}={a}^{2}-{b}^{2}.\,$Solving for$\,c,\,$we have:
$\begin{array}{l}c=±\sqrt{{a}^{2}-{b}^{2}}\hfill \\ \,\,\,=±\sqrt{9-4}\hfill \\ \,\,\,=±\sqrt{5}\hfill \end{array}$
Therefore, the coordinates of the foci are$\,\left(\text{5}-\sqrt{5},-2\right)\,$and$\,\left(\text{5+}\sqrt{5},-2\right).$
Next we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse as shown in (Figure).
### Try It
Express the equation of the ellipse given in standard form. Identify the center, vertices, co-vertices, and foci of the ellipse.
$4{x}^{2}+{y}^{2}-24x+2y+21=0$
$\,\frac{{\left(x-3\right)}^{2}}{4}+\frac{{\left(y+1\right)}^{2}}{16}=1;\,$center:$\,\left(3,-1\right);\,$vertices:$\,\left(3,-\text{5}\right)\,$and$\,\left(3,\text{3}\right);\,$co-vertices:$\,\left(1,-1\right)\,$and$\,\left(5,-1\right);\,$foci:$\,\left(3,-\text{1}-2\sqrt{3}\right)\,$and$\,\left(3,-\text{1+}2\sqrt{3}\right)$
### Solving Applied Problems Involving Ellipses
Many real-world situations can be represented by ellipses, including orbits of planets, satellites, moons and comets, and shapes of boat keels, rudders, and some airplane wings. A medical device called a lithotripter uses elliptical reflectors to break up kidney stones by generating sound waves. Some buildings, called whispering chambers, are designed with elliptical domes so that a person whispering at one focus can easily be heard by someone standing at the other focus. This occurs because of the acoustic properties of an ellipse. When a sound wave originates at one focus of a whispering chamber, the sound wave will be reflected off the elliptical dome and back to the other focus. See (Figure). In the whisper chamber at the Museum of Science and Industry in Chicago, two people standing at the foci—about 43 feet apart—can hear each other whisper.
### Locating the Foci of a Whispering Chamber
The Statuary Hall in the Capitol Building in Washington, D.C. is a whispering chamber. Its dimensions are 46 feet wide by 96 feet long as shown in (Figure).
1. What is the standard form of the equation of the ellipse representing the outline of the room? Hint: assume a horizontal ellipse, and let the center of the room be the point$\,\left(0,0\right).$
2. If two senators standing at the foci of this room can hear each other whisper, how far apart are the senators? Round to the nearest foot.
1. We are assuming a horizontal ellipse with center$\,\left(0,0\right),$ so we need to find an equation of the form$\,\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,\,$where$\,a>b.\,$We know that the length of the major axis,$\,2a,\,$is longer than the length of the minor axis,$\,2b.\,$So the length of the room, 96, is represented by the major axis, and the width of the room, 46, is represented by the minor axis.
• Solving for$\,a,$ we have$\,2a=96,$ so$\,a=48,$ and$\,{a}^{2}=2304.$
• Solving for$\,b,$ we have$\,2b=46,$ so$\,b=23,$ and$\,{b}^{2}=529.$
Therefore, the equation of the ellipse is$\,\frac{{x}^{2}}{2304}+\frac{{y}^{2}}{529}=1.$
2. To find the distance between the senators, we must find the distance between the foci,$\,\left(±c,0\right),\,$where$\,{c}^{2}={a}^{2}-{b}^{2}.\,$Solving for$\,c,$we have:
$\begin{array}{ll}{c}^{2}={a}^{2}-{b}^{2}\hfill & \hfill \\ {c}^{2}=2304-529\hfill & \begin{array}{cccc}& & & \end{array}\text{Substitute using the values found in part (a)}.\hfill \\ \,\,\,c=±\sqrt{2304-529}\hfill & \begin{array}{cccc}& & & \end{array}\text{Take the square root of both sides}.\hfill \\ \,\,\,c=±\sqrt{1775} \hfill & \begin{array}{cccc}& & & \end{array}\text{Subtract}.\hfill \\ \,\,\,c\approx ±42\hfill & \begin{array}{cccc}& & & \end{array}\text{Round to the nearest foot}.\hfill \end{array}$
The points$\,\left(±42,0\right)\,$represent the foci. Thus, the distance between the senators is$\,2\left(42\right)=84\,$feet.[/hidden-answer]
### Try It
Suppose a whispering chamber is 480 feet long and 320 feet wide.
1. What is the standard form of the equation of the ellipse representing the room? Hint: assume a horizontal ellipse, and let the center of the room be the point$\,\left(0,0\right).$
2. If two people are standing at the foci of this room and can hear each other whisper, how far apart are the people? Round to the nearest foot.
1. $\frac{{x}^{2}}{57,600}+\frac{{y}^{2}}{25,600}=1$
2. The people are standing 358 feet apart.
Access these online resources for additional instruction and practice with ellipses.
### Key Equations
Horizontal ellipse, center at origin $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1,\text{ }a>b$ Vertical ellipse, center at origin $\frac{{x}^{2}}{{b}^{2}}+\frac{{y}^{2}}{{a}^{2}}=1,\text{ }a>b$ Horizontal ellipse, center $\,\left(h,k\right)$ $\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1,\text{ }a>b$ Vertical ellipse, center$\,\left(h,k\right)$ $\frac{{\left(x-h\right)}^{2}}{{b}^{2}}+\frac{{\left(y-k\right)}^{2}}{{a}^{2}}=1,\text{ }a>b$
### Key Concepts
• An ellipse is the set of all points$\,\left(x,y\right)\,$in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci).
• When given the coordinates of the foci and vertices of an ellipse, we can write the equation of the ellipse in standard form. See (Figure) and (Figure).
• When given an equation for an ellipse centered at the origin in standard form, we can identify its vertices, co-vertices, foci, and the lengths and positions of the major and minor axes in order to graph the ellipse. See (Figure) and (Figure).
• When given the equation for an ellipse centered at some point other than the origin, we can identify its key features and graph the ellipse. See (Figure) and (Figure).
• Real-world situations can be modeled using the standard equations of ellipses and then evaluated to find key features, such as lengths of axes and distance between foci. See (Figure).
### Section Exercises
#### Verbal
Define an ellipse in terms of its foci.
An ellipse is the set of all points in the plane the sum of whose distances from two fixed points, called the foci, is a constant.
Where must the foci of an ellipse lie?
What special case of the ellipse do we have when the major and minor axis are of the same length?
This special case would be a circle.
For the special case mentioned above, what would be true about the foci of that ellipse?
What can be said about the symmetry of the graph of an ellipse with center at the origin and foci along the y-axis?
It is symmetric about the x-axis, y-axis, and the origin.
#### Algebraic
For the following exercises, determine whether the given equations represent ellipses. If yes, write in standard form.
$2{x}^{2}+y=4$
$4{x}^{2}+9{y}^{2}=36$
yes;$\,\frac{{x}^{2}}{{3}^{2}}+\frac{{y}^{2}}{{2}^{2}}=1$
$4{x}^{2}-{y}^{2}=4$
$4{x}^{2}+9{y}^{2}=1$
yes;$\frac{{x}^{2}}{{\left(\frac{1}{2}\right)}^{2}}+\frac{{y}^{2}}{{\left(\frac{1}{3}\right)}^{2}}=1$
$4{x}^{2}-8x+9{y}^{2}-72y+112=0$
For the following exercises, write the equation of an ellipse in standard form, and identify the end points of the major and minor axes as well as the foci.
$\frac{{x}^{2}}{4}+\frac{{y}^{2}}{49}=1$
$\frac{{x}^{2}}{{2}^{2}}+\frac{{y}^{2}}{{7}^{2}}=1;\,$Endpoints of major axis$\,\left(0,7\right)\,$and$\,\left(0,-7\right).\,$Endpoints of minor axis$\,\left(2,0\right)\,$and$\,\left(-2,0\right).\,$Foci at$\,\left(0,3\sqrt{5}\right),\left(0,-3\sqrt{5}\right).$
$\frac{{x}^{2}}{100}+\frac{{y}^{2}}{64}=1$
${x}^{2}+9{y}^{2}=1$
$\frac{{x}^{2}}{{\left(1\right)}^{2}}+\frac{{y}^{2}}{{\left(\frac{1}{3}\right)}^{2}}=1;\,$Endpoints of major axis$\,\left(1,0\right)\,$and$\,\left(-1,0\right).\,$Endpoints of minor axis$\,\left(0,\frac{1}{3}\right),\left(0,-\frac{1}{3}\right).\,$Foci at$\,\left(\frac{2\sqrt{2}}{3},0\right),\left(-\frac{2\sqrt{2}}{3},0\right).$
$4{x}^{2}+16{y}^{2}=1$
$\frac{{\left(x-2\right)}^{2}}{49}+\frac{{\left(y-4\right)}^{2}}{25}=1$
$\frac{{\left(x-2\right)}^{2}}{{7}^{2}}+\frac{{\left(y-4\right)}^{2}}{{5}^{2}}=1;\,$Endpoints of major axis$\,\left(9,4\right),\left(-5,4\right).\,$Endpoints of minor axis$\,\left(2,9\right),\left(2,-1\right).\,$Foci at$\,\left(2+2\sqrt{6},4\right),\left(2-2\sqrt{6},4\right).$
$\frac{{\left(x-2\right)}^{2}}{81}+\frac{{\left(y+1\right)}^{2}}{16}=1$
$\frac{{\left(x+5\right)}^{2}}{4}+\frac{{\left(y-7\right)}^{2}}{9}=1$
$\frac{{\left(x+5\right)}^{2}}{{2}^{2}}+\frac{{\left(y-7\right)}^{2}}{{3}^{2}}=1;\,$Endpoints of major axis$\,\left(-5,10\right),\left(-5,4\right).\,$Endpoints of minor axis$\,\left(-3,7\right),\left(-7,7\right).\,$Foci at$\,\left(-5,7+\sqrt{5}\right),\left(-5,7-\sqrt{5}\right).$
$\frac{{\left(x-7\right)}^{2}}{49}+\frac{{\left(y-7\right)}^{2}}{49}=1$
$4{x}^{2}-8x+9{y}^{2}-72y+112=0$
$\frac{{\left(x-1\right)}^{2}}{{3}^{2}}+\frac{{\left(y-4\right)}^{2}}{{2}^{2}}=1;\,$Endpoints of major axis$\,\left(4,4\right),\left(-2,4\right).\,$Endpoints of minor axis$\,\left(1,6\right),\left(1,2\right).\,$Foci at$\,\left(1+\sqrt{5},4\right),\left(1-\sqrt{5},4\right).$
$9{x}^{2}-54x+9{y}^{2}-54y+81=0$
$4{x}^{2}-24x+36{y}^{2}-360y+864=0$
$\frac{{\left(x-3\right)}^{2}}{{\left(3\sqrt{2}\right)}^{2}}+\frac{{\left(y-5\right)}^{2}}{{\left(\sqrt{2}\right)}^{2}}=1;\,$Endpoints of major axis$\,\left(3+3\sqrt{2},5\right),\left(3-3\sqrt{2},5\right).\,$Endpoints of minor axis$\,\left(3,5+\sqrt{2}\right),\left(3,5-\sqrt{2}\right).\,$Foci at$\,\left(7,5\right),\left(-1,5\right).$
$4{x}^{2}+24x+16{y}^{2}-128y+228=0$
$4{x}^{2}+40x+25{y}^{2}-100y+100=0$
$\frac{{\left(x+5\right)}^{2}}{{\left(5\right)}^{2}}+\frac{{\left(y-2\right)}^{2}}{{\left(2\right)}^{2}}=1;\,$Endpoints of major axis$\,\left(0,2\right),\left(-10,2\right).\,$Endpoints of minor axis$\,\left(-5,4\right),\left(-5,0\right).\,$Foci at$\,\left(-5+\sqrt{21},2\right),\left(-5-\sqrt{21},2\right).$
${x}^{2}+2x+100{y}^{2}-1000y+2401=0$
$4{x}^{2}+24x+25{y}^{2}+200y+336=0$
$\frac{{\left(x+3\right)}^{2}}{{\left(5\right)}^{2}}+\frac{{\left(y+4\right)}^{2}}{{\left(2\right)}^{2}}=1;\,$Endpoints of major axis$\,\left(2,-4\right),\left(-8,-4\right).\,$Endpoints of minor axis$\,\left(-3,-2\right),\left(-3,-6\right).\,$Foci at$\,\left(-3+\sqrt{21},-4\right),\left(-3-\sqrt{21},-4\right).$
$9{x}^{2}+72x+16{y}^{2}+16y+4=0$
For the following exercises, find the foci for the given ellipses.
$\frac{{\left(x+3\right)}^{2}}{25}+\frac{{\left(y+1\right)}^{2}}{36}=1$
Foci$\,\left(-3,-1+\sqrt{11}\right),\left(-3,-1-\sqrt{11}\right)$
$\frac{{\left(x+1\right)}^{2}}{100}+\frac{{\left(y-2\right)}^{2}}{4}=1$
${x}^{2}+{y}^{2}=1$
Focus$\,\left(0,0\right)$[/hidden-answer]
${x}^{2}+4{y}^{2}+4x+8y=1$
$10{x}^{2}+{y}^{2}+200x=0$
Foci$\,\left(-10,30\right),\left(-10,-30\right)$
#### Graphical
For the following exercises, graph the given ellipses, noting center, vertices, and foci.
$\frac{{x}^{2}}{25}+\frac{{y}^{2}}{36}=1$
$\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=1$
Center$\,\left(0,0\right),\,$Vertices$\,\left(4,0\right),\left(-4,0\right),\left(0,3\right),\left(0,-3\right),\,$Foci$\,\left(\sqrt{7},0\right),\left(-\sqrt{7},0\right)$
$4{x}^{2}+9{y}^{2}=1$
$81{x}^{2}+49{y}^{2}=1$
Center$\,\left(0,0\right),\,$Vertices$\,\left(\frac{1}{9},0\right),\left(-\frac{1}{9},0\right),\left(0,\frac{1}{7}\right),\left(0,-\frac{1}{7}\right),\,$Foci$\,\left(0,\frac{4\sqrt{2}}{63}\right),\left(0,-\frac{4\sqrt{2}}{63}\right)$
$\frac{{\left(x-2\right)}^{2}}{64}+\frac{{\left(y-4\right)}^{2}}{16}=1$
$\frac{{\left(x+3\right)}^{2}}{9}+\frac{{\left(y-3\right)}^{2}}{9}=1$
Center$\,\left(-3,3\right),\,$Vertices$\,\left(0,3\right),\left(-6,3\right),\left(-3,0\right),\left(-3,6\right),\,$Focus$\,\left(-3,3\right)\,$
Note that this ellipse is a circle. The circle has only one focus, which coincides with the center.
$\frac{{x}^{2}}{2}+\frac{{\left(y+1\right)}^{2}}{5}=1$
$4{x}^{2}-8x+16{y}^{2}-32y-44=0$
Center$\,\left(1,1\right),\,$Vertices$\,\left(5,1\right),\left(-3,1\right),\left(1,3\right),\left(1,-1\right),\,$Foci$\,\left(1,1+4\sqrt{3}\right),\left(1,1-4\sqrt{3}\right)$
${x}^{2}-8x+25{y}^{2}-100y+91=0$
${x}^{2}+8x+4{y}^{2}-40y+112=0$
Center$\,\left(-4,5\right),\,$Vertices$\,\left(-2,5\right),\left(-6,4\right),\left(-4,6\right),\left(-4,4\right),\,$Foci$\,\left(-4+\sqrt{3},5\right),\left(-4-\sqrt{3},5\right)$
$64{x}^{2}+128x+9{y}^{2}-72y-368=0$
$16{x}^{2}+64x+4{y}^{2}-8y+4=0$
Center$\,\left(-2,1\right),\,$Vertices$\,\left(0,1\right),\left(-4,1\right),\left(-2,5\right),\left(-2,-3\right),\,$Foci$\,\left(-2,1+2\sqrt{3}\right),\left(-2,1-2\sqrt{3}\right)$
$100{x}^{2}+1000x+{y}^{2}-10y+2425=0$
$4{x}^{2}+16x+4{y}^{2}+16y+16=0$
Center$\,\left(-2,-2\right),\,$Vertices$\,\left(0,-2\right),\left(-4,-2\right),\left(-2,0\right),\left(-2,-4\right),\,$Focus$\,\left(-2,-2\right)$
For the following exercises, use the given information about the graph of each ellipse to determine its equation.
Center at the origin, symmetric with respect to the x– and y-axes, focus at$\,\left(4,0\right),$ and point on graph$\,\left(0,3\right).$
Center at the origin, symmetric with respect to the x– and y-axes, focus at$\,\left(0,-2\right),$ and point on graph$\,\left(5,0\right).$
$\frac{{x}^{2}}{25}+\frac{{y}^{2}}{29}=1$
Center at the origin, symmetric with respect to the x– and y-axes, focus at$\,\left(3,0\right),$ and major axis is twice as long as minor axis.
Center$\,\left(4,2\right)$; vertex$\,\left(9,2\right)$; one focus:$\,\left(4+2\sqrt{6},2\right)$.
$\frac{{\left(x-4\right)}^{2}}{25}+\frac{{\left(y-2\right)}^{2}}{1}=1$
Center$\,\left(3,5\right)$; vertex$\,\left(3,11\right)$; one focus:$\,\left(3,\text{ 5+4}\sqrt{\text{2}}\right)$
Center$\,\left(-3,4\right)$; vertex$\,\left(1,4\right)$; one focus:$\,\left(-3+2\sqrt{3},4\right)$
$\frac{{\left(x+3\right)}^{2}}{16}+\frac{{\left(y-4\right)}^{2}}{4}=1$
For the following exercises, given the graph of the ellipse, determine its equation.
$\frac{{x}^{2}}{81}+\frac{{y}^{2}}{9}=1$
$\frac{{\left(x+2\right)}^{2}}{4}+\frac{{\left(y-2\right)}^{2}}{9}=1$
#### Extensions
For the following exercises, find the area of the ellipse. The area of an ellipse is given by the formula$\,\text{Area}=a\cdot b\cdot \pi .$
$\frac{{\left(x-3\right)}^{2}}{9}+\frac{{\left(y-3\right)}^{2}}{16}=1$
$\text{Area = 12π}\,\text{square}\,\text{units}$
$\frac{{\left(x+6\right)}^{2}}{16}+\frac{{\left(y-6\right)}^{2}}{36}=1$
$\frac{{\left(x+1\right)}^{2}}{4}+\frac{{\left(y-2\right)}^{2}}{5}=1$
$\text{Area = 2}\sqrt{\text{5}}\text{π}\,\text{square}\,\text{units}$
$4{x}^{2}-8x+9{y}^{2}-72y+112=0$
$9{x}^{2}-54x+9{y}^{2}-54y+81=0$
$\text{Area = 9π}\,\text{square}\,\text{units}$
#### Real-World Applications
Find the equation of the ellipse that will just fit inside a box that is 8 units wide and 4 units high.
Find the equation of the ellipse that will just fit inside a box that is four times as wide as it is high. Express in terms of$\,h,$ the height.
$\frac{{x}^{2}}{4{h}^{2}}+\frac{{y}^{2}}{\frac{1}{4}{h}^{2}}=1$
An arch has the shape of a semi-ellipse (the top half of an ellipse). The arch has a height of 8 feet and a span of 20 feet. Find an equation for the ellipse, and use that to find the height to the nearest 0.01 foot of the arch at a distance of 4 feet from the center.
An arch has the shape of a semi-ellipse. The arch has a height of 12 feet and a span of 40 feet. Find an equation for the ellipse, and use that to find the distance from the center to a point at which the height is 6 feet. Round to the nearest hundredth.
$\frac{{x}^{2}}{400}+\frac{{y}^{2}}{144}=1$. Distance = 17.32 feet
A bridge is to be built in the shape of a semi-elliptical arch and is to have a span of 120 feet. The height of the arch at a distance of 40 feet from the center is to be 8 feet. Find the height of the arch at its center.
A person in a whispering gallery standing at one focus of the ellipse can whisper and be heard by a person standing at the other focus because all the sound waves that reach the ceiling are reflected to the other person. If a whispering gallery has a length of 120 feet, and the foci are located 30 feet from the center, find the height of the ceiling at the center.
Approximately 51.96 feet
A person is standing 8 feet from the nearest wall in a whispering gallery. If that person is at one focus, and the other focus is 80 feet away, what is the length and height at the center of the gallery?
### Glossary
center of an ellipse
the midpoint of both the major and minor axes
conic section
any shape resulting from the intersection of a right circular cone with a plane
ellipse
the set of all points$\,\left(x,y\right)\,$in a plane such that the sum of their distances from two fixed points is a constant
foci
plural of focus
focus (of an ellipse)
one of the two fixed points on the major axis of an ellipse such that the sum of the distances from these points to any point$\,\left(x,y\right)\,$on the ellipse is a constant
major axis
the longer of the two axes of an ellipse
minor axis
the shorter of the two axes of an ellipse
1. Architect of the Capitol. http://www.aoc.gov. Accessed April 15, 2014. | 16,095 | 46,046 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2023-40 | longest | en | 0.935882 |
https://www.geeksforgeeks.org/data-structures-and-algorithms-set-38/?ref=rp | 1,623,662,685,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487611641.26/warc/CC-MAIN-20210614074543-20210614104543-00390.warc.gz | 682,717,396 | 20,562 | Related Articles
Data Structures and Algorithms | Set 38
• Last Updated : 21 Feb, 2019
This topic contains basic questions of Algorithm which can be helpful for GATE CS Preparation. So, it is recommended to solve each of these questions if you are preparing for GATE.
Ques-1: Which one of the following correctly determines the solution of the recurrence relation given below with T(1) = 1 ?
`T(n)= 2T(n/4) + n1/2 `
(A) O(n2)
(B) O(n)
(C) O(n1/2 log n)
(D) O(log n)
Explanation:
According Master Theorem,
`T(n)= 2T(n/4) + n1/2 `
Applying Masters Theorem,
Here,
```a = 2, b = 4, K = 1/2, and p = 0
So, bK = 41/2 = 2
Thus, a = bK and (p > -1) ```
So, the formula is,
`T(n)= O(nlogba log(P+1)n)`
Therefore,
`T(n) = O(nlog 42 log(0 + 1)n) = O(n1/2 log n)`
So, option (C) is correct.
Ques-2: For merging two unsorted list of size p and q into sorted list of size (p + q). The time complexity in terms of number of comparisons is:
(A) O(log p + log q)
(B) O(p log p) + q log q)
(C) O(p + q)
(D) None
Explanation:
For sorting the array of size p individually it takes O(p log p) and the array of size q takes O(q log q) time, then merging will take O(m + n) time.
Therefore, total number of comparisons
```= O(p log p) + O(q log q) + p + q
= O(p log p) + O(q log q) ```
So, option (B) is correct.
Ques-3: Which of the following sorting algorithms has the highest best case time complexity using array data structure ?
(A) Heap sort
(B) Insertion sort
(C) Bubble sort
(D) Selection sort
Explanation:
Best case time complexity of Heap sort is O(n log n)
Best case time complexity of Insertion sort is O(n)
Best case time complexity of Bubble sort is O(n)
Best case time complexity of selection sort is O(n2).
So, option (D) is correct.
Ques-4: Which of the following input gives the best case time for selection sort ?
(A) 1 2 3 4 5 6 7 8 9
(B) 2 3 1 5 9 7 8 6
(C) 9 8 7 6 5 4 3 2 1
(D) All of the above take same amount of time.
Explanation:
Selection sort in worst case and best case takes same time.
So, option (D) is correct.
Ques-5: What is the time complexity of recursive function given below:
`T(n)= 4T(n/2) + n2 `
(A) O(n2)
(B) O(n)
(C) O(n2 log n)
(D) O(n log n)
Explanation:
According Master Theorem,
Here,
```a = 4, b = 2, k = 2, p = 0
So, bk = 4 i.e., a = bk ```
Therefore, the formula is
```T(n) = O(nlog ba log(P+1)n)
So, T(n)= O(nlog 24 log(0 + 1)n) = O(n2 log n)```
So, option (C) is correct.
Attention reader! Don’t stop learning now. Learn all GATE CS concepts with Free Live Classes on our youtube channel.
My Personal Notes arrow_drop_up | 846 | 2,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2021-25 | latest | en | 0.816781 |
https://modelingcommons.org/browse/one_model/1264 | 1,675,175,510,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499871.68/warc/CC-MAIN-20230131122916-20230131152916-00601.warc.gz | 426,974,540 | 8,062 | # Sierpinski Simple
### 1 collaborator
Uri Wilensky (Author)
### Tags
mathematics
Tagged by Reuven M. Lerner over 9 years ago
Model group CCL | Visible to everyone | Changeable by group members (CCL)
Model was written in NetLogo 5.0.4 • Viewed 215 times • Downloaded 22 times • Run 0 times
## WHAT IS IT?
The fractal that this model produces was discovered by the great Polish mathematician Waclaw Sierpinski in 1916. Sierpinski was a professor at Lvov and Warsaw. He was one of the most influential mathematicians of his time in Poland and had a worldwide reputation. One of the moon's craters is named after him.
## HOW IT WORKS
The basic geometric construction of the Sierpinski tree goes as follows. We begin with a single point on the plane and then apply a repetitive scheme of operations to it. Grow a "spider" centered at this point by drawing three equal line segments directed to the vertices of an equilateral triangle. Then at each vertex of the triangle repeat the construction --- grow a similar "spider" only scale it down by the factor of two.
`````` . Step 0: Start with a point
|
|
| Step 1: Grow a spider
/ \
/ \
/ \
|
|
/|\
/ | \
| Step 2: Repeat step 1
/ \
| / \ |
|/ \|
/ \ / \
/ \ / \
``````
The Sierpinski tree is closely related to the class of fractals called Sierpinski Carpets which includes the famous Sierpinski Triangle or as it is usually called The Sierpinski Gasket.
The features that characterize the Sierpinski tree are self-similarity and connectedness. It is not always easy to determine if a fractal is connected. It took almost a decade to prove the connectedness of the famous Mandelbrot set. However connectedness is apparent from the way Sierpinski tree is generated; at each iteration the set is connected.
## HOW TO USE IT
Push the SETUP button to clear the world and initialize globals. Press repeatedly on the GO ONCE button to perform iterations of the Sierpinski algorithm.
## THINGS TO NOTICE
Notice the use of `hatch` primitive which makes it so simple to generate fractals like Sierpinski tree.
## THINGS TO TRY
Try to write a program that draws other self-similar shapes. For instance try the rule below
`````` . Step 0
|
|
|
______________ Step 1
|
|
|
|
__|__
|
|
__|___________|__ Step 2
| | |
|
__|__
|
``````
The resulting fractal is known in Algebraic Topology as a Universal Covering of the Figure Eight.
## NETLOGO FEATURES
Notice how the curves are formed using several agents following the same rules. Also, take note of the use of the `hatch` command.
## RELATED MODELS
L-System Fractals
## HOW TO CITE
If you mention this model in a publication, we ask that you include these citations for the model itself and for the NetLogo software:
This model was created as part of the project: CONNECTED MATHEMATICS: MAKING SENSE OF COMPLEX PHENOMENA THROUGH BUILDING OBJECT-BASED PARALLEL MODELS (OBPML). The project gratefully acknowledges the support of the National Science Foundation (Applications of Advanced Technologies Program) -- grant numbers RED #9552950 and REC #9632612.
This model was converted to NetLogo as part of the projects: PARTICIPATORY SIMULATIONS: NETWORK-BASED DESIGN FOR SYSTEMS LEARNING IN CLASSROOMS and/or INTEGRATED SIMULATION AND MODELING ENVIRONMENT. The project gratefully acknowledges the support of the National Science Foundation (REPP & ROLE programs) -- grant numbers REC #9814682 and REC-0126227. Converted from StarLogoT to NetLogo, 2001.
Click to Run Model
```turtles-own [ modulus ]
; create a turtle and set its initial location and modulus
to setup
clear-all
crt 1
[
setxy 0 -3
set modulus 0.5 * max-pycor
pen-down
]
reset-ticks
end
; ask the turtles to go forward by modulus, create a new turtle to
; draw the next iteration of sierpinski's tree, and return to its place
to grow
hatch 1
[ fd modulus
set modulus (0.5 * modulus) ] ; new turtle's modulus is half its parent's
end
; draw the sierpinski tree
to go
[
repeat 3
[
grow
right 120 ; turn counter-clockwise to draw more legs
]
die ; kill all the living turtles
]
tick
end
```
There are 10 versions of this model.
Uri Wilensky over 9 years ago Updated to NetLogo 5.0.4 Download this version
Uri Wilensky over 10 years ago Updated version tag Download this version
Uri Wilensky over 10 years ago Updated to version from NetLogo 5.0.3 distribution Download this version
Uri Wilensky over 12 years ago Updated from NetLogo 4.1 Download this version
Uri Wilensky over 12 years ago Updated from NetLogo 4.1 Download this version
Uri Wilensky over 12 years ago Updated from NetLogo 4.1 Download this version
Uri Wilensky over 12 years ago Updated from NetLogo 4.1 Download this version
Uri Wilensky over 12 years ago Model from NetLogo distribution Download this version | 1,248 | 4,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2023-06 | latest | en | 0.936886 |
https://www.physicsforums.com/threads/triangle-question.611736/ | 1,513,097,451,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948517350.12/warc/CC-MAIN-20171212153808-20171212173808-00088.warc.gz | 785,743,934 | 15,969 | # Triangle question
1. Jun 6, 2012
### praharmitra
Hey guys,
This is NOT homework. I remember solving this question many years ago (at least 10 years ago). I am trying to recall the solution again and am just not able to. The question is -
In a triangle ABC, AD is the angle bisector of angle BAC. AB = CD. Prove that angle BAC = 72 degrees.
A diagram is attached.
NOTE - I am actually not even sure the question has enough information as given. I am just recalling from memory. If you think more information is required give an appropriate reason as to why you think that is true.
Usually, the way I see if the question has enough information is that I try to construct a triangle given the above properties. If I can construct a unique triangle, then of course the information is enough. Else not.
Anyway, give it a try.
https://www.dropbox.com/s/wpvei01hw1nr25l/2012-06-06%2000.39.06.jpg
2. Jun 6, 2012
### HallsofIvy
Staff Emeritus
Looks like non-sense to me. Do this: draw a right triangle, ABC, with right angle at A. At vertex C, use a protractor with radius set at length AB to mark point D on BC. D will be between B and C because the length of a leg of right triangle is always less than the length of the hypotenuse. You now have exactly the situation shown in your picture but the angle is 90 degrees, not 72. Clearly that can be done taking any acute angle at A.
3. Jun 6, 2012
### praharmitra
Well, but with that construction AD will not be the angle bisector of angle A.
4. Jun 6, 2012
### DaveC426913
72 degrees is 1/5 of a circle, or the complementary angle of a pentagram. My guess is that the solution will involve a pentagram or star of some sort.
5. Jun 6, 2012
### coolul007
this sounds very much like golden ratio
6. Jun 6, 2012
### praharmitra
Hi guys,
I figured out the problem with the question. There is not enough information to solve this problem. You can see this by doing the following construction. Draw the line AD first (This can be any length, for this argument atleast). Now draw the two equal angles BAD and DAC on either side of AD. Again, the angle could take any value. Extend the sides AB and AC such that BDC forms a straight line and BD = DC.
Now, in this triangle, AB > CD. Now rotate the side BDC around the point D such that the length of CD increases and that of AB decreases. We can always do this. Since the rotation is continuous, we will always have some angle of rotation at which AB = CD.
We then have constructed exactly the triangle described above! Since the angle is arbitrary, the question is incomplete.
I now recall what the original question was. There was an additional given -
angle ABC = 2 X angle BCA
Try and solve it now! | 686 | 2,718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-51 | longest | en | 0.917708 |
https://metanumbers.com/1078812 | 1,638,991,819,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363520.30/warc/CC-MAIN-20211208175210-20211208205210-00370.warc.gz | 455,411,135 | 7,703 | # 1078812 (number)
1,078,812 (one million seventy-eight thousand eight hundred twelve) is an even seven-digits composite number following 1078811 and preceding 1078813. In scientific notation, it is written as 1.078812 × 106. The sum of its digits is 27. It has a total of 7 prime factors and 48 positive divisors. There are 308,016 positive integers (up to 1078812) that are relatively prime to 1078812.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 7
• Sum of Digits 27
• Digital Root 9
## Name
Short name 1 million 78 thousand 812 one million seventy-eight thousand eight hundred twelve
## Notation
Scientific notation 1.078812 × 106 1.078812 × 106
## Prime Factorization of 1078812
Prime Factorization 22 × 33 × 7 × 1427
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 7 Total number of prime factors rad(n) 59934 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,078,812 is 22 × 33 × 7 × 1427. Since it has a total of 7 prime factors, 1,078,812 is a composite number.
## Divisors of 1078812
48 divisors
Even divisors 32 16 8 8
Total Divisors Sum of Divisors Aliquot Sum τ(n) 48 Total number of the positive divisors of n σ(n) 3.19872e+06 Sum of all the positive divisors of n s(n) 2.11991e+06 Sum of the proper positive divisors of n A(n) 66640 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1038.66 Returns the nth root of the product of n divisors H(n) 16.1887 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,078,812 can be divided by 48 positive divisors (out of which 32 are even, and 16 are odd). The sum of these divisors (counting 1,078,812) is 3,198,720, the average is 66,640.
## Other Arithmetic Functions (n = 1078812)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 308016 Total number of positive integers not greater than n that are coprime to n λ(n) 12834 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 84019 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 308,016 positive integers (less than 1,078,812) that are coprime with 1,078,812. And there are approximately 84,019 prime numbers less than or equal to 1,078,812.
## Divisibility of 1078812
m n mod m 2 3 4 5 6 7 8 9 0 0 0 2 0 0 4 0
The number 1,078,812 is divisible by 2, 3, 4, 6, 7 and 9.
## Classification of 1078812
• Arithmetic
• Abundant
### Expressible via specific sums
• Polite
• Practical
• Non-hypotenuse
## Base conversion (1078812)
Base System Value
2 Binary 100000111011000011100
3 Ternary 2000210212000
4 Quaternary 10013120130
5 Quinary 234010222
6 Senary 35042300
8 Octal 4073034
10 Decimal 1078812
12 Duodecimal 440390
20 Vigesimal 6eh0c
36 Base36 n4f0
## Basic calculations (n = 1078812)
### Multiplication
n×y
n×2 2157624 3236436 4315248 5394060
### Division
n÷y
n÷2 539406 359604 269703 215762
### Exponentiation
ny
n2 1163835331344 1255559521477883328 1354512678484598268846336 1461264531701326427610653432832
### Nth Root
y√n
2√n 1038.66 102.561 32.2282 16.0912
## 1078812 as geometric shapes
### Circle
Diameter 2.15762e+06 6.77838e+06 3.6563e+12
### Sphere
Volume 5.25928e+18 1.46252e+13 6.77838e+06
### Square
Length = n
Perimeter 4.31525e+06 1.16384e+12 1.52567e+06
### Cube
Length = n
Surface area 6.98301e+12 1.25556e+18 1.86856e+06
### Equilateral Triangle
Length = n
Perimeter 3.23644e+06 5.03955e+11 934279
### Triangular Pyramid
Length = n
Surface area 2.01582e+12 1.47969e+17 880846
## Cryptographic Hash Functions
md5 c2b16d93fb950e8a964fa53b2766677c e447e545e6e41ff1b414299f554d39d5c03a2024 52e0cf054bb704b8f98da3d9d7a4561c5e530c07059893c7999e1ce06af7dd7a f9f9679b6465cd0b71dcaf54433464eda5ba116b1390442c2f59b9d6d29729ca227331f9c666fcb61313b0ef53256e5853b262c5df079642dc3227273f9a8cc8 4b44e1bcb7c9fbb3a8a3dfa180c9df226ca2c946 | 1,563 | 4,361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-49 | latest | en | 0.79745 |
http://mathhelpforum.com/calculus/191674-derivatives-inverse-trig-function.html | 1,527,248,166,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867085.95/warc/CC-MAIN-20180525102302-20180525122302-00289.warc.gz | 188,465,870 | 9,513 | # Thread: Derivatives of inverse trig function
1. ## Derivatives of inverse trig function
I need help taking the derivative of y = arctan(x+1)
This is what I do
y = arctan(x+1)
x=tan(y+1)
cos^2(y+1) = y'
cos^2(arctan(x+1)+1) = y'
cos^2(((1-sec^2x)^0.5)+2))=y'
I am suppose to get a polynomial but I cannot take the derivative of it to get one. I just get stuck at the last step. Any help?
2. ## Re: Derivatives of inverse trig function
Originally Posted by Barthayn
I need help taking the derivative of y = arctan(x+1)
This is what I do
y = arctan(x+1)
x=tan(y+1)
cos^2(y+1) = y'
cos^2(arctan(x+1)+1) = y'
cos^2(((1-sec^2x)^0.5)+2))=y'
I am suppose to get a polynomial but I cannot take the derivative of it to get one. I just get stuck at the last step. Any help?
The tan and arctan functions don't work that way. You can only take the tangent of the whole expression
$\displaystyle \tan(y) = \tan(\arctan(x+1))$
$\displaystyle \tan(y) = x+1$
$\displaystyle x = \tan(y) - 1$
When you differentiate wrt x you should get: $\displaystyle 1 = \sec^2(y)y'$ | 362 | 1,062 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-22 | latest | en | 0.84176 |
https://www.jiskha.com/display.cgi?id=1397010247 | 1,503,352,924,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109670.98/warc/CC-MAIN-20170821211752-20170821231752-00077.warc.gz | 912,242,242 | 3,686 | # math
posted by .
Name the position of the first digit of the quotient.
1. 832/4
2. 217/7
I don't understand. Someone please explain to me.
• math -
since 832 >= 400, 832/400 > 1, and the 1st digit is in the hundreds position
217 < 700, but 217 >= 70, so 1st digit of quotient is in the tens position.
• math -
348/4
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More Similar Questions | 443 | 1,761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-34 | latest | en | 0.843494 |
https://www.popflock.com/learn?s=margin_of_error | 1,627,220,159,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151672.96/warc/CC-MAIN-20210725111913-20210725141913-00407.warc.gz | 998,168,573 | 26,246 | Margin of Error
Get Margin of Error essential facts below. View Videos or join the Margin of Error discussion. Add Margin of Error to your PopFlock.com topic list for future reference or share this resource on social media.
Margin of Error
Probability densities of polls of different sizes, each color-coded to its 95% confidence interval (below), margin of error (left), and sample size (right). Each interval reflects the range within which one may have 95% confidence that the true percentage may be found, given a reported percentage of 50%. The margin of error is half the confidence interval (also, the radius of the interval). The larger the sample, the smaller the margin of error. Also, the further from 50% the reported percentage, the smaller the margin of error.
The margin of error is a statistic expressing the amount of random sampling error in the results of a survey. The larger the margin of error, the less confidence one should have that a poll result would reflect the result of a survey of the entire population. The margin of error will be positive whenever a population is incompletely sampled and the outcome measure has positive variance, which is to say, the measure varies.
The term margin of error is often used in non-survey contexts to indicate observational error in reporting measured quantities. It is also used in colloquial speech to refer to the amount of space or amount of flexibility one might have in accomplishing a goal. For example, it is often used in sports by commentators when describing how much precision is required to achieve a goal, points, or outcome. A bowling pin used in the United States is 4.75 inches wide, and the ball is 8.5 inches wide, therefore one could say a bowler has a 21.75 inch margin of error when trying to hit a specific pin to earn a spare (e.g.,1 pin remaining on the lane).
## Concept
Consider a simple yes/no poll ${\displaystyle P}$ as a sample of ${\displaystyle n}$ respondents drawn from a population ${\displaystyle N{\text{, }}(n< reporting the percentage ${\displaystyle p}$ of yes responses. We would like to know how close ${\displaystyle p}$ is to the true result of a survey of the entire population ${\displaystyle N}$, without having to conduct one. If, hypothetically, we were to conduct poll ${\displaystyle P}$ over subsequent samples of ${\displaystyle n}$ respondents (newly drawn from ${\displaystyle N}$), we would expect those subsequent results ${\displaystyle p_{1},p_{2},\ldots }$ to be normally distributed about ${\displaystyle {\overline {p}}}$. The margin of error describes the distance within which a specified percentage of these results is expected to vary from ${\displaystyle {\overline {p}}}$.
According to the 68-95-99.7 rule, we would expect that 95% of the results ${\displaystyle p_{1},p_{2},\ldots }$ to fall within about two standard deviations (${\displaystyle \pm 2\sigma _{P}}$) either side of the true mean ${\displaystyle {\overline {p}}}$. This interval is called the confidence interval, and the radius (half the interval) is called the margin of error, corresponding to a 95% confidence level.
Generally, at a confidence level ${\displaystyle \gamma }$, a sample sized ${\displaystyle n}$ of a population having expected standard deviation ${\displaystyle \sigma }$ has a margin of error
${\displaystyle MOE_{\gamma }=z_{\gamma }\times {\sqrt {\frac {\sigma ^{2}}{n}}}}$
where ${\displaystyle z_{\gamma }}$ denotes the quantile (also, commonly, a z-score), and ${\displaystyle {\sqrt {\frac {\sigma ^{2}}{n}}}}$ is the standard error.
## Standard deviation and standard error
We would expect the normally distributed values ${\displaystyle p_{1},p_{2},\ldots }$ to have a standard deviation which somehow varies with ${\displaystyle n}$. The smaller ${\displaystyle n}$, the wider the margin. This is called the standard error ${\displaystyle \sigma _{\overline {p}}}$.
For the single result from our survey, we assume that ${\displaystyle p={\overline {p}}}$, and that all subsequent results ${\displaystyle p_{1},p_{2},\ldots }$ together would have a variance ${\displaystyle \sigma _{P}^{2}=P(1-P)}$.
${\displaystyle {\text{Standard error}}=\sigma _{\overline {p}}\approx {\sqrt {\frac {\sigma _{P}^{2}}{n}}}\approx {\sqrt {\frac {p(1-p)}{n}}}}$
Note that ${\displaystyle p(1-p)}$ corresponds to the variance of a Bernoulli distribution.
## Maximum margin of error at different confidence levels
For a confidence level ${\displaystyle \gamma }$, there is a corresponding confidence interval about the mean ${\displaystyle \mu \pm z_{\gamma }\sigma }$, that is, the interval ${\displaystyle [\mu -z_{\gamma }\sigma ,\mu +z_{\gamma }\sigma ]}$ within which values of ${\displaystyle P}$ should fall with probability ${\displaystyle \gamma }$. Precise values of ${\displaystyle z_{\gamma }}$ are given by the quantile function of the normal distribution (which the 68-95-99.7 rule approximates).
Note that ${\displaystyle z_{\gamma }}$ is undefined for ${\displaystyle |\gamma |\geq 1}$, that is, ${\displaystyle z_{1.00}}$ is undefined, as is ${\displaystyle z_{1.10}}$.
${\displaystyle \gamma }$ ${\displaystyle z_{\gamma }}$ ${\displaystyle \gamma }$ 0.68 0.999 0.90 0.9999 0.95 1.959963984540 0.99999 0.98 0.999999 0.99 0.9999999 0.995 0.99999999 0.997 0.999999999
Log-log graphs of ${\displaystyle MOE_{\gamma }(0.5)}$ vs sample size n and confidence level ?. The arrows show that the maximum margin error for a sample size of 1000 is ±3.1% at 95% confidence level, and ±4.1% at 99%. The inset parabola ${\displaystyle \sigma _{p}^{2}=p-p^{2}}$ illustrates the relationship between ${\displaystyle \sigma _{p}^{2}}$ at ${\displaystyle p=.0.71}$ and ${\displaystyle \sigma _{max}^{2}}$ at ${\displaystyle p=.0.5}$
Since ${\displaystyle \max \sigma _{P}^{2}=\max P(1-P)=0.25}$ at ${\displaystyle p=0.5}$, we can arbitrarily set ${\displaystyle p={\overline {p}}=0.5}$, calculate ${\displaystyle \sigma _{P}}$, ${\displaystyle \sigma _{\overline {p}}}$, and ${\displaystyle z_{\gamma }\sigma _{\overline {p}}}$ to obtain the maximum margin of error for ${\displaystyle P}$ at a given confidence level ${\displaystyle \gamma }$ and sample size ${\displaystyle n}$, even before having actual results. With ${\displaystyle p=0.5,n=1013}$
${\displaystyle MOE_{95}(0.5)=z_{0.95}\sigma _{\overline {p}}\approx z_{0.95}{\sqrt {\frac {\sigma _{P}^{2}}{n}}}=1.96{\sqrt {\frac {.25}{n}}}=0.98/{\sqrt {n}}=\pm 3.1\%}$
${\displaystyle MOE_{99}(0.5)=z_{0.99}\sigma _{\overline {p}}\approx z_{0.99}{\sqrt {\frac {\sigma _{P}^{2}}{n}}}=2.58{\sqrt {\frac {.25}{n}}}=1.29/{\sqrt {n}}=\pm 4.1\%}$
Also, usefully, for any reported ${\displaystyle MOE_{95}}$
${\displaystyle MOE_{99}={\frac {z_{0.99}}{z_{0.95}}}MOE_{95}\approx 1.3\times MOE_{95}}$
## Specific margins of error
If a poll has multiple percentage results (for example, a poll measuring a single multiple-choice preference), the result closest to 50% will have the highest margin of error. Typically, it is this number that is reported as the margin of error for the entire poll. Imagine poll ${\displaystyle P}$ reports ${\displaystyle p_{a},p_{b},p_{c}}$ as ${\displaystyle 71\%,27\%,2\%,n=1013}$
${\displaystyle MOE_{95}(P_{a})=z_{0.95}\sigma _{\overline {p_{a}}}\approx 1.96{\sqrt {\frac {p_{a}(1-p_{a})}{n}}}=0.89/{\sqrt {n}}=\pm 2.8\%}$ (as in the figure above)
${\displaystyle MOE_{95}(P_{b})=z_{0.95}\sigma _{\overline {p_{b}}}\approx 1.96{\sqrt {\frac {p_{b}(1-p_{b})}{n}}}=0.87/{\sqrt {n}}=\pm 2.7\%}$
${\displaystyle MOE_{95}(P_{c})=z_{0.95}\sigma _{\overline {p_{c}}}\approx 1.96{\sqrt {\frac {p_{c}(1-p_{c})}{n}}}=0.27/{\sqrt {n}}=\pm 0.8\%}$
As a given percentage approaches the extremes of 0% or 100%, its margin of error approaches ±0%.
## Comparing percentages
Imagine multiple-choice poll ${\displaystyle P}$ reports ${\displaystyle p_{a},p_{b},p_{c}}$ as ${\displaystyle 46\%,42\%,12\%,n=1013}$. As described above, the margin of error reported for the poll would typically be ${\displaystyle MOE_{95}(P_{a})}$, as ${\displaystyle p_{a}}$is closest to 50%. The popular notion of statistical tie or statistical dead heat, however, concerns itself not with the accuracy of the individual results, but with that of the ranking of the results. Which is in first?
If, hypothetically, we were to conduct poll ${\displaystyle P}$ over subsequent samples of ${\displaystyle n}$ respondents (newly drawn from ${\displaystyle N}$), and report result ${\displaystyle p_{w}=p_{a}-p_{b}}$, we could use the standard error of difference to understand how ${\displaystyle p_{w_{1}},p_{w_{2}},p_{w_{3}},\ldots }$ is expected to fall about ${\displaystyle {\overline {p_{w}}}}$. For this, we need to apply the sum of variances to obtain a new variance, ${\displaystyle \sigma _{P_{w}}^{2}}$,
${\displaystyle \sigma _{P_{w}}^{2}=\sigma _{P_{a}-P_{b}}^{2}=\sigma _{P_{a}}^{2}+\sigma _{P_{b}}^{2}-2\sigma _{P_{a},P_{b}}=p_{a}(1-p_{a})+p_{b}(1-p_{b})+2p_{a}p_{b}}$
where ${\displaystyle \sigma _{P_{a},P_{b}}=-P_{a}P_{b}}$ is the covariance of ${\displaystyle P_{a}}$and ${\displaystyle P_{b}}$.
Thus (after simplifying),
${\displaystyle {\text{Standard error of difference}}=\sigma _{\overline {w}}\approx {\sqrt {\frac {\sigma _{P_{w}}^{2}}{n}}}={\sqrt {\frac {p_{a}+p_{b}-(p_{a}-p_{b})^{2}}{n}}}=0.029,P_{w}=P_{a}-P_{b}}$
${\displaystyle MOE_{95}(P_{a})=z_{0.95}\sigma _{\overline {p_{a}}}\approx \pm {3.1\%}}$
${\displaystyle MOE_{95}(P_{w})=z_{0.95}\sigma _{\overline {w}}\approx \pm {5.8\%}}$
Note that this assumes that ${\displaystyle P_{c}}$ is close to constant, that is, respondents choosing either A or B would almost never chose C (making ${\displaystyle P_{a}}$and ${\displaystyle P_{b}}$ close to perfectly negatively correlated). With three or more choices in closer contention, choosing a correct formula for ${\displaystyle \sigma _{P_{w}}^{2}}$ becomes more complicated.
## Effect of finite population size
The formulae above for the margin of error assume that there is an infinitely large population and thus do not depend on the size of population ${\displaystyle N}$, but only on the sample size ${\displaystyle n}$. According to sampling theory, this assumption is reasonable when the sampling fraction is small. The margin of error for a particular sampling method is essentially the same regardless of whether the population of interest is the size of a school, city, state, or country, as long as the sampling fraction is small.
In cases where the sampling fraction is larger (in practice, greater than 5%), analysts might adjust the margin of error using a finite population correction to account for the added precision gained by sampling a much larger percentage of the population. FPC can be calculated using the formula[1]
${\displaystyle \operatorname {FPC} ={\sqrt {\frac {N-n}{N-1}}}}$
...and so if poll ${\displaystyle P}$ were conducted over 24% of, say, an electorate of 300,000 voters
${\displaystyle MOE_{95}(0.5)=z_{0.95}\sigma _{\overline {p}}\approx {\frac {0.98}{\sqrt {72,000}}}=\pm 0.4\%}$
${\displaystyle MOE_{95_{FPC}}(0.5)=z_{0.95}\sigma _{\overline {p}}{\sqrt {\frac {N-n}{N-1}}}\approx {\frac {0.98}{\sqrt {72,000}}}{\sqrt {\frac {300,000-72,000}{300,000-1}}}=\pm 0.3\%}$
Intuitively, for appropriately large ${\displaystyle N}$,
${\displaystyle \lim _{n\to 0}{\sqrt {\frac {N-n}{N-1}}}\approx 1}$
${\displaystyle \lim _{n\to N}{\sqrt {\frac {N-n}{N-1}}}=0}$
In the former case, ${\displaystyle n}$ is so small as to require no correction. In the latter case, the poll effectively becomes a census and sampling error becomes moot.
## Notes
1. ^ Isserlis, L. (1918). "On the value of a mean as calculated from a sample". Journal of the Royal Statistical Society. Blackwell Publishing. 81 (1): 75-81. doi:10.2307/2340569. JSTOR 2340569. (Equation 1)
## References
• Sudman, Seymour and Bradburn, Norman (1982). Asking Questions: A Practical Guide to Questionnaire Design. San Francisco: Jossey Bass. ISBN 0-87589-546-8
• Wonnacott, T.H. and R.J. Wonnacott (1990). Introductory Statistics (5th ed.). Wiley. ISBN 0-471-61518-8. | 3,588 | 12,095 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 103, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2021-31 | latest | en | 0.911754 |
https://socratic.org/questions/how-do-you-solve-x-2-6-5x-1 | 1,726,338,794,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.73/warc/CC-MAIN-20240914161327-20240914191327-00404.warc.gz | 491,525,336 | 6,063 | # How do you solve x^2+6=5x?
Jun 12, 2016
$x = 3$ or $x = 2$
#### Explanation:
${x}^{2} + 6 = 5 \cdot x$
${x}^{2} - 5 \cdot x + 6 = 0$
It can be written as two factors in the form:
$\left(x + a\right) \cdot \left(x + b\right) = 0$.
Where a, b are two integers
such that :
$a + b = - 5$
and
$a \cdot b = 6$
Thus, $a = - 2$ and $b = - 3$
Substituting the values of a and b we have:
${x}^{2} - 5 \cdot x + 6 = 0$
$\left(x - 3\right) \cdot \left(x - 2\right) = 0$
$x - 3 = 0$
Or
$x - 2 = 0$
so,
$x = 3$
or
$x = 2$ | 240 | 526 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-38 | latest | en | 0.602587 |
https://julianmilford.net/qa/question-is-17-and-68-a-coprime-number.html | 1,620,798,590,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991252.15/warc/CC-MAIN-20210512035557-20210512065557-00283.warc.gz | 362,577,761 | 7,208 | Question: Is 17 And 68 A Coprime Number?
Are 12 and 17 Coprime numbers?
Calculate the greatest (highest) common factor (divisor), gcf, hcf, gcd.
Two methods used below.12 = 22 × 3; 12 is not a prime, is a composite number;17 is a prime number, it cannot be broken down to other prime factors;.
What is the number of Coprime in 2048?
The prime factorization of 2,048 is 211. Since it has a total of 11 prime factors, 2,048 is a composite number.
Is 18 a Coprime number?
Yes, 18 and 35 are co-prime numbers. The factors of 18 are 1, 2, 3, 6, 9, and 18 while the factors of 35 are 1, 5, 7, and 35. Since the HCF is 1, they are coprime.
Are 15 and 37 Coprime numbers?
Yes, 15 and 37 are Co- prime numbers.
Is 18 and 25 are Coprime numbers?
Two integers a and b are said to be coprimes if they have no common factors other than 1 and the number itself. … 18 and 25 have no common factor other than 1. =>HCF(18,25)=1. Yes, (18,25) are coprime numbers.
Are 4 and 5 Coprime numbers?
Another example is 4 and 5: 4 = 2*2*1; 5 = 5*1 (Prime). The only common factor is 1, so they are coprime.
Are 2 and 3 twin primes?
Properties. Usually the pair (2, 3) is not considered to be a pair of twin primes. Since 2 is the only even prime, this pair is the only pair of prime numbers that differ by one; thus twin primes are as closely spaced as possible for any other two primes.
What are the co prime numbers between 1 to 100?
Coprime Numbers From 1 to 100 1 and 31 only. 29 and 31 are prime numbers. They have only one common factor 1. Thus they are co-prime.
Is 18 and 35 a Coprime number?
factors of 18: 1,2,3,6,9,18. factors of 35: 1,5,7,35. yes, they are co-prime numbers because only one factor is common that is 1.
Is 17 a co prime number?
As every prime number has only two factors 1 and the number itself, the only common factor of two prime numbers will be 1. For example, 11 and 17 are two prime numbers. … The only common factor is 1 and hence is co-prime.
What is 1 called if it is not a prime?
A prime number (or a prime) is a natural number greater than 1 that is not a product of two smaller natural numbers. A natural number greater than 1 that is not prime is called a composite number.
Is 3 and 5 a Coprime number?
For eg: (2,3), (11,12), (99,100) and so on; they have 1 as their HCF. The sum of any two coprime numbers is always coprime with the product of the same two coprime numbers. For eg: The sum of 2 and 3 is 5 and the product of 2 and 3 is 6. Hence, 5 and 6 are coprime numbers.
Which is the smallest composite number?
4All even number except 2 are composite numbers. 4 is the smallest composite number.
What is a Coprime number?
When two numbers have no common factors other than 1. In other words there is no whole number that you could divide them both by exactly (without any remainder). 21 and 22 are coprime: • The factors of 21 are 1, 3, 7 and 21.
What is the GCF of 17 and 68?
The greatest factor on the two lists that they have in common is the GCF of 17 and 68. Therefore, the GCF of 17 and 68 is 17. | 933 | 3,053 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2021-21 | latest | en | 0.949772 |
https://knowt.com/flashcards/0076c10d-b8e7-4d42-9ee3-71cab94db02c | 1,726,597,304,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00708.warc.gz | 318,800,491 | 34,622 | # Unit 4: Electric Circuits
Studied by 5 people
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Current
1 / 31
# Earn XP
## Description and Tags
### 32 Terms
1
Current
The continuous flow of charge
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2
Average current
Iavg = change in charge/ change in time
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3
Battery
A battery is a device that maintains an electric potential difference between the two terminals.
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4
Direct current
The flow is from higher potential to lower potential. The electricity also flows in that direction called direct current.
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5
Resistance
It is the impedance to the flow of electricity through a material. Asa charge moves through a material, it eventually hits a non-moving nucleus in the material.
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6
Resistivity
It can be thought of as the density of nuclei the electrons may strike.
R = ρ l / A
• R = resistance of the circuit
• ρ = resistivity
• l = length
• A = cross-sectional area
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7
low resistivity
conductors
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8
high resistivity
insulators
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9
resistors in series
R eq = R1 + R2
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10
resistors in parallel
1/Req = 1/R1 + 1/R2
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11
Ammeter
An ammeter is a device with a very low resistance that measures the current.
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12
Voltmeter
It measures the electric potential called potential drop.
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13
Ohm’s Law
V = IR
• R is the resistance in the circuit.
• V is the potential difference in the circuit
• I is the electric current
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14
Power dissipation
P = VIP = I^2 RP = V^2 IR
• P is the power
• V is the potential difference in the circuit.
• I is the electric current.
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15
Kirchhoff’s rules
• The loop rule states that the voltage drop across any complete loop in the circuit is 0V.
• This statement follows from the conservation of energy when applied to circuits.
• The junction rule states that the sum of all current flowing into any junction is equal to the current flowing out of the junction.
• This statement follows from the conservation of charge.
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16
Capacitance
C = QV
• C = refers to the capacitance that we measure in farads
• Q = refers to the equal charge that we measure in coulombs
• V = refers to the voltage that we measure in volts
• Besides, there is another formula that appears like this:
• C = refers to the capacitance
• K = refers to the relative permittivity
• ε0 = refers to the permittivity of free space
• A = refers to the surface area of the plates
• d = refers to the distance between places measured
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17
Capacitors in parallel
Cp = C1 + C2
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18
Capacitors in series
1/Cs = 1/C1 + 1/C2
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19
Name three possible energy sources for a circuit
• Battery
• Photoelectric cell
• Generator
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20
A circuit is opened and closed using a:
Switch
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21
What will happen to the charges in a circuit when a switch is closed?
The charges will flow through the circuit.
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22
What will happen to the charges in a circuit when a switch is open?
The circuit is broken and the charges stop flowing through the circuit.
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23
What is the standard voltage per branch in a home in the United States?
120 V
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24
A switch that automatically opens if the current is too high is a ________________
Circuit Breaker
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25
The symbol used to represent resistance in a schematic diagram is
a zigzag line
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26
A closed circuit is a circuit in which charge
can flow
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27
When two light bulbs are connected in series, the
same amount of current always flows through each bulb
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28
When resistors are put in parallel with each other their overall resistance is
smaller than the resistance of any of the resistors
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29
As more lamps are put into a series circuit, the overall current in the circuit
decreases
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30
As more lamps are put into a parallel circuit, the overall current in the circuit
increases
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31
When one light bulb in a parallel circuit containing several light bulbs burns out, the other light bulbs
burn the same as before
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32
Electrical devices in our homes are connected in
parallel
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null | null | null | null | null | null | THIS IS NOT ->Delawareliberal
Tuesday, October 24, 2006
The "party of personal responsibility" where art thou?
They need to read this:
The fact that you want to see Spivack debate anyone is hilarious!
I still have about a dozen questions Castle has ducked for 8 months. I'm sure Spivack has a few.
Let me see....a guy cares for his family and wife during surgery and a guy has health issues and you question their personal responsibility???????!!!!! Little disconnect here....come back to orbit!
Smith has more ducks than Lums pond. Castle has more dodges than I.G Burton. Why can't Castle return my calls and multiple emails? Too sick? Riiiiiight. Why can't Wenk?
Why can't Smith make up for lost time in th next 14 days?
One sentence explains everything. Their records are indefensible.
I think you forgot a couple of personal responsibility quotes:
"I'm a bitch." --Dennis Spivack on why he wouldn't debate Karen Hartley Nagle in the Democratic Primaries.
"I hate democracy." --Dennis Spivack on why he doesn't want to debate Michael Berg and Karen Hartley Nagle in the general election.
can we get in to see this debate??????
"Why can't Castle return my calls and multiple emails? Too sick? Riiiiiight. Why can't Wenk?"
Perhaps the creepiest statement you've ever made. I know, I know, you're a constituent. You're also batshit insane.
Clever. You should lend that wit to the service of the country rather than the service of one woman's vanity.
Remember, Castle is the bad guy. 14 days to go. Stay focused and keep your eyes on the prize. Do the right thing and I'll be a big booster of KHN next time provided she wins a primary.
You're also batshit insane.
Send me your email and I'll send you the emails. If you still think the questions are insane then - I'll stop.
In the meantime you are going to have to come up with some better put downs.
Yes, Jason. Having a wife recuperate from surgery and tending to four kids while she is doing so is NOT a legitimate reason for missing a few functions.
You are a complete and utter putz. But, we all knew that already.
(BTW, you also missed the part where Smith said he'd debate Colantuono "anytime, anywhere, any place.")
I wonder if you put it to the public: Which would they say is more virtuous -- tending to your wife and children, or skipping out on them to tend to your political career?
Please Hube. If it is "anytime, anyplace" why can't Smith get it together to get on stage with Coluntouno? What is the hold up?
Dude, are you being totally obtuse on purpose? Have they scheduled a debate? If so, when? Where? Can you read? C'mon, use both hemispheres of your brain, if that's what you call it.
And answer my question as to which is more virtuous.
question: are the debates on Thursday between Spivak and McCastle for the public?
question #2 - in regards of question above, will this be shown on WHYY?
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null | null | null | null | null | null | House GOP weighs reviving contentious ObamaCare fix bill
Speaker John Boehner (R-Ohio) left open the possibility that the House will resurrect a bill to fix ObamaCare decried by staunch conservatives.
The Speaker told reporters on Tuesday morning, “There’s still conversation going on over how best to deal with that issue.”
GOP leaders had to pull the bill from consideration on the House floor two weeks ago due to the lack of GOP votes for the measure that would transfer funding around to help individuals with pre-existing conditions.
Pressed as to whether leaders will, in fact, hold a vote on the measure, Boehner responded, "conversation underway.”
The Speaker added, “I think there's a lot more conversation about all of ObamaCare that needs to take place.”
House GOP leaders drew the ire of a number of their colleagues when they opted to hold a vote on a measure that would fix a funding depleted program in ObamaCare for individuals with pre-existing conditions.
House Majority Leader Eric Cantor (R-Va.) was behind the push to vote on a measure that would pay for the program by moving money from a so-called “slush” fund created in ObamaCare to the high-risk insurance pools.
Influential conservative interest groups decried the effort as a ploy to bolster a program that the GOP was elected to repeal in its entirety. One group castigated the GOP measure as “CantorCare,” as it was part of his effort to show a kinder, gentler side to the Republican brand.
On Friday, Cantor released a planning memo for the summer months in which he highlighted that the House will vote on a measure to repeal ObamaCare.
The House is set to vote on a different bill this week included under the kinder, gentler umbrella: the Working Families Flexibility Act.
Sophomore Rep. Martha Roby (R-Ala.) sponsored the bill designed to give hourly employees in the private sector the option of choosing to earn time off instead of earning overtime if they work more than 40 hours per week.
Cantor told reporters on Tuesday that “it puts parents over politics. There is no reason in the world for anyone to object to this bill.”
Yet there are objections, including a veto threat issued by the White House.
Democrats call the bill a "hit on women's paychecks."
Cantor is holding an event in Virginia Tuesday afternoon at a private company to showcase the importance of passing the Roby bill. | null | null | null | null | null | null | null | null | null |
books4purpose.com | 1,652,869,634,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662521883.7/warc/CC-MAIN-20220518083841-20220518113841-00404.warc.gz | 9,964,911 | 16,750 | Warning: A non-numeric value encountered in /home/customer/www/books4purpose.com/public_html/wp-content/themes/Divi/functions.php on line 5560
949.391.3560 [email protected]
Select Page
Basic Math Operators
All spreadsheet programs use standard operators for formulas, such as a plus sign for addition (+), a minus sign for subtraction (-), an asterisk for multiplication (*), a forward slash for division (/), and a caret (^) for exponents.
All formulas must begin with an equals sign (=). This is because you want the cell to contain, or equal to, the formula and the value it will calculate.
What is a cell reference?
You can create simple formulas manually (for example, =8+2 or = 100*5). However, most of the time you will use cell addresses to create a formula. Otherwise known as making a cell reference. Using cell references will ensure that your formulas are always accurate, because you can change the value of referenced cells without having to rewrite the formula.
By combining a mathematical operator within cell references, you can create a variety of simple formulas in Excel. Formulas can also include a combination of cell references and numbers, as in the examples below:
### Creating a formula:
In the following example, I’ll use a simple formula and cell references to calculate the sum of a budget.
Select the cell that will contain the formula.
Type the equals sign (=). Notice how it appears in both the cell and the formula bar.
Type the cell address of the cell you wish to reference first in the formula.
Type the mathematical operator you wish to use. Enter the cell address of the cell you wish to reference second in the formula, cell B2 in this example.
Press Enter. The formula will then be calculated, and the value will be displayed in the cell.
#### Changing values with cell references
The real advantage of cell references is that they allow you to update data in your spreadsheet without having to rewrite formulas. In the example below, we have changed the value of cell B1 from \$1,200 to \$1,800. The formula in B3 will automatically recalculate and display the new value in cell B3.
To create a formula using your mouse:
Instead of typing cell addresses manually, you can use your mouse to point and click on the cells you wish to include in your formula. This can save a lot of time and effort when creating formulas. In the following example below, we will create a formula to calculate the cost of ordering several boxes of plastic silverware.
Select the cell that will contain the formula.
• Type the equals sign (=).
• Select the cell you wish to reference first in the formula, cell B3 in our example. The cell address will appear in the formula.
• Type the mathematical operator you wish to use. In our example, we are using the multiplication sign (*).
• Click on the cell you wish to reference second in the formula, cell C3 in our example. The cell address will appear in the formula.
• Press The formula will be calculated, and the value will be displayed in the cell.
Formulas can also be copied to adjacent cells with the fill handle, which can save a lot of time and effort if you need to perform the same calculation multiple times in a spreadsheet.
Let’s edit a formula:
Sometimes you may want to modify an existing formula. In the example below, we’ve entered an incorrect cell address in our formula, so we’ll need to correct it.
Select the cell containing the formula you wish to edit.
Click the formula bar to edit the formula. You can also double-click the cell to view and edit the formula directly within the cell.
A border will appear around any referenced cells. In our example, we’ll change the second part of the formula to reference cell B2 instead of cell C2.
When finished, press Enter
The formula will be updated, and the new value will be displayed in the cell.
If you change your mind, you can press the Esc key on your keyboard to avoid accidentally making changes to your formula.
### Now you try it…
1. Open an existing Excel workbook.
2. Create a simple addition formula using cell references. If you are using the example, create the formula in cell B4 to calculate the total budget.
3. Try modifying the value of a cell referenced in a formula. If you are using the example, change the value of cell B2 to \$2,000. Notice how the formula in cell B4 recalculates the total.
4. Edit a formula using the formula bar. If you are using the example, edit the formula in cell B9 to change the division sign (/) to a minus sign (-). | 963 | 4,541 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2022-21 | longest | en | 0.855964 |
http://www.purplemath.com/learning/viewtopic.php?f=9&t=2387&p=6927 | 1,498,418,149,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320570.72/warc/CC-MAIN-20170625184914-20170625204914-00002.warc.gz | 616,440,774 | 6,445 | ## Negative Exponent Problem
Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
maroonblazer
Posts: 51
Joined: Thu Aug 12, 2010 11:16 am
Contact:
### Negative Exponent Problem
Hi,
The problem states:
"Write the following expression with only positive exponents and in simplest form. Variables are not equal to zero".
$x^{-1} + x^{-5}$
The answer is:
$\frac{x^4+1}{x^5}$
From the initial expression I get:
$\frac{1}{x} + \frac{1}{x^5}$
It's here that I get stuck. The next step seems like it would be:
$\frac{1}{x}+\frac{1}{x(x^5)}$
From there you'd go:
$\frac{1}{2x(x^4)}$
But that seems to go nowhere.
Help?
Thanks,
mb
stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
$\frac{1}{x} + \frac{1}{x^5}$
It's here that I get stuck. The next step seems like it would be:
$\frac{1}{x}+\frac{1}{x(x^5)}$
This would be like trying to convert "1/2 + 1/32" to a common denominator by multiplying 32 by 2, but "2" and "64" aren't "common". Instead, you'd need to convert the "2" to "32": (1/2)(1/16) + 1/32 = 16/32 + 1/32 = (16 + 1)/32.
Use that same process here.
maroonblazer
Posts: 51
Joined: Thu Aug 12, 2010 11:16 am
Contact:
### Re: Negative Exponent Problem
Instead, you'd need to convert the "2" to "32":
I love you!
$\frac{1}{x}*\frac{x^4}{x^4} + \frac{1}{x^5}=\frac{x^4+1}{x^5}$
Thank you,
mb
Return to “Advanced Algebra ("pre-calculus")” | 494 | 1,416 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 8, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-26 | longest | en | 0.861683 |
https://convertoctopus.com/110-cubic-meters-to-deciliters | 1,590,699,648,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347400101.39/warc/CC-MAIN-20200528201823-20200528231823-00517.warc.gz | 305,801,282 | 7,935 | ## Conversion formula
The conversion factor from cubic meters to deciliters is 10000, which means that 1 cubic meter is equal to 10000 deciliters:
1 m3 = 10000 dL
To convert 110 cubic meters into deciliters we have to multiply 110 by the conversion factor in order to get the volume amount from cubic meters to deciliters. We can also form a simple proportion to calculate the result:
1 m3 → 10000 dL
110 m3 → V(dL)
Solve the above proportion to obtain the volume V in deciliters:
V(dL) = 110 m3 × 10000 dL
V(dL) = 1100000 dL
The final result is:
110 m3 → 1100000 dL
We conclude that 110 cubic meters is equivalent to 1100000 deciliters:
110 cubic meters = 1100000 deciliters
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 deciliter is equal to 9.0909090909091E-7 × 110 cubic meters.
Another way is saying that 110 cubic meters is equal to 1 ÷ 9.0909090909091E-7 deciliters.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one hundred ten cubic meters is approximately one million one hundred thousand deciliters:
110 m3 ≅ 1100000 dL
An alternative is also that one deciliter is approximately zero times one hundred ten cubic meters.
## Conversion table
### cubic meters to deciliters chart
For quick reference purposes, below is the conversion table you can use to convert from cubic meters to deciliters
cubic meters (m3) deciliters (dL)
111 cubic meters 1110000 deciliters
112 cubic meters 1120000 deciliters
113 cubic meters 1130000 deciliters
114 cubic meters 1140000 deciliters
115 cubic meters 1150000 deciliters
116 cubic meters 1160000 deciliters
117 cubic meters 1170000 deciliters
118 cubic meters 1180000 deciliters
119 cubic meters 1190000 deciliters
120 cubic meters 1200000 deciliters | 508 | 1,871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2020-24 | latest | en | 0.786804 |
null | null | null | null | null | null | HP Pre 3
The HP Pre 3 is the latest in a line of similarly-styled handsets from the now defunct Palm division of HP. Each model has improved the original's excellent design and tweaked both the hardware quality and the way the operating system works too. There has also been a significant improvement in both battery and processor performance.
When HP bought Palm in April 2010 there was much excitement. Could this be a new era of investment and opportunity for the much-loved PDA company? Sadly, it seems not, as HP recently announced that it was shutting down production of both the Pre 3 and the only-recently launched Touchpad.
So this review comes to you at an interesting time. It's certainly going to be possible for you to get a Pre 3 from one of several suppliers. HP had announced that it would sell the phone at a loss for just £50, but as we're writing this there have been virtually no retailers actually offering the handset at this price. It is, however, possible to get it on a pretty cheap monthly tariff via Carphone Warehouse or buying it direct from online retailers.
So, the question has to be, if you can find one for a good price, should you invest in a phone that will have no successor and an operating system that is very likely on the decline.
Originally announced around the time the iPhone was starting to make waves, the Palm Pre was always a lovely looking device. With a pebble aesthetic, it offered a small size along with a beautiful big screen and a full QWERTY keyboard which slid out from the bottom of the handset.
In early versions, this mechanism was clumsy and flawed. To get the slider to work smoothly, and not get caught half-way out, you needed to press the middle of the screen and use your thumb to slide the screen upward. This was fine if the phone was locked, but if it wasn't then you would end up messing things up in the quest to access the keyboard. Happily, HP's Pre 3 has fixed this entiely. The phone now has a smooth, beautiful action that works wherever you press and push. This means you can flip it out using the space at the bottom of the screen.
Also gone are the horrible, sharp, edges of the original that surrounded the keyboard tray. These were very unpleasant, and it was one thing that made the original Pre (and indeed, the Pre 2) feel cheap in comparison with rivals from Android manufacturers, and of course Apple's high-end iPhone.
Around the outside edge of the phone there is a USB socket for charging and data transfer, a screen lock button on the top-right corner of the phone, a silent mode slider and a 3.5mm headphone jack. On the left-hand side, there are two volume buttons too, which we found impossible to locate and operate during a call.
On the back there's a tricky cover which snaps off to reveal the battery and SIM card socket. You can remove the battery if you want, and HP has made this really easy indeed, providing a pull chord that lifts the battery out of its home. Very well thought-out indeed. There is also the option of getting the Touchstone which allows you to charge the Pre 3 without wires. Handy if you are used to dumping your phone on your desk.
The Pre 3's keyboard is a contentious issue. Some people like it, and enjoy typing on it, and some people find it too small to use comfortably. However, when you learn the trick to using this phone's keyboard it's actually one of the nicest we've ever used on any phone. It's raised keys make it easy to locate the key you're looking for, and the rubber material means that you can dig a fingernail in to get some traction as you type.
We suspect that the Blackberry keyboard allows for faster typing long-term, but for a enjoyable process we actually think the Pre can't be beaten. We judge it this way: if a phone with a QWERTY keyboard makes you want to email more, on the move, then it's surely a success. And that's the case with the Pre. We found ourselves replying to more email, and in more detail, than we would on a touchscreen handset. The Pre is then, a success in its main field, which is as a rival to Blackberry's multitude of very capable business handsets.
We like the logic behind the keyboard too. There are 36 keys on the Pre 3. Mostly, these are letters with only backspace, return, full stop, space, shift, alt and sym taking up any other space. The alt button switches letters to a either numbers or symbols while the 'sym' button brings up a touch panel on-screen which gives more obscure offerings. There even emoticons included, in case you can't be bothered to type them in.
Interestingly, there's no on-screen keyboard. So if you want to type without the keyboard being extended, you're bang out of luck.
When it comes to the eye-phone interface, HP has got it licked. The screen on the Pre is an impressive 480 x 800 resolution squeezed into a 3.58-inch area. This means that text is crystal clear and incredibly easy to read. Web pages, in particular look incredible with massive detail. We even found browsing the Internet to be one of the best mobile experiences we've had.
The screen also has the advantage that it has been mounted as close to the surface of the phone as possible. This gives it a feeling of high-technology. It's easy to forget it's just an LCD screen, and assume it's some sort of surface projection from a brand new type of technology.
The touchscreen is responsive too. WebOS users will know how important gestures are in this OS, and they only work on a nice sensitive screen. Flicking around cards, or apps as most people call them, is super-fast and amazingly smooth. To get rid of an app, you just swipe up from the bottom of the screen, and it flies off never to be seen again. All of this works with at least as much style and finesse as the iPhone.
And like the iPhone will have in iOS5, deleting things or dismissing notifications is just a matter of swiping to the right. It's logical, and works brilliantly.
The camera
The Pre can shoot video at resolutions of 720p and the 5 megapixel camera produces some very nice photographs. On the built-in screen, images look stunning, and happily when you email them to friends, they still look great too.
What's more, the Pre's camera app is incredible simple. There aren't a lot of customisable options, so it's really just a matter of pointing and shooting. You can turn the flash off or switch to video mode, but that's about it.
Even better though, the camera has the little to no lag. Press the shutter button, and the photo gets taken straight away. This is a much more important feature than an array of idiotic picture modes, and we have to applaud HP for spending the time getting this phone to work as it should.
In many ways, the camera app is a good example of what's good and bad about the whole phone. It's simple, and works smoothly and as you'd expect. The problem is, it's not got much room for growth, and keen users will struggle to get the control they desire. Deep stuff.
Notifications and multi-tasking
It's worth mentioning the HP notification system too, because it's a mixed bag. On the one hand, information about incoming mail, tweets and social media updates all appear at the bottom of the screen as a little icon. This works well, as you can see everything that's happening.
The problem with notifications is that they generally aren't updated on a push basis. This means that they only arrive when the application polls the server. This is far from a major issue, but email junkies will want their notifications to appear as soon as the mail hits the server, and it's not our experience that the Pre can deliver this.
The thing we really like about notifications is that there's real consistency with how they are presented. On Android, apps can provide notifications in seemingly very random ways. The dropdown bar on those phones can often become a mess of things you're being told about. And with iOS 4 the iPhone system is horrible, popping up messages in the middle of the screen is very unsophisticated compared to the HP WebOS system. The good news for Apple fans is that notifications are soon going to get a lot better.
The other strength of WebOS is that it does multi-tasking properly. And by properly, we mean the way most people expect it to work. In Android, apps run much like they do in a Linux environment. This means that closing an app doesn't necessarily close it. Google's apps, on Android for example don't close when you switch out of them. On WebOS it's handled slightly differently. Apps can be closed by the user and when they are closed they really are closed. It's more like a PC or Mac really, where you are in charge of what is open and closed.
Lag and running out of memory
And there's one clear reason that the company has done this. Memory. The phone has just 512Mb of RAM. That means that having too much open at once can cause real problems. So much so that if you have too many "cards" or applications open, the phone will moan at you, and demand you get rid of some open apps.
This happens quite inconsistently though, so you can't really plan to only have a certain amount open at any one time. This is a minor issue though, as for the most part the phone is quite good at managing its limited memory.
Another infrequent, but annoying, problem is that of lag. We notice that web pages with Flash embedded were a particular source of trouble for the plucky Pre 3. Several times the phone just locked up, and took several minutes before allowing us to close anything. This slowdown, once started seemed hard to reverse too. In the end, when the handset gets like this, you need to consider restarting it. It's also worth pointing out that restarting the Pre 3 takes longer than the creation of the entire universe. We have no idea what the phone is doing during boot, but we can only assume it's really difficult.
Email is excellent
The Pre excels when it comes to email, although it doesn't have as many service provider options as Android seems to. If you're a Hotmail user, then you'll also have to jump through hoops to get that service to deliver email to your phone. Unlike other handsets, there don't seem to be many pre-loaded presets for other mail providers.
On the plus side, Gmail works out of the box, and you can have multiple accounts too. We use both personal and Google Apps accounts, so this is very helpful indeed. You can also add Exchange accounts which makes it ideal for those in corporate environments. Yahoo mail is also provided for, which is nice, although makes us wonder why Hotmail isn't offered the same luxury.
The unified inbox works well too, as you'd hope, and is far better than any we've seen on Android. The one feature missing is the ability to exclude an account from this combined inbox. This isn't a major pain, but it's a feature coming to the 7.5 update of Windows Phone 7, and it makes a nice change.
Like other smartphones, there's the ability to sync contacts, calendars and mail. We found this system elegant, and the way the Pre handles your calendar is unique too. Although the layout is quite compact, the company has managed to extract a lot of mileage from it, and your complicated calendar is greatly simplified by the Pre 3.
It would be remiss of us not to mention apps here, so allow us to explain the situation. There aren't any. There aren't likely to be any and the few there are either cost a lot of money, or just aren't that good.
The disgracefully named 'Spaz' is a must-have, purely because it appears to be the platform's only Twitter client. It's an okay piece of software that has several quirks. It is, however, better in every way than the official Twitter Android client, so that's good news.
Other than that, a few games and a pretty decent Facebook app you won't feel the need to fire up the app catalogue all that often. With HP withdrawing all its WebOS devices, we'd wager app support will get even worse from now on, unless another company licences the OS for its phones.
By far the best option for owners, or potential owners, of this phone is to buy it and assume that it's feature complete. We often call the Pre a feature phone, rather than a smartphone, which is a little mean. In fact, apps really only supplement functionality missing from the device, it could be argued that the Pre does everything most people need, so what's the point of an app store, but try telling that to an iPhone or Android user.
Music and media
It's fair to say that most people won't be buying this handset for music or video. Its storage capacity is really sub-standard at 8GB. The biggest let-down is that there's no external storage option. You can't simply add a 16GB microSD card to increase storage capacity, because there's no mircoSD card slot.
And, with a normally configured phone, with a few apps and some photos taken on the built-in camera, we only had around 6GB of usable storage. This isn't anywhere near enough for most music-loving smartphone owners.
Copying files over USB is quick though, and it's hassle-free. When you plug the phone in, it asks you if you'd like to switch to USB mode, or simply charge the phone. In USB mode, the phone shows up as an external drive, like any other USB storage device. The problem is, when it's in this mode, you can't make or receive calls or text messages. In fact, all you can do is copy files, and look at a USB logo.
Once you've got music on the phone though, it does sound pretty good. There's no EQ settings, but the default sound is pretty flat, which means you'll pretty much hear the music as it was recorded. The only problem with that is, MP3s are often sub-standard quality. We listened to some Jennifer Lopez (it's best you don't ask) and it had a good range, although it was a little too bright without much low-end. The mid-range was fine though, but nothing much to write home about.
There's no point pretending that the HP Pre 3 is a contender in the smartphone arena. Apple and Google have that sewn up with worth contributors in the form of BlackBerry and Windows Phone 7. What the Pre 3, and WebOS can offer is one of the simplest, most beautifully designed user interfaces, a nice likable keyboard and for the first time, reasonable battery life.
If you buy the Pre accepting that it's got little or no real developer support, the apps are pretty much universally rubbish and you'll end up using the phone pretty much in the state it came out of the box, then you'll relax and start to enjoy the handset a lot more.
For phone calls, email and text messaging it's just about the best all-rounder we've used. Social network applications cover Linkedin, Facebook and Twitter, which, for the time being, is good enough. There are a smattering of games and apps in the App Catalogue, but we found ourselves not really caring all that much. This isn't a phone designed at gamers, it's a phone aimed at business users who want something with terrific email support and that works well as a phone and don't want to be just another BlackBerry user.
It's hard to really moan about much on the Pre, as long as you walk into buying one with your eyes open. We've used ours instead of an Android handset, and the honest truth is we don't actually miss the dozens of apps we had installed. Obviously, a time will come when we do, but for most use, the Pre has it all.
So it's with a tinge of sadness that we conclude that this worthy, pleasant little phone will be the last of its kind. HP has absorbed Palm for its patents rather than for its products, and that means a good opeating system is probably going to die a death.
Get one while you can, even if its just for posterity.
The Palm Pre 3 was kindly loaned to us by Handtec.co.uk | null | null | null | null | null | null | null | null | null |
https://excelweez.com/how-to-calculate-percentiles-in-excel/ | 1,708,786,761,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474541.96/warc/CC-MAIN-20240224144416-20240224174416-00371.warc.gz | 246,948,255 | 38,745 | # How to Calculate Percentiles in Excel
Excel is a beast when it comes to calculations. This means you can use it to calculate percentiles. If you are looking for a way to calculate Percentiles quickly, then this guide is for you.
First, I will take you through what a percentile means and then show you three functions you can use.
## What is Percentile?
A percentile tells you how well you did compared to everyone else. For example, if you’re in the 75th percentile, it means you did better than 75% of the students. So, if there were 100 students, you’d be among the top 25. It’s a way to see where you stand in a group based on your performance.
### Percentile Function in Excel
Excel has 3 variations of the percentile function.
PERCENTILE: This is the function that was mostly used in the older Excel versions. It’s still available for compatibility. However, the modern functions have been split to accommodate the two types of percentiles.
PERCENTILE.INC: In this function, if you got the exact same score as someone else, it still counts you both as that percentile. It includes you in the count. PERCENTILE.INC is the most used Percentile function in Excel
PERCENTILE.EXC: If you and someone else got the exact same score, you’re not both counted in that percentile. It excludes the shared score.
## How to calculate percentile in Excel?
To calculate percentile, you need to use the following syntax.
=PERCENTILE.INC(array,k)
Array in this formula is the range of cells containing the values that you want to calculate its kth percentile.
On the other hand, K stands for the value between 0 and 1 that results in kth percentile value.. For instance, if you wish to find the 80th percentile, the k value is 0.8 or 80%. If you wish to find the 30th percentile, then the k value will be 0.3 or 30%
Here are the steps:
Assuming you wish to find the 60th percentile of the following data.
1. Open the spreadsheet containing the values you want to find values of
2. Type the following formula in an empty cell and press enter. For this example, I will use cell E5
=PERCENTILE.INC(B5:B14,60%)
As you can see from the image above, Excel has computed the 60th percentile automatically.
## How do I calculate the 90th percentile in Excel?
To calculate the 90th percentile in Excel for given values. Type the formula =PERCENTILE.INC(Values,90%)in a new cell. Note that values can be actual values separated by a comma or a range of values, such as A2:A10. Press enter, and Excel will compute the 90th percentile of your values.
Alternatively, you can use the formula =PERCENTILE.INC(Values,0.9). If the values are in the range A2:A15, then the final formula should appear as =PERCENTILE.INC(A2:A15,0.9). Now press enter to get the results
## How do I calculate the 95% percentile in Excel?
You can calculate the 95th percentile by using the following formula. Note that this formula assumes the values are in the range B5:B15. Therefore, the formula should be. =PERCENTILE.INC(B2:B15,95%)
## How do I calculate the 75th percentile in Excel?
To calculate the 75th percentile in Excel, use any of the following formulas.=PERCENTILE.INC(B2:B14,75%) or =PERCENTILE.INC(B2:B14,0.75). Replace the range B2:B14 with the actual range of your data.
## How do I calculate the 25th percentile in Excel?
You can calculate 25th percentile by using the formula =PERCENTILE.INC(B2:B15,25%) or =PERCENTILE.INC(B2:B15,0.25) | 850 | 3,435 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-10 | longest | en | 0.924674 |
https://web2.0calc.com/questions/help_31284 | 1,590,586,505,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347394074.44/warc/CC-MAIN-20200527110649-20200527140649-00019.warc.gz | 616,515,602 | 5,948 | +0
# help
0
102
1
Mr. McBride ordered 6 cheese pizzas and 7 pepperoni pizzas and paid \$153.50 for his son's birthday party. He realized he was going to run out of pizza and made a second order. This time, he ordered 2 cheese and 2 pepperoni pizzas and paid \$47.00. What is the cost of one pepperoni pizza?
Mar 28, 2020
#1
+4569
0
Let c be the number of cheese pizzas and p be the number of pepperoni pizzas.
By systems of equations, we have:
6c+7p=153.50
2c+2p=47.00
...........................................
c+p=23.50, 6c+6p=141.00
Thus, the cost of one pepperoni pizza is \$12.50.
Mar 28, 2020 | 195 | 614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2020-24 | latest | en | 0.89653 |
null | null | null | null | null | null | Is the SIFT algorithm scalable? Most questions I've seen asked about SIFT seem focussed on a simple comparison between two images. Instead of determining how similar two images are, would it be practical to use SIFT to find the N closest matching images out of a collection of thousands of images?
For example, would it be practical to use SIFT to generate keypoints for a batch of images, store the keypoints in a database, and then find the ones that have the shortest Euclidean distance to the keypoints generated for a "query" image?
When calculating the Euclidean distance, would you ignore the x, y, scale, and orientation parts of the keypoints, and only look at the descriptor?
asked Mar 02 '11 at 08:55
Cerin's gravatar image
edited Mar 02 '11 at 08:56
I am interested in this as well. Particularly I am interested in doing this with Lucene/Solr.
(Mar 02 '11 at 12:54) Oscar Täckström
Since Lucene/Solr are focussed on full text search, wouldn't they be inappropriate for this domain? The descriptors may be largely unique, but have very small Euclidean distances to "similar" descriptors. I was looking into a RDMS supported KNN solution, such as PostGIS or Mahout Taste.
(Mar 02 '11 at 13:13) Cerin
I found, but I have no idea about its scalability. My experience with Lucene/Solr is very limited, the reason I asked for this is that all textual data in the system is indexed by this.
I looked quickly at PostGIS, but to me it seems that this will only support simple geometry and is most concerned with finding intersections/encapsulation of polygons.
(Mar 02 '11 at 13:29) Oscar Täckström
I think that if you do some sort of discretized feature extraction such as the one mentioned in my answer you might be able to use lucene/solr as black boxes to do queries on your image database; the main downside being that more complex queries (i.e., that depend for example on the relative positions of features) won't be supported (unless you are willing to add features that consider the relative position of the original features).
(Mar 02 '11 at 13:33) Alexandre Passos ♦
4 Answers:
Searching with SIFT is still an open area of research, but it is indeed possible. The simplest and fastest search algorithms for SIFT in a large database involve computing the descriptors for images as soon as they enter the database, using a clustering algorithm do to vector quantization on the descriptors to get a set of discrete features for each image, and finally use those features as one would use text features to search the database, maybe using tf-idf feature vectors or something similar.
This, as you mentioned, ignored the x, y, scale, and orientation of the keypoints. Taking those things in consideration is a lot harder, however, as you'd have to search not only for matches between the images' descriptors but for spatially consistent matches, and things like partial occlusion, 3d point-of-view differences, etc, can make this a very hard problem. One possible solution is to first filter the images in the database by presence/absence of the vector-quantized features and then do more expensive matching algorithms on the returned list of images to get a better precision. I can't refer you to any papers in this area as I've only watched a couple of seminars and didn't catch the references.
answered Mar 02 '11 at 09:06
Alexandre%20Passos's gravatar image
Alexandre Passos ♦
Would inverted indices really scale for SIFT-features used in this way? I thought you had to use some fast approximate nearest neighbour search approach in order to get this to scale.
(Mar 02 '11 at 13:33) Oscar Täckström
Inverted indices will work if you discretize the features. You do lose some resolution.
The problem with nearest-neighbor-search is that an image is a bag of nearly-unique sift features, so unless you define a distance metric between bags of continuous vectors it doesn't make sense to do nearest neighbor search, and I don't know of any obviously right way of doing this.
(Mar 02 '11 at 13:35) Alexandre Passos ♦
i would assume that someone has tried a straight LSH ( approach. do you know of any papers where this has been tried?
(Mar 24 '11 at 16:31) Travis Wolfe
Can you recommend any tools for clustering high-dimensional data?
(Apr 07 '11 at 09:52) Cerin
How about kmeans? If you have a NVidia GPU, thy this ;)
(Apr 09 '11 at 05:29) Andreas Mueller
Some work in this direction was done by Sivic: Have a look at this series of paper and related literature. Basically, as Alexandre says: This is still an area of open research but it seems feasible.
answered Mar 03 '11 at 07:18
Andreas%20Mueller's gravatar image
Andreas Mueller
I'm not much versed in this area, but I've heard of attempts using kd-trees for image matching based on sift. Here some papers I found looking for "sift and kd-trees":
answered Mar 22 '11 at 13:25
Breno's gravatar image
The work by Torralba, Fergus, and Weiss, "Small Codes and Large Image Databases for Recognition" (CVPR 2008), describes an efficient SIFT-based image retrieval method capable of finding the nearest neighbors in a database of millions of images in a fraction of a second on a single PC.
For each image, a gist descriptor is computed, which is then reduced to a short binary code (tens to hundreds of bits). Similar images will have a small Hamming distance between their codes. Since the codes are short, a brute force search over the database is fast.
The mapping from the gist descriptor to the binary code is learned. One of the methods presented is based on a multiple-layer restricted Boltzmann machine, and uses the data set's image category labels during the "fine tuning" phase of learning.
answered Mar 30 '11 at 22:58
Louis%20Howe's gravatar image
Louis Howe
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# CHP 1
## Description: jemh101
### Read the Text Version
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r, 0 ≤ r < b. Step 2 : If r = 0, the HCF is b. If r ≠ 0, apply Euclid’s lemma to b and r. Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be HCF (a, b). Also, HCF(a, b) = HCF(b, r). 3. The Fundamental Theorem of Arithmetic : Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur. 4. If p is a prime and p divides a2, then p divides a, where a is a positive integer.5. To prove that 2, 3 are irrationals.6. Let x be a rational number whose decimal expansion terminates. Then we can express x p in the form q , where p and q are coprime, and the prime factorisation of q is of the form 2n5m, where n, m are non-negative integers.7. Let x = p be a rational number, such that the prime factorisation of q is of the form 2n5m, q where n, m are non-negative integers. Then x has a decimal expansion which terminates.8. Let x = p be a rational number, such that the prime factorisation of q is not of the form q 2n 5m, where n, m are non- negative integers. Then x has a decimal expansion which is non-terminating repeating (recurring). REAL NUMBERS 19 A NOTE TO THE READERYou have seen that :HCF ( p, q, r) × LCM (p, q, r) ≠ p × q × r, where p, q, r are positive integers(see Example 8). However, the following results hold good for three numbersp, q and r : p ⋅q ⋅r ⋅HCF(p, q, r) LCM (p, q, r) = HCF( p, q) ⋅ HCF(q,r) ⋅ HCF( p, r) p ⋅q ⋅r⋅ LCM(p, q, r) HCF (p, q, r) = LCM( p, q) ⋅ LCM(q, r) ⋅ LCM( p, r)not©toNbCeErReTpublished | 585 | 1,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2023-06 | latest | en | 0.849513 |
https://socratic.org/questions/how-do-you-find-e-1-given-e-1-1-1-1#286133 | 1,675,766,331,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500456.61/warc/CC-MAIN-20230207102930-20230207132930-00199.warc.gz | 534,631,761 | 5,793 | How do you find E^-1 given E=((1, -1), (1, 1)) ?
Jul 8, 2016
$\frac{1}{2} \left(\begin{matrix}1 & 1 \\ - 1 & 1\end{matrix}\right)$
Explanation:
${\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)}^{-} 1 = \frac{1}{a d - b c} \left(\begin{matrix}d & - b \\ - c & a\end{matrix}\right)$
here that's
$\frac{1}{1 \cdot 1 - \left(- 1\right) \cdot 1} \left(\begin{matrix}1 & 1 \\ - 1 & 1\end{matrix}\right)$
$\frac{1}{2} \left(\begin{matrix}1 & 1 \\ - 1 & 1\end{matrix}\right)$ | 214 | 481 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2023-06 | latest | en | 0.204943 |
https://web2.0calc.com/questions/help-algebra_10 | 1,618,364,711,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038076454.41/warc/CC-MAIN-20210414004149-20210414034149-00105.warc.gz | 706,872,826 | 5,586 | +0
# help algebra
0
82
1
The sum of two numbers is 16 and their product is 36. Find the reciprocal of the sum of the two numbers.
Dec 31, 2020
#1
+814
0
Hello Guest!
Let's set the 2 numbers as a and b.
a + b = 16
a*b = 36
I'm a bit confused about your question because the reciprocal of the sum of the two numbers is just 1/16. 1/(a+b)
The sum being 16, and the reciprocal of 16 is 1/16.
I've seen many problems about the sum of the reciprocals of the numbers. 1/a+1/b
Which is equal to (a+b)/(a*b) = 16/36 = 4/9
However, that's not what the problem asks for.
I hope this helped. :))))
=^._.^=
Dec 31, 2020 | 215 | 623 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2021-17 | latest | en | 0.956227 |
https://www.nickzom.org/blog/2019/06/25/how-to-calculate-and-solve-for-cumulative-gor-of-reservoir-fluid-flow-the-calculator-encyclopedia/ | 1,611,053,670,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703518240.40/warc/CC-MAIN-20210119103923-20210119133923-00683.warc.gz | 930,867,253 | 29,327 | # How to Calculate and Solve for Cumulative GOR of Reservoir Fluid Flow | The Calculator Encyclopedia
The image above represents the cumulative GOR.
To compute for the cumulative GOR, two essential parameters are needed and these parameters are cumulative gas production (Gp) and cumulative oil production (Np).
The formula for calculating the cumulative GOR:
Rp = Gp / Np
Where;
Rp = Cumulative GOR
Gp = Cumulative Gas Production
Np = Cumulative Oil Production
Let’s solve an example;
Find the cumulative GOR with a cumulative gas production of 32 and a cumulative oil production of 44.
This implies that;
Gp = Cumulative Gas Production = 32
Np = Cumulative Oil Production = 44
Rp = Gp / Np
Rp = 32 / 44
Rp = 0.72
Therefore, the cumulative GOR is 0.72.
Calculating the Cumulative Gas Production when the Cumulative GOR and the Cumulative Oil Production is Given
Gp = Rp x Np
Where;
Gp = Cumulative Gas Production
Rp = Cumulative GOR
Np = Cumulative Oil Production
Let’s solve an example;
Find the cumulative gas production when the cumulative GOR is 56 and the cumulative oil production is 19.
This implies that;
Rp = Cumulative GOR = 56
Np = Cumulative Oil Production = 19
Gp = Rp x Np
Gp = 56 x 19
Gp = 1064
Therefore, the cumulative gas production is 1064.
Calculating the Cumulative Oil Production when the Cumulative GOR and the Cumulative Gas Production is Given
Np = Gp / Rp
Where;
Np = Cumulative Oil Production
Rp = Cumulative GOR
Gp = Cumulative Gas Production
Let’s solve an example;
Find the cumulative oil production when the cumulative GOR is 18 and the cumulative gas production is 72.
This implies that;
Rp = Cumulative GOR = 18
Gp = Cumulative Gas Production = 72
Np = Gp / Rp
Np = 72 / 18
Np = 4
Therefore, the cumulative oil production is 4.
Nickzom Calculator – The Calculator Encyclopedia is capable of calculating the cumulative GOR.
To get the answer and workings of the cumulative GOR using the Nickzom Calculator – The Calculator Encyclopedia. First, you need to obtain the app.
You can get this app via any of these means:
To get access to the professional version via web, you need to register and subscribe for NGN 1,500 per annum to have utter access to all functionalities.
You can also try the demo version via https://www.nickzom.org/calculator
Apple (Paid) – https://itunes.apple.com/us/app/nickzom-calculator/id1331162702?mt=8
Once, you have obtained the calculator encyclopedia app, proceed to the Calculator Map, then click on Petroleum under Engineering
Now, Click on Reservoir Fluid Flow under Petroleum
Now, Click on Cumulative GOR under Reservoir Fluid Flow
The screenshot below displays the page or activity to enter your value, to get the answer for the cumulative GOR according to the respective parameter which are the cumulative gas production (Gp) and cumulative oil production (Np).
Now, enter the value appropriately and accordingly for the parameter as required by the cumulative gas production (Gp) is 32 and cumulative oil production (Np) is 44.
Finally, Click on Calculate
As you can see from the screenshot above, Nickzom Calculator– The Calculator Encyclopedia solves for the cumulative GOR and presents the formula, workings and steps too. | 839 | 3,235 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-04 | latest | en | 0.73985 |
null | null | null | null | null | null | ‘With Them or With Us’
Almost three decades ago, a group of radical Islamist students, dressed in army fatigues or covered in scarves and black chadors, forced their way into the American embassy in Tehran. According to some accounts, Mahmoud Ahmadinejad, then a student at a second-tier technical college in Tehran, was invited to join the hostage takers. He declined, saying he would join only if they would also occupy the Soviet embassy in Tehran. “No to the West, No to the East” was in those days the much-touted slogan of the regime.
The students, we now know, initially planned their action as political theater, no more than a daring stunt. After a few days, and some publicity, they intended to end the occupation. Regime leader Ayatollah Ruhollah Khomeini, however, had other ideas. Sensing the political gains of prolonging the crisis, he called the take-over of the embassy the Second Revolution and used it to consolidate his hold on power. Every year since, the regime has organized mass demonstrations to celebrate this brazen breach of diplomatic decorum as a way to bolster its own popularity.
Yesterday, the regime tried to repeat this ritual--but something else happened on the way to the embassy. From weeks before, sources close to the Iranian Revolutionary Guards Corps (IRGC) and the Basij declared their intention to make the day into a showdown with the opposition, a day the regime would bring out its forces in full. In the words of Sobhe-Sadeq, the political organ of the IRGC, the world would be forced to accept that “the people” are with the regime, and that the democratic opposition is nothing but a silly handful of effeminate upper-middle-class sissies.
But the rifts within the regime were on full display during yesterday’s ceremonies. Days earlier, Ahmadinejad had declared his “satisfaction” with the results of the negotiations with the United States in Geneva, and promised more direct contact. In response, Hossein Shariat-Madari, the infamous editor of the daily Keyhan--nowseen asthe semi-official voice of Supreme Leader Ayatollah Ali Khamenei--wrote in a surprisingly sharp-tongued editorial that only Khamenei sets Iranian nuclear and foreign policy; the editorial went so far as to even deny there had ever been direct negotiations between the United States and Iran in Geneva. Leaving no doubt as to his true feelings, Khamenei began yesterday’s commemoration with a scathing attack on America. Obama has been all talk and no action, Khamenei said, and his talks and his smiles have been so transparently bereft of real substance that they would not even “fool a child.”
Before regime supporters and critics took to the streets yesterday, another key development took place. Ayatollah Hossein Montazeri, now clearly the spiritual father of the Green movement, issued a remarkable statement, apologizing for the fact that, 30 years ago, as one of the leaders of the clerical regime, he had supported the occupation of the American embassy. It was, Montazeri declared, a foolish, costly decision, tantamount to a declaration of war. He went on to lambaste the regime for now selling the country cheaply to China and Russia. What is the difference between Russia and America, he asked, and why is it acceptable to give the former all kinds of concessions, while refusing to even negotiate with the latter?
As the day wore on, it became clear that the regime’s attempt to intimidate people into silence or inaction had failed again. Regime critics showed up in impressive numbers. Hundreds were beaten by security forces, including opposition leaders Mir Hussein Moussavi and Mehdi Karubi.
One of the most clever and telling slogans of yesterday’s demonstrators was a play on Barack Obama’s name. Obama, if written in Persian, is composed of three words--ou, ba, ma--meaning, “He is with us.” During yesterday’s protests, people tore down posters of Khamenei, and shouted “Obama, Obama, ya ba oona, ya ba ma”--“Obama, Obama, either with them, or with us.”
President Obama’s statement remembering the anniversary of the hostage crisis invited Iran and America to put the past behind them, and look to a better tomorrow. But his reference to the democratic aspirations of the Iranian people was too oblique. “The world,” he said, “continues to bear witness to their powerful calls for justice, and their courageous pursuit of universal rights.” In Tehran and several other major Iranian cities yesterday, people risked life and limb demanding these rights. Confirming that the world is watching leaves the question Iranians are asking of Obama--ba ma ya, ba anha--still without a clear answer.
Abbas Milani is the Hamid and Christina Moghadam Director of Iranian Studies at Stanford, where he is the co-director of the Iran Democracy Project. His latest book is Eminent Persian: The Men and Women who Made Modern Iran, 1941-1979 (Syracuse University Press).
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https://math.stackexchange.com/questions/1312639/difficulty-proving-multivariate-limit-involving-sin2x-does-not-exist | 1,618,427,900,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038078021.18/warc/CC-MAIN-20210414185709-20210414215709-00460.warc.gz | 478,649,362 | 38,642 | # Difficulty proving multivariate limit involving sin^2(x) does not exist
I am attempting to prove that $$\lim_{(x,y) \to (0,0)} \frac{x \sin^{2}(y)}{x^{2} + y^{2}}$$ does not exist, but am having issues. I can't seem to find a path for which the value is not zero (to prove that the limit value depends on the given path) , however I do believe that it does not exist, as I can't seem to prove that it does either with the Squeeze theorem.
I have attempted approaching along $y=0$, $x=0$, $y=mx$, $y=mx^{2}$ to name a few, and the usual strategies of attempting to cancel out factors or splitting up my equation don't seem to help as they all just give me zero. Am I on the wrong track here with believing the limit does not exist in the first place?
Hint: What if it were $y^2$ instead of $\sin^2 y$ in the numerator?
• Ahh, right the limit would approach zero. Thanks. – HavelTheGreat Jun 4 '15 at 22:58
• Small suggestion: The limit doesn't approach anything, either it exists or it doesn't, and if it does, it's just a number and doesn't dance around or approach anything. – zhw. Jun 4 '15 at 23:05
• Right, complete misuse of 'approach' on my part. Thanks again. – HavelTheGreat Jun 4 '15 at 23:11
$x^2+y^2 \geq 2|x||y| \to \left|\dfrac{x\sin^2y}{x^2+y^2}\right| \leq \dfrac{|x|\sin^2y}{2|x||y|} \leq \dfrac{1}{2}|\sin y|\cdot \left|\dfrac{\sin y}{y}\right| \to$ limit $= 0$ by squeeze theorem.
How about using polar coordinates? Define $$x^{2}+y^{2}=r$$ , $$x=r\cos(\beta)$$, $$y=r\sin(\beta)$$, Substitute to get $$\lim_{r\to 0}\frac{r\cos(\beta)\sin^{2}((r\sin(\beta))}{r}$$ The outer $$r$$ cancels out with the denominator and the other $$r$$ cause the function approaching to $$0$$. | 544 | 1,699 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2021-17 | latest | en | 0.889506 |
https://gedexamhire.com/what-kind-of-math-is-on-the-ged-test | 1,726,457,605,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.29/warc/CC-MAIN-20240916012328-20240916042328-00338.warc.gz | 247,056,593 | 31,600 | # What Kind Of Math Is On The Ged Test?
What Kind Of Math Is On The Ged Test? (Novel) Math is the number of squares in a two-dimensional (2D) graph. It is a well-known fact that math is not on your blog. It’s been found across the board in many different places. For example, a full-page news service site is definitely not on your blog because of writing heavy columns about math specifically. Math in Science Math functions as you would expect in the sense of mathematical reality. Usually, the shape of a graph is not pretty. It has a tiny set of triangles which, in some places, can be turned around and turned into multiple rectangles. This is just bad math then. In many cases, a graph is made out of this set. But in some cases, a graph is made out of polygons (or, in the case of a small city-size rectangle, triangles). That’s all great because for a diagram, the graph is essentially a rectangle but even better since at the edges there is not a big need for it to be painted or printed without a lot of geometry. You may wonder what kind of math is on the basis of the shapes of polygons. Quite often, there are two ways that a graph is drawn. The first is a way for the background color space to be transparent (similar to “blur” in printing), then text with color mapping onto the base. The second is a way for the background color space to be adjusted so that the colors are like b/w for the same font size (by default 20px) and similar to the base color coordinate system. This means that the background color space can support graphic layout solutions that make use of multiple layers of different colors. Examples include the so-called blader, which displays the vertical details (such as the vertices of three or more edges and lines) and shows that it’s able to completely scale well. But in some instances, you’re interested in how some of these solutions look, and your objective might be to plot them. Is your goal to work with a background color? If its your goal to work on a structure, is making a drawing of multi-element shapes? And not just with a single container. The Art of Drawing Algorithms Of course, there many algorithmic approaches that you can use instead.
## Take My Online Math Class
There are many—or can be—theories you can use when you can draw some content: But in the end, you have to understand how simple examples make sense. As, maybe, most code examples are long, and not yet well understood. There are a couple of things to keep in mind. First of all, there are well-known, well-established frameworks for creating code in such a way. Take, for example, PHP. If you are keen on learning the PHP frameworks, it’s probably best to read at least one or two tutorials available on the web, most commonly in the context of PHP classes (like classes in C#, C#.NET, LINQ.NET ). If you are a developer not interested in PHP, you may want to read at least at one of the tutorials mentioned above. Here are some examples: You will have to learn how to use C#. PHP and C#. Although theWhat Kind Of Math Is On The Ged Test? I do indeed take it out of my math and am guessing my findings for some reason are odd as it seems it will take the same amount of time as most of the other tests related to math. I know that before the internet however most of what I receive is that there is a large amount of no knowledge about math. I have been using the same test I believe they are using actually since my kids were tiny with no advanced skills and the term “bluetooth” has expired. My kids when they started learning is too young to do a “serious math test”. How weird is it any easier to choose a particular test that has the main features being that you can take an almost everything on the graph as an example and then compare it to the “just another math class”, i.e. whatever you need. I do not honestly call it “mine testing” to be so obvious but I will take some time to make that a point. When I took the exam for just the “I have an E” they preferred the “I hate to code this stuff but I am not sure how to do so please vote for it too”.
## I Need Someone To Do My Online Classes
My kids were making it out of an e, where i got some different stuff by doing everything and looking at them. Not making sense, I would just remove it as i’m not there now. Usually my kids are saying “ok so because a lot of you are “like that” are there (just as you are) don’t you see that though? It’s more my kids think for the average math person around me though and I can appreciate it but they know just as clearly as me that other people don’t. I do not honestly call it “mine testing”. I honestly share on the “I hate to code this stuff but I am not sure how to do so please vote for it too” sign on is easy because most of my common sense students and others are the same way. I don’t want to ask as many people as possible about what you think, then make really sure your kids are different. It certainly took me quite a while to get to know this all by myself, but probably right now it’s one of the only thing you can judge that they will (for you) add. Finally I am sorry to say that I have no experience with test testing, but I do know in real life that in my career I have been told that you are not “mindful”. There’s also a lot of knowledge content you cannot make seem “fake”. There is a point where people may use it to “match out” but it is something new to learn. I’d imagine that you’ve learned something new. Let me know if you find out otherwise, I’ve had an opportunity to do so. Thanks for sharing some of your experiences on this. We really need to learn as much as we can 🙂 No offence, but your statement just comes across that you have a really obvious way to feel when reading someone else’s book. Only that how you feel is much more of a reason for me to avoid writing things. Oh my god!! – I’m a kid again now- I think I did actually do it) – I never signed up for the test. I just did not get to google the info (and I totally forgot that it was random lol) or if we’re taughtWhat Kind Of Math Is On The Ged Test? At this point, I’d like a big thanks for the latest Gedtests that come out in testing school, testing real math. I wrote them for a very small set of test questions in it. But, not everybody is having fun with them. Either way you can add all the fun things you like in the Gedtests (which is pretty much ALL fun in the world, by the way).
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Which is why if you could use Google Apps to test your test one, you will find that GedTest gives you lots of great things to keep it going at all times. The main thing is that some people who have not used one of these types are starting to throw out very specific crap. The rest of the test you can call “test,” or even you can pull back and up the number of days you have. I have covered what I think are the main points that you should care about… My test and more related ones. My real test and blog: Math (here: what you needed (at least first!)): Math Test-a.t. (this) is a test of how people are thinking about math. It is meant to explain things about number systems. There is an awesome group of tutorials where you can do math test-a.t. by walking, by hand, around mathematical objects and figure this question around yourself and all your problems. Although it may seem difficult, it requires a few very important skills such as you are required not to take one test at a time. Not able to do it! (yes, that’s right, one of the most painful aspects of math.) You have to say something with up-votes. And the result is always; the test here starts out with 60 points, but that’s the point! Math Test-b.t. (this) is the end goal of math these days. If you go back and now don’t know what Math is, you will quickly come to realize that you need an useful source of 60 points to be a good one. I (often) refer to Mutation Tackles, as any natural and practical way of testing numerology. It sets a lot of testing focus, almost zero or more.
## Take Online Courses For Me
The others are math tests, math calculators, math like numbers and even mathematical rules like y-axis(11). It is very important to know this. Although you can never know how quickly you are going to and from these new and even better ones, it simply means you need to be able to pay attention and follow a good set of guidelines before you run away with the results. Math and Psychology and Science and Chemistry and Physics and Math and Math! In most of these test, you will need the test and other skills to justify anything you put it in the above class. Plus, you don’t need to necessarily have any experience with those. I run this one anyway. http://www.youtube.com/watch?v=7D3LwUqo4L Now, in the third question, try to have a fun discussion about why all these Math aren’t useful, why they are but pointless, and why now is the time to have a test of its own. And so on! Ok, that’s that too. This was my top 5 place
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My Gedmani by the Town Voice. by Philip Shibley, Editor, Town Voice Town and Haruna | 2,337 | 10,103 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-38 | latest | en | 0.952978 |
https://learnbps.bismarckschools.org/mod/glossary/showentry.php?eid=48831&displayformat=dictionary | 1,720,883,398,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514510.7/warc/CC-MAIN-20240713150314-20240713180314-00010.warc.gz | 301,451,355 | 12,849 | #### MAT-01.DPS.D.02
1st Grade (MAT) Targeted Standard (DPS) Data Probability and Statistics (D) Data Learners will represent and interpret data.
## Progressions
Data Analysis
• MAT-01.DPS.D.02 Analyze data by answering descriptive questions.
• MAT-02.DPS.D.03 Analyze data and interpret the results to solve one-step comparison problems using information from the graphs.
• MAT-03.DPS.D.03 Analyze data and make simple statements to solve one- and two-step problems using information from the graphs.
• MAT-04.DPS.D.02 Generate data and create line plots to display a data set of fractions of a unit (1/2, 1/4, 1/8). Solve problems involving addition and subtraction of fractions by using information presented in line plots.
• MAT-04.DPS.D.03 Utilize graphs and diagrams to represent and solve word problems using the four operations involving whole numbers, benchmark fractions, and decimals.
• MAT-05.DPS.D.01 Generate data and create line plots to display a data set of fractions of a unit (1/2, 1/4, 1/8). Use grade-level operations for fractions to solve problems involving information presented in line plots.
• MAT-05.DPS.D.02 Utilize graphs and diagrams to represent, analyze, and solve authentic problems using information presented in one or more tables or line plots, including whole numbers, fractions, and decimals.
• MAT-06.DPS.DA.02 Calculate measures of center (median and mean) and variability (range and mean absolute deviation) to answer a statistical question. Identify mode(s) if they exist.
• MAT-06.DPS.DA.03 Identify outliers by observation and describe their effect on measures of center and variability. Justify which measures would be appropriate to answer a statistical question.
• MAT-06.DPS.DA.04 Display numerical data in plots on a number line, including dot plots and histograms. Describe any overall patterns in data, such as gaps, clusters, and skews.
• MAT-07.DPS.DA.02 Analyze and draw inferences about a population using single and multiple random samples by using given measures of center and variability for the numerical data set.
• MAT-08.DPS.DA.01 Interpret scatter plots for bivariate measurement data to investigate patterns such as clustering, outliers, positive or negative association, linear association, and nonlinear association.
• MAT-08.DPS.DA.02 Draw a trend line on a given scatter plot with a linear association and justify its fit by describing the closeness of the data points to the line.
• MAT-08.DPS.DA.03 Solve authentic problems in the context of bivariate measurement data by interpreting the slope and intercept(s) and making predictions using a linear model.
• MAT-08.DPS.DA.04 Construct and interpret a two-way table summarizing bivariate categorical data collected from the same subjects.
• MAT-10.DPS.02 Compare the center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets using statistics appropriate to the shape of the data distribution.
• MAT-10.DPS.03 Represent data on two quantitative variables on a scatter plot and describe how the variables are related.
• MAT-10.DPS.04 Distinguish between correlation and causation.
• MAT-10.DPS.10 Construct and interpret two-way frequency tables of data for two categorical variables. Use the two-way table as a sample space to decide if events are independent and approximate conditional probabilities.
• MAT-12.DPS.01 Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers).
• MAT-12.DPS.02 Use the mean and standard deviation of a data set to fit it to a normal distribution and estimate population percentages. Recognize that there are data sets for which such a procedure is not appropriate.
• MAT-12.DPS.03 Evaluate reports based on data.
• MAT-12.DPS.04 Represent data on a scatter plot for two quantitative variables and describe how the variables are related.
• MAT-12.DPS.05 Informally assess the fit of a function by plotting and analyzing residuals.
• MAT-12.DPS.06 Use data from a sample survey to estimate a population means or proportion; develop a margin of error through the use of simulation models for random sampling.
• MAT-12.DPS.07 Understand the process of making inferences about population parameters based on a random sample from that population.
• MAT-12.DPS.08 Decide if a specified model is consistent with results from a given data-generating process (e.g., using simulation). | 992 | 4,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-30 | latest | en | 0.824761 |
http://nrich.maths.org/public/leg.php?code=5001&cl=1&cldcmpid=8062 | 1,506,372,509,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818693363.77/warc/CC-MAIN-20170925201601-20170925221601-00694.warc.gz | 248,335,962 | 9,549 | # Search by Topic
#### Resources tagged with Odd and even numbers similar to Odd Times Even:
Filter by: Content type:
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### There are 61 results
Broad Topics > Numbers and the Number System > Odd and even numbers
### Two Numbers Under the Microscope
##### Stage: 1 Challenge Level:
This investigates one particular property of number by looking closely at an example of adding two odd numbers together.
### Odd Times Even
##### Stage: 1 Challenge Level:
This problem looks at how one example of your choice can show something about the general structure of multiplication.
### Always, Sometimes or Never? Number
##### Stage: 2 Challenge Level:
Are these statements always true, sometimes true or never true?
### Always, Sometimes or Never?
##### Stage: 1 and 2 Challenge Level:
Are these statements relating to odd and even numbers always true, sometimes true or never true?
### Pairs of Legs
##### Stage: 1 Challenge Level:
How many legs do each of these creatures have? How many pairs is that?
### Diagonal Trace
##### Stage: 2 Challenge Level:
You can trace over all of the diagonals of a pentagon without lifting your pencil and without going over any more than once. Can the same thing be done with a hexagon or with a heptagon?
### Crossings
##### Stage: 2 Challenge Level:
In this problem we are looking at sets of parallel sticks that cross each other. What is the least number of crossings you can make? And the greatest?
### Square Subtraction
##### Stage: 2 Challenge Level:
Look at what happens when you take a number, square it and subtract your answer. What kind of number do you get? Can you prove it?
### What Do You Need?
##### Stage: 2 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### What Number?
##### Stage: 1 Short Challenge Level:
I am less than 25. My ones digit is twice my tens digit. My digits add up to an even number.
### The Thousands Game
##### Stage: 2 Challenge Level:
Each child in Class 3 took four numbers out of the bag. Who had made the highest even number?
### Largest Even
##### Stage: 1 Challenge Level:
How would you create the largest possible two-digit even number from the digit I've given you and one of your choice?
### How Odd
##### Stage: 1 Challenge Level:
This problem challenges you to find out how many odd numbers there are between pairs of numbers. Can you find a pair of numbers that has four odds between them?
### Multiplication Series: Number Arrays
##### Stage: 1 and 2
This article for teachers describes how number arrays can be a useful reprentation for many number concepts.
### Becky's Number Plumber
##### Stage: 2 Challenge Level:
Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
### Number Round Up
##### Stage: 1 Challenge Level:
Arrange the numbers 1 to 6 in each set of circles below. The sum of each side of the triangle should equal the number in its centre.
### I Like ...
##### Stage: 1 Challenge Level:
Mr Gilderdale is playing a game with his class. What rule might he have chosen? How would you test your idea?
### The Set of Numbers
##### Stage: 1 Challenge Level:
Can you place the numbers from 1 to 10 in the grid?
### Three Spinners
##### Stage: 2 Challenge Level:
These red, yellow and blue spinners were each spun 45 times in total. Can you work out which numbers are on each spinner?
### Take One Example
##### Stage: 1 and 2
This article introduces the idea of generic proof for younger children and illustrates how one example can offer a proof of a general result through unpacking its underlying structure.
### Cube Bricks and Daisy Chains
##### Stage: 1 Challenge Level:
Daisy and Akram were making number patterns. Daisy was using beads that looked like flowers and Akram was using cube bricks. First they were counting in twos.
### Make 37
##### Stage: 2 and 3 Challenge Level:
Four bags contain a large number of 1s, 3s, 5s and 7s. Pick any ten numbers from the bags above so that their total is 37.
### Grouping Goodies
##### Stage: 1 Challenge Level:
Pat counts her sweets in different groups and both times she has some left over. How many sweets could she have had?
### Lots of Biscuits!
##### Stage: 1 Challenge Level:
Help share out the biscuits the children have made.
### More Numbers in the Ring
##### Stage: 1 Challenge Level:
If there are 3 squares in the ring, can you place three different numbers in them so that their differences are odd? Try with different numbers of squares around the ring. What do you notice?
### More Carroll Diagrams
##### Stage: 2 Challenge Level:
How have the numbers been placed in this Carroll diagram? Which labels would you put on each row and column?
### Light the Lights
##### Stage: 1 Challenge Level:
Investigate which numbers make these lights come on. What is the smallest number you can find that lights up all the lights?
### Carroll Diagrams
##### Stage: 1 Challenge Level:
Use the interactivities to fill in these Carroll diagrams. How do you know where to place the numbers?
### Magic Vs
##### Stage: 2 Challenge Level:
Can you put the numbers 1-5 in the V shape so that both 'arms' have the same total?
### Number Differences
##### Stage: 2 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
### Number Tracks
##### Stage: 2 Challenge Level:
Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
### Odds and Threes
##### Stage: 2 Challenge Level:
A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.
### Take Three Numbers
##### Stage: 2 Challenge Level:
What happens when you add three numbers together? Will your answer be odd or even? How do you know?
### Part the Piles
##### Stage: 2 Challenge Level:
Try to stop your opponent from being able to split the piles of counters into unequal numbers. Can you find a strategy?
### Sets of Four Numbers
##### Stage: 2 Challenge Level:
There are ten children in Becky's group. Can you find a set of numbers for each of them? Are there any other sets?
### What Could it Be?
##### Stage: 1 Challenge Level:
In this calculation, the box represents a missing digit. What could the digit be? What would the solution be in each case?
### Play to 37
##### Stage: 2 Challenge Level:
In this game for two players, the idea is to take it in turns to choose 1, 3, 5 or 7. The winner is the first to make the total 37.
### Down to Nothing
##### Stage: 2 Challenge Level:
A game for 2 or more people. Starting with 100, subratct a number from 1 to 9 from the total. You score for making an odd number, a number ending in 0 or a multiple of 6.
### Numbers as Shapes
##### Stage: 1 Challenge Level:
Use cubes to continue making the numbers from 7 to 20. Are they sticks, rectangles or squares?
### Break it Up!
##### Stage: 1 and 2 Challenge Level:
In how many different ways can you break up a stick of 7 interlocking cubes? Now try with a stick of 8 cubes and a stick of 6 cubes.
### Odd Tic Tac
##### Stage: 1 Challenge Level:
An odd version of tic tac toe
### One of Thirty-six
##### Stage: 1 Challenge Level:
Can you find the chosen number from the grid using the clues?
### Seven Flipped
##### Stage: 2 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
### Curious Number
##### Stage: 2 Challenge Level:
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
### Pairs of Numbers
##### Stage: 1 Challenge Level:
If you have ten counters numbered 1 to 10, how many can you put into pairs that add to 10? Which ones do you have to leave out? Why?
### Ring a Ring of Numbers
##### Stage: 1 Challenge Level:
Choose four of the numbers from 1 to 9 to put in the squares so that the differences between joined squares are odd.
### Red Even
##### Stage: 2 Challenge Level:
You have 4 red and 5 blue counters. How many ways can they be placed on a 3 by 3 grid so that all the rows columns and diagonals have an even number of red counters?
### Sets of Numbers
##### Stage: 2 Challenge Level:
How many different sets of numbers with at least four members can you find in the numbers in this box?
### Domino Sorting
##### Stage: 1 Challenge Level:
Try grouping the dominoes in the ways described. Are there any left over each time? Can you explain why?
### Venn Diagrams
##### Stage: 1 and 2 Challenge Level:
Use the interactivities to complete these Venn diagrams. | 2,094 | 9,044 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-39 | latest | en | 0.864033 |
https://www.physicsforums.com/threads/question-regarding-change-in-enthalpy.733399/ | 1,531,815,315,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589618.52/warc/CC-MAIN-20180717070721-20180717090721-00578.warc.gz | 954,914,559 | 15,265 | # Homework Help: Question regarding change in enthalpy
1. Jan 18, 2014
### Marshillboy
This isn't a formal homework question so much as a conceptual question for my own edification.
I'm reading my textbook's section on enthalpy and energy, and given the expression:
ΔH=nCpΔT
It states that, "we can use this expression to represent the change in enthalpy when n moles of an ideal gas are heated, regardless of any conditions on pressure or volume."
I know that the ideal gas law stats that PV = nRT, and thus T is proportional to PV.
How can it be, then, that enthalpy change is only affected by temperature change and not affected by changes in pressure and/or volume?
2. Jan 18, 2014
### Staff: Mentor
First, that equation for enthalpy holds for constant pressure. At const. P, the volume will change when T changes of course. But it is taken care of by considering only T because P is fixed.
What the book mentions are "conditions" on P and V. This means that you do not need to know what P is, or what the initial and final volumes are. So long as you know ΔT, you can calculate ΔH.
3. Jan 18, 2014
### Staff: Mentor
What they are saying it that the enthalpy of an ideal gas is independent of pressure. If we regard the enthalpy (per unit mass) of a pure substance to be a function of pressure and temperature, the we can write:
H = H(T,P)
From this it follows that:
$$dH=\frac{\partial H}{\partial T}dT+\frac{\partial H}{\partial P}dP$$
But, by definition,
$$C_p=\frac{\partial H}{\partial T}$$
so
$$dH=C_pdT+\frac{\partial H}{\partial P}dP$$
For real gases in the limit of low pressures, it has been found experimentally that:
$$\frac{\partial H}{\partial P}→0$$
But real gases approach ideal gas behavior in the limit of low pressures. So, for ideal gases, the enthalpy is independent of pressure.
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https://www.coursehero.com/file/6733473/285hw6/ | 1,490,465,768,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189031.88/warc/CC-MAIN-20170322212949-00623-ip-10-233-31-227.ec2.internal.warc.gz | 877,970,184 | 130,679 | # 285hw6 - HW 61 15 1. Sec.3.4:3. We have 15 = k 0:2 ) k =...
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HW 6 1 1. Sec.3.4:3. We have 15 = k & 0 : 2 ) k = 15 0 : 2 = 75 . x 0 = 0 and v 0 = ± 10 m= sec . Then we have the following problem: x 00 + ! 2 0 x = 0 ; ! 0 = r k m = r 75 3 = 5; x 0 = 0 ; v 0 = ± 10 . Then x ( t ) = A cos 5 t + B sin 5 t . x (0) = 0 = A . Hence x ( t ) = B sin 5 t . x 0 (0) = ± 10 ) ± 10 = 5 B ) B = ± 2 . Hence x ( t ) = ± 2 sin !t = 2 sin ( !t ± ) . Amplitude = 2 , period = 2 ! 0 = 2 5 sec; frequency = 5 rad= sec , or 1 perod = 5 2 = : 796 hz: 2. Sec.3.4:7. We are going to use the formula p 1 p 2 = R 1 p L 1 R 2 p L 2 : We have 1 = 3960 p 100 : 1 (3960 + r ) p 100 ; hence r = 1 : 9795 mi or r = 10 ; 450 ft. F = ma gives 2 hg ± 2 xg = 2 hx 00 ; where x = x ( t ) is the depth of the bottom of the buoy beneath the surface at time t . After simpli±cation we get x 00 + g ±h x = g . Besides we have initial conditions x (0) = x 0 (0) = 0 . Let us consider homogeneous 1 Note to the grader: please grade the following problems: Sec.3.4:7, Sec.3.4:10, Sec. 3.5: 17, Sec. 3.5: 49, Sec. 3.5: 55 (20pts for each problem). 1
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equation x 00 + g x = 0 ; then the characteristic equation is r 2 + g = 0 = ) r = r g i: We shall look for a particular solution of the non-homogeneous equation in the form x p = c since right-hand side is constant. Thus 0 + g c = g = ) c = The general solution of the given non-homogeneous equation will be x ( t ) = + c 1 cos r g t + c 2 sin r g t ; so we have x 0 ( t ) = ± c 1 r g sin r g t + c 2 r g cos r g t: Now we apply the initial conditions x (0) = + c 1 = 0 = ) c 1 = ± &h; x 0 (0) = c 2 r g = 0 = ) c 2 = 0 : Thus x ( t ) = ± cos q g t; that is x ( t ) = (1 ± cos ! 0 t ) ; where ! 0 = r g : Thus, the buoy undergoes simple harmonic motion about the equilib- rium position x e = with period p = 2 ± ! 0 = 2 ± s g : If = 0 : 5 g cm 3 ; h = 200 cm ; g = 980 cm s 3 then the amplitude of oscillation is = 100 cm and the period isgiven by p = 2 ± s g = 2 ± r 0 : 5 ² 200 980 ³ 2 : 01 s: 2
m = 1 2 ; c = 3 ; k = 4 ; x 0 = 2 ; v 0 = 0 . 1 2 x 00 + 3 x 0 + 4 x = 0 ; x (0) = 2 ; x 0 (0) = 0 : We have c 2 4 km = 9 8 > 0 , and this is an overdamped motion . 1 2 r 2 + 3 r + 4 = 0 ) r 2 + 6 r + 8 = 0 ) r = 2 ; r = 4 : General solution: x ( t ) = C 1 e 2 t + C 2 e 4 t . Using initial conditions, C 1 + C 2 = 2 2 C 1 4 C 2 = 0 ) C 1 = 4 ; C 2 = 2 : Answer: x ( t ) = 4 e 2 t 2 e 4 t 4 e 2 t 2 e 4 t 0 1 2 3 4 5 6 7 8 9 10 0.0 0.5 1.0 1.5 2.0 x y 5. Sec.3.4:17. Because c 2 = 4 km (8 2 = 4 ± 16 ± 1 = 64) we have critically damped case. x
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## This note was uploaded on 01/26/2012 for the course MATH 285 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.
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285hw6 - HW 61 15 1. Sec.3.4:3. We have 15 = k 0:2 ) k =...
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Ask a homework question - tutors are online | 1,370 | 3,277 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-13 | longest | en | 0.789075 |
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```GCSE Maths Starter 13
1. Write down the reciprocal of 3
2. Write 42 as a product of its prime factors
3. Find the perimeter and the shaded area of the
shape below.
4. Expand: 5a(3a+2)
5. 123 × 26
Lesson 13 Ratio
Mathswatch clip (61/94).
• To simplify a ratio (Grade D )
• To solve ratio problems using the unitary
I like simple
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tastes
Two
goes
into
both
4
But
the
cake
will
notin
be
They
can
be
simplified
Repeat
the
process
exactly
the
same
!! the
and
2
big
enough
!
same way as fractions are
simplified.
Flour
2
2
4
:
::
1 Sugar
21
I’m going to try a cake using flour and
But now my cake has used a ratio of 4 : 6 !!!
sugar in a ratio of 2 : 3
But
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But
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iscake
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Lesson 13 Ratio
Mathswatch clip (61/94).
1)
4)
7)
10)
1:2
4:8
2:3
6:9
1:4
2:8
1:3
7 : 21
Write the following ratios in their simplest
form.
2:3
1:3
2)
8 : 12
3)
4 : 12
1:2
4:5
5)
10 : 20
6)
8 : 10
7:3
3:2
8)
14 : 6
9)
30 : 20
4 : 11
3:2
11)
8 : 22
12)
9:6
Lesson 13 Ratio
Mathswatch clip (61/94).
1. Helen has 16 DVDs and 24 videos.
Write down the ratio of the number of DVDs to the number of videos.
Give your ratio in its simplest form.
2. There are 40 students on a school trip.
12 of the students are girls.
The rest are boys.
Find the ratio of the number of girls to the number of boys.
Give your ratio in its simplest form.
3. There are 50 sweets in a packet.
24 of the sweets are mints.
The rest of the sweets are toffees.
Write down the ratio of the number of mints to the number of toffees.
Give your ratio in its simplest form.
Lesson 13 Ratio
Mathswatch clip (61/94).
1. Helen has 16 DVDs and 24 videos.
Write down the ratio of the number of DVDs to the number of videos.
Give your ratio in its simplest form. 16:24 = 2:3
2. There are 40 students on a school trip.
12 of the students are girls.
The rest are boys.
Find the ratio of the number of girls to the number of boys.
Give your ratio in its simplest form. 12:28 = 3:7
3. There are 50 sweets in a packet.
24 of the sweets are mints.
The rest of the sweets are toffees.
Write down the ratio of the number of mints to the number of toffees.
Give your ratio in its simplest form. 24:26 = 12:13
Dividing in a given ratio
Divide £40 in the ratio 2 : 3.
A ratio is made up of parts.
We can write the ratio 2 : 3 as
2 parts : 3 parts
The total number of parts is
2 parts + 3 parts = 5 parts
We need to divide £40 by the total number of parts.
£40 ÷ 5 = £8
8 of 67
Dividing in a given ratio
Divide £40 in the ratio 2 : 3.
Each part is worth £8 so
2 parts = 2 × £8 = £16
and
3 parts = 3 × £8 = £24
£40 divided in the ratio 2 : 3 is
£16 : £24
Always check that the parts add up to the original amount.
£16 + £24 = £40
9 of 67
Dividing in a given ratio
A citrus twist cocktail contains orange juice, lemon juice
and lime juice in the ratio 6 : 3 : 1.
How much of each type of juice is contained
in 750 ml of the cocktail?
First, find the total number of parts in the ratio.
6 parts + 3 parts + 1 part = 10 parts
Next, divide 750 ml by the total number of parts.
750 ml ÷ 10 = 75 ml
10 of 67
Dividing in a given ratio
A citrus twist cocktail contains orange juice, lemon juice
and lime juice in the ratio 6 : 3 : 1.
How much of each type of juice is contained
in 750 ml of the cocktail?
Each part is worth 75 ml so,
6 parts of orange juice = 6 × 75 ml = 450 ml
3 parts of lemon juice = 3 × 75 ml = 225 ml
1 part of lime juice = 75 ml
Check that the parts add up to 750 ml.
450 ml + 225 ml + 75 ml = 750 ml
11 of 67
Lesson 13 Ratio
Mathswatch clip (61/94).
Share in the given ratios
1. Share £400 between Courtney and Adele in the ratio 2:3
2. Share £280 between April and Jack in the ratio 2:5
3. Share £500 between Molly and Toni in the ratio 3:7
Extension work
1. Share £240 between Brad and Emma and Louis in the ratio
3:5:12
2. Share £600 between Kera, Chloe and Raven in the ratio 1:5:6
3. Share £5 between Jake, Adele and Lucretia in the ratio 7:10:8
Lesson 13 Ratio
Mathswatch clip (61/94).
1. Courtney = £160 and Adele = £240
2. April = £80 and Jack = £200
3. Molly =£150 and Toni = £350
Extension work
Emma = £60
2. Kera = £50
Chloe = £250
3. Jake = £1.40 Adele = £2.00
£1.60
Louis = £144
Raven = £300
Lucretia =
Lesson 13 Ratio
Mathswatch clip (61/94).
Exam questions
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null | null | null | null | null | null | Edition: U.S. / Global
Stern but Steady on Panama
Published: May 10, 1989
The year's prize for brazenness goes to Gen. Manuel Antonio Noriega of Panama. With all the world watching, including two former U.S. Presidents, he turned a 3-to-1 victory for the opposition into a triumph for his own hand-picked presidential candidate, Carlos Duque. According to Jimmy Carter and surveys coordinated by the Roman Catholic Church, the winner by a landslide was the challenger, Guillermo Endara.
Plainly, Panama's arrogant godfather cares nothing for foreign opinion. Having survived the Reagan Administration's effort to destroy him with economic sanctions, General Noriega may believe he's invulnerable. Yet over time he assuredly cannot ignore what Panamanians think. They have done something courageous. The Bush Administration would repay them poorly by doing something dumb, like sending additional troops and talking about abrogating the Panama Canal treaties.
The first instinct for many in the United States would be to seek instant retaliation for General Noriega's insult to democracy. But unilateral sanctions haven't worked, and extralegal means to destroy his grip are likely to produce a new dictator, not a democratic Panama. For that reason it's hard to imagine a more harebrained idea than armed intervention and the abrogation of the treaties, as suggested by Senators Connie Mack and Bob Graham. This time, President Bush's wait-and-see caution seems right on target.
Any response has to take account of Panama's size and history. Panama's two million citizens live in two cities separated by jungles and joined by a waterway. The country was all but created in 1903 by the U.S., which wrested territorial rights to the canal from the tiny republic. Thus the treaty negotiated by Mr. Carter, giving Panama control of the canal by the year 2000, is seen by most Latin Americans as an act of restitution.
Treaties are made with countries, not regimes, and a crooked vote is not grounds for abrogation. Further, armed intervention would risk sabotage of the canal's intricate and priceless locks. That's one reason the Pentagon has long opposed military moves, which would also embroil the 10,000 U.S. troops stationed in Panama. Meanwhile, those who would encourage one or another of General Noriega's ambitious rivals forget that Washington once looked on the general himself as its client in Panama and helped consolidate his power.
Further economic sanctions are pointless. They have already caused great hardship to Panamanians without toppling their strongman. Indeed, it would make sense to offer to ease them as part of a regional strategy meant to help encourage a democratic challenge, in the Philippine mode. Like Ferdinand Marcos, General Noriega could well be undone by his contemptuous disregard for even the appearance of legitimacy.
Central Americans once habitually manipulated elections, and North Americans didn't much care. Uniformed caudillos repeatedly rigged ballots in Guatemala, El Salvador, Honduras and Nicaragua.
Starting with the leftist threat in Nicaragua a decade ago, however, Washington began to show real concern for democratic elections. North Americans can now credibly cry foul over General Noriega's theft and seek justice for his people. But Washington cannot achieve these laudable ends if it emulates his disdain for law and civilized opinion. | null | null | null | null | null | null | null | null | null |
https://gomathanswerkey.com/texas-go-math-grade-6-lesson-16-4-answer-key/ | 1,713,081,702,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816875.61/warc/CC-MAIN-20240414064633-20240414094633-00791.warc.gz | 258,147,416 | 56,506 | # Texas Go Math Grade 6 Lesson 16.4 Answer Key Solving Volume Equations
Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 16.4 Answer Key Solving Volume Equations.
## Texas Go Math Grade 6 Lesson 16.4 Answer Key Solving Volume Equations
Example 1
A rectangular swimming pool is 25 meters long and 17$$\frac{1}{2}$$ meters wide. It has an average depth of 1$$\frac{1}{2}$$ meters. What is the volume of the pool?
Label the rectangular prism to represent the pool.
l = 25 meters; w = 17$$\frac{1}{2}$$ meters; h = 1$$\frac{1}{2}$$ meters
Use the formula to write an equation.
v = lwh
Texas Go Math Grade 6 Lesson 16.4 Question 1.
Miguel has a toolbox that measures 18$$\frac{1}{2}$$ inches by 12$$\frac{1}{2}$$ inches by 4 inches. What is the volume of the toolbox?
V = _________ cubic inches
Determine the volume of the toolbox.
V = 18$$\frac{1}{2}$$ ∙ 12$$\frac{1}{2}$$ ∙ 4 substitute for the given values
V = $$\frac{37}{2}$$ ∙ $$\frac{25}{2}$$ ∙ 4 write mixed numbers as fractions
V = $$\frac{3,700}{4}$$ simplify
V = 925 cubic inches volume of the toolbox
The volume of the toolbox is 925 cubic inches
Example 2
Samuel has an ant farm with a volume of 375 cubic inches. The width of the ant farm is 2.5 inches and the length is 15 inches. What is the height of Samuel’s ant farm?
The height of the ant farm is 10 inches.
Reflect
Question 2.
Communicate Mathematical Ideas Explain how you would find the solution to Example 2 using the formula V = 8h.
Here B = l × w can be evaluated and then the given volume can be divided by B to evaluate h.
Question 3.
Find the height of this shape, which has a volume of $$\frac{15}{16}$$ cubic feet.
Data:
l = $$\frac{3}{4}$$ = 0.75
w = $$\frac{1}{2}$$ = 0.5
Volume = $$\frac{15}{16}$$ = 0.9375
Write equation of volume of a rectangular prism:
VoLume = l × w × h
Solve for h:
h = $$\frac{\text { Volume }}{l \times w}$$
Substitute values:
h = $$\frac{0.9375}{0.75 \times 0.5}$$
Evaluate:
h = 2.5
Height of the given rectangular prism is equal to 2.5 = 2$$\frac{1}{2}$$ feet.
Example 3
The classroom aquarium holds 30 gallons of water. It is 0.8 feet wide and has a height of 2 feet. Find the length of the aquarium.
Reflect
Persevere in Problem Solving Flow much does the water in the classroom aquarium weigh? Explain.
The volume of the classroom aquarium is 4 cubic feet and the equivalent of 1 cubic foot is approximately 62.43 pounds. Therefore, multiply the volume by the estimated weight.
62.43 × 4 = 249.72 pounds
The water in the aquarium weighs about 249.72 pounds.
Question 5.
An aquarium holds 33.75 gallons of water. It has a length of 2 feet and a height of 1.5 feet. What is the volume of the aquarium? What is the width of the aquarium? Explain.
Determine the volume and width of the aquarium.
V = $$\frac{33.75}{7.5}$$ volume of the aquarium in cubic feet
V = 4.5 cubic feet
V = lwh formula for the volume of the aquarium
4.5 = 2 ∙ w ∙ 1.5 substitute for the given values
4.5 = 3w simplify
$$\frac{4.5}{3}=\frac{3 w}{3}$$ divide both sides of the equation by 3
1.5 = w width of the aquarium
The volume of the aquarium is 4.5 cubic feet and the width is 1.5 feet.
Question 1.
Find the volume of this rectangular prism. (Example 1)
The volume of the rectangular prism is _________ cubic inches.
Determine the volume.
The volume of the rectangular prism is 15$$\frac{3}{4}$$ cubic inches.
Question 2.
Write an equation to find the width of the rectangular prism. Show your work. (Example 2)
Determine the width of the rectangular prism
V = lwh formula for the volume of the rectangular prism
w = $$\frac{V}{l h}$$ equation for the width of the prism
w = $$\frac{6,336 \mathrm{~cm}^{3}}{16 \mathrm{~cm} \cdot 18 \mathrm{~cm}}$$ substitute for the given values
w = $$\frac{6,336 \mathrm{~cm}^{3}}{288 \mathrm{~cm}^{2}}$$ simplify
w = 22 cm width of the rectangular prism
The width of the rectangular prism is 22 cm.
One red clay brick weighs 5.76 pounds. The brick is 8 inches long and 2 inches wide. If the clay weighs 0.08 pounds per cubic inch, what is the volume of the brick? Write an equation to find the height of the brick. Show your work. (Example 3)
Determine the volume and height of the brick.
V = $$\frac{5.76}{0.08}$$ volume of the brick in cubic inches
V = 72 cubic inches
V = lwh formula for the volume of the brick
72 = 8 ∙ 2.25 ∙ h substitute for the given values
72 = 18h simplify
$$\frac{72}{18}=\frac{18 h}{18}$$ divide both sides of the equation by 18
4 inches = h height of the brick
The bricks is 4 inches in height.
Essential Question Check-In
Question 4.
How do you solve problems about volume of right rectangular prisms?
Solving problems involving the volume of right rectangular prisms, identify the length, width, and height of the rectangular prism then get its product.
Question 5.
Jala has an aquarium in the shape of a rectangular prism with the dimensions shown. What is the height of the aquarium?
Height = ___________
Solution to this example is given beLow
l = 24.25 inches; w = 12.5 inches; h = ? ; V = 3758.75 in3
V = lwh Write the formula.
3758.75 = 24.25 × 12.5 × h Use the formula to write an equation
3758.75 = (303.125)h Multiply
$$\frac{3758.75}{303.125}=\frac{303.125 h}{303.125}$$ Divide both sides of the equation by 303125.
12.4 = h
Height of the given aquarium is 12.4 inches
Question 6.
Find the volume of a juice box that is 3 in. by 1$$\frac{1}{2}$$ in. by 4 in.
Volume = ______________
Determine the voLume of the juice box.
V = 3 ∙ 1.5 ∙ 4 substitute for the given values
V = 18 in3 volume of the juice box
The volume of the juice box is 18 cubic inches.
Question 7.
Find the width of a cereal box that has a volume of 3,600 cm3 and is 20 cm long and 30 cm high.
Width = ____________
Determine the width of the box.
V = lwh formula for the volume of box
3, 600 = 20 ∙ w ∙ 30 substitute for the given values
3,600 = 600w simplify
$$\frac{3,600}{600}=\frac{600 w}{600}$$ divide both sides of the equation by 600
6 cm = w width of the cereal box
The width of the cereal box is 6 cm.
Question 8.
Bill has a box of markers that has a base of 8 cm by 20 cm and a height of 6 cm. Martin’s pencil box has a height of 4 cm and a base that is 15 cm by 16 cm. Bill says his marker box has the same volume as Martin’s pencil box. Is Bill right? Explain.
Determine the volume of both objects then compare.
V = 8 ∙ 20 ∙ 6 substitute for the dimensions of the box of markers
V = 960 cm3 volume of the box of markers
V = 4 ∙ 15 ∙ 16 substitute for the dimensions of the pencil box
V = 960 cm3 volume of the pencil box
Yes, Bill is right with his conjecture because the volume of the box of marker and pencil box are both the same.
Question 9.
Physical Science A small bar of gold measures 40 mm by 25 mm by 2 mm. One cubic millimeter of gold weighs about 0.0005 ounces. Find the volume in cubic millimeters and the weight in ounces of this small bar of gold.
Data:
Density = 0.0005
l = 40
w = 25
h = 2
Write equation of volume of a rectangular prism:
VoLume = l × w × h
Substitute vaLues:
Volume = 40 × 25 × 2
Evaluate:
Volume = 2000
Volume of the bLock is 2000 cubic millimeters.
Write equation of density of a substance:
Density = $$\frac{\text { mass }}{\text { Volume }}$$
Solve for mass:
mass = Density × Volume
Substitute values:
mass = 0.0005 × 2000
Evaluate:
mass = 1
Mass of this block is 1 ounce.
History The average stone on the lowest level of the Great Pyramid in Egypt was a rectangular prism 5 feet long by 5 feet high by 6 feet deep and weighed 15 tons. What was the volume of the average stone? How much did one cubic foot of this stone weigh?
Data:
mass = 15
l = 5
w = 5
h = 6
Write equation of volume of a rectangular prism:
Volume = l × w × h
Substitute values:
VoLume = 5 × 5 × 6
Evaluate:
Volume = 150
Volume of the block is 150 cubic feet.
Write equation of density of a substance:
Density = $$\frac{\text { mass }}{\text { Volume }}$$
Substitute values:
Density = $$\frac{15}{150}$$
Evaluate:
Density = 0.1
Density of this block is 0.1 ton per cubic feet.
Question 11.
A freshwater fish is healthiest when there is at least one gallon of water for every inch of its body length. Roshel wants to put a goldfish that is about 2$$\frac{1}{2}$$ inches long in her tank. Roshel’s tank is 7 inches long, 5 inches wide, and 7 inches high. The volume of 1 gallon of water is about 231 cubic inches.
a. How many gallons of water would Roshel need for the fish?
a. According to the given information, the amount of water needed by the goldfish is 2$$\frac{1}{2}$$ × 1 = 2$$\frac{1}{2}$$ = 2.5 gallons of water. This is equal to 2.5 × 231 = 577.5 cubic inches.
b. What is the volume of Roshel’s tank?
Volume of the tank is 7 × 5 × 7 = 245 cubic inches.
c. Is her fish tank large enough for the fish? Explain.
Since the volume of water required is 577.5 > 245, her tank is not large enough to store the amount of water required by the fish.
Question 12.
A box of crackers is a rectangular box with the dimensions shown. The box is one-fourth full. What is the volume of crackers in the box? _______________________________
Determine the volume of the crackers in the box.
V = 8 ∙ 4 ∙ 4 substitute for the given values
V = 128 in3 volume of the box
V = $$\frac{128}{4}$$ divide the volume by 4
V = 32 in3 volume of the crackers in the box
The volume of the crackers in the box is 32 cubic inches.
H.O.T. Focus on Higher Order Thinking
Question 13.
Multistep Larry has a clay brick that is 7 inches long, 3.5 inches wide, and 1.75 inches thick, the same size as the gold stored in Ft. Knox in the form of gold bars. Find the volume of this brick. If the weight of the clay in the brick is 0.1 pound per cubic inch and the weight of the gold is 0.7 pounds per cubic inch, find the weight of the brick and the gold bar. Round all answers the nearest tenth.
Volume of the brick or bar = __________ cubic inches
Weight of the brick = __________ pounds
Weight of the gold bar = __________ pounds
VoLume of the brick or bar is 7 × 3.5 × 1.75 = 42.875 cubic inches.
Weight of the brick in pounds is the product of its given density and its volume, therefore its weight is 42.875 × 0.1 = 4.2875 pounds.
Weight of the bar in pounds is the product of its given density and its voLume, therefore its weight is 42.875 × 0.7 = 30.0125 pounds.
Question 14.
Represent Real-World Problems Luisa’s toaster oven, which is in the shape of a rectangular prism, has a base that is 55 cm long by 40 cm wide. It is 30 cm high. Luisa wants to buy a different oven with the same volume but a smaller length, so it will fit better on her kitchen counter. What is a possible set of dimensions for this different oven?
VoLume of this toaster is 55 × 40 × 30 = 66000 cubic centimeters.
The volume of the new toaster must be same but with different dimensions provided that l < 55 centimeters, therefore:
Volume = 66000 = 48 × 42 × 32.74
Note that the length of the new toaster is less than 55 but to maintain the same volume, the width and height of the toaster has increased.
Multiple Representations Use the formula V = Bh to write a different version of this formula that you could use to find the area of the base B of a rectangular prism if you know the height h and the volume V. Explain what you did to find this equation.
Write equation of volume of a rectangular prism in terms of base Area B:
Volume = B × h
Divide both sides of this equation with h to solve for B:
This is the equation of the same formula for B.
Question 16.
Communicate Mathematical Ideas Explain how you could find the volume of a cube that has an edge of e.
Determine the volume of cube.
V = e ∙ e ∙ e multiply the edge of the cube by itself three times
V = e3 volume of the cube
Tne volume of the cube is e3.
Question 17.
Justify Reasoning Mariel says that a jewelry box that is 3 inches high, 4$$\frac{1}{2}$$ inches wide, and 5 inches long has a volume of 67$$\frac{1}{2}$$ inches. Katy says that answer is not quite correct. What is the error in Mariel’s answer? | 3,568 | 12,149 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-18 | latest | en | 0.804504 |
http://slideplayer.com/slide/3793173/ | 1,521,459,724,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257646875.28/warc/CC-MAIN-20180319101207-20180319121207-00665.warc.gz | 261,662,368 | 18,441 | # Study Guide. a. 1 of 2 numbers multiplied together to get a product b. The answer to a division problem c. The answer to a multiplication problem d. None.
## Presentation on theme: "Study Guide. a. 1 of 2 numbers multiplied together to get a product b. The answer to a division problem c. The answer to a multiplication problem d. None."— Presentation transcript:
Study Guide
a. 1 of 2 numbers multiplied together to get a product b. The answer to a division problem c. The answer to a multiplication problem d. None of the above
a) A multiple of 2 b) 2, 4, 6, 8, 10, 12, … c) Divisible by 2 d) All of the above
a) Please Excuse My Dear Aunt Sally b) Parenthesis, Exponents, Multiplication and Division, Addition and Subtraction c) PEMDAS d) Both B and C
a) The product of two whole numbers b) Numbers found by skip counting c) 2, 4, 6, 8, 10, … are multiples of 2 d) All of the above
a) Only even numbers b) A number that only has two factors, 1 and itself c) A number that has many factors d) All numbers found on the number line
a) All the factors of a number b) All the factors of a number expect the number itself c) The answer to a multiplication problem d) None of the above
a) A conversation you have with a friend b) 1 of 2 numbers multiplied together to get a product c) A mathematical guess based on observations d) This is not a vocabulary word we have had
a) The answer to an addition problem b) The answer to a multiplication problem c) The answer to a subtraction problem d) The answer to a division problem
a) 6 and 7 is a factor pair for 42 b) 2 numbers that when multiplied together get a specific product c) 1 of 2 numbers multiplied together to get a product d) Both A and B
a) All numbers found on the number line b) A number that has many factors c) Only even numbers d) A number that only has two factors, 1 and itself
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# Answer to Question #11929 in Algebra for Elise Ruby
Question #11929
As part of a flight training, a student pilot was required to fly to an airport and then return. the average speed on the way to the airport was 125 mph, and the average speed returning was 218.75 mph. Find the distance between the two airports if the total flying time was 8.25 hours.
d = 125 * t1 = 218.75 * t2
t1 + t2 = 8.25, so t1 = 8.25 - t2. Getting:
125
* (8.25 - t2) = 218.75 * t2, 1031.25 - 125 * t2 = 218.75 * t2,
(218.75 + 125)
* t2 = 1031.25, t2 = 3 h
d = 218.75 * t2 = 218.75 * 3 = 656.25 miles
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null | null | null | null | null | null | Emphatically no!
Son: Mum (who's 93), would you have a pig's valve put in your heart?
Mum: No fear!
Beküldő: pseldomymous 2010. június 25.
- Something you say to sound like a hard-ass before doing something that your terrified of
Jim( about to jump into the shark pool): Oh shit, oh shit, oh shit...ok, gotta calm down. No fear!
Jim: (jumps in, dies.)
Beküldő: Ohsnap!!!! 2009. december 11.
A member of the Billingham Bede College social area. He is a wild animal, untamed by man. Not many have seen his provocotive and unaturual state of living. His habits include, dancing to such classics as "Cotton Eye Joe" aswell as chasing Richard around the social area. He is often partnered up with "Matrix" and "FP" (floating ponce) as they prove themselves to be an absolute bunch of sick cunts.
"NoFear, stop fucking dancing you poncy cunt"
"shit, no fear and FP are coming over, fuck off nf, sick cunt"
"nf stop fucking shagging matrix you big tit."
"aye, sound nf, buy us a litre, daft cunt"
Beküldő: Michael Crooks 2007. március 9.
(n). the state of being afraid of nothing. Also, being marcus allen.
"Yo, No Fear marcus, your a peice of gremlin shit"
Beküldő: Schulte 2005. december 19.
He that shall soon pwn the ruling Doggus!
No Fear is a wonderful fella
Beküldő: Urmom 2003. június 15.
Acronym for the National Organization For the Education of the Aryan Race.
Employed by a clothing brand marketed to white, middle-class, suburban youth. Few are aware that a significant portion of their profits go to grassroots neo-Nazi organizations.
Unbeknownst to many, the company behind the NO FEAR line of textiles is adamantly racist.
Beküldő: anonymous 2004. november 20.
Pr0n King
I am No Fear, I am the king of pr0n!
Beküldő: SpearsHairyLover 2003. augusztus 4.
Ingyenes Napi Email
| null | null | null | null | null | null | null | null | null |
https://www.gradesaver.com/textbooks/math/other-math/thinking-mathematically-6th-edition/chapter-5-number-theory-and-the-real-number-system-5-4-the-irrational-numbers-exercise-set-5-4-page-298/93 | 1,576,244,264,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540555616.2/warc/CC-MAIN-20191213122716-20191213150716-00465.warc.gz | 703,662,197 | 13,102 | ## Thinking Mathematically (6th Edition)
Published by Pearson
# Chapter 5 - Number Theory and the Real Number System - 5.4 The Irrational Numbers - Exercise Set 5.4 - Page 298: 93
#### Answer
I would like to stay on earth because I want to live with my family and friends because when space traveler would come back to an unknown futuristic world the friends and loved ones would be long gone.
#### Work Step by Step
After a two-year journey to a futuristic world in which loved ones and friends are long gone. Einstein’s special relativity equation is as follows: ${{R}_{a}}={{R}_{f}}\sqrt{1-{{\left( \frac{v}{c} \right)}^{2}}}$ Here, ${{R}_{a}}$ represents rate by which the age of traveler increases,${{R}_{f}}$ represents the relative aging rate of a friend on earth, v is the speed of astronauts, and c is the speed of light. The equation described that the time would pass speedily on the Earth than time would pass in space. As v approaches the speed of light,then \begin{align} & {{R}_{a}}={{R}_{f}}\sqrt{1-{{\left( \frac{c}{c} \right)}^{2}}} \\ & ={{R}_{f}}\sqrt{1-{{\left( 1 \right)}^{2}}} \\ & ={{R}_{f}}\sqrt{0} \\ & =0 \end{align} Thus, nearby to the speed of light, the astronaut’s aging rate relative to a friend on Earth is zero. This implies when the people on Earth would be getting older the space traveler would barely get older. When space traveler would come back to an unknown futuristic world the friends and loved ones would be long gone. Hence, I would like to stay on earth because I want to live with my family and friends.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 444 | 1,718 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-51 | latest | en | 0.912255 |
http://mathhelpforum.com/math-puzzles/152497-number-theory-easy-proof.html | 1,475,160,109,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661833.70/warc/CC-MAIN-20160924173741-00057-ip-10-143-35-109.ec2.internal.warc.gz | 163,788,018 | 12,897 | Thread: Number Theory , an easy proof .
1. Number Theory , an easy proof .
Let's first recall some Pythagorean Triples :
$(3,4,5) ~,~ (5,12,13) ~,~ (8,15,17) ~,~ (9,40,41) ~,~ (11,60,61)$ etc .
Can you see a pattern ? In every triple , we must find one integer from $(x,y,z)$ , such that it is the mutiple of $5$ .
$5,5,15,40,60$ etc .
I would like you to prove this fact , the proof shouldn't be so long , i believe ...
2. Let (a,b,c) be a pythagorean triple. Observe that $x^2 \equiv 0,1,4$ (mod) 5.
Suppose neither $a,b,c\equiv 0$ (mod) 5. Then $a^2,b^2,c^2\equiv 1,4$ (mod) 5
But then we can never have $a^2+b^2\equiv c^2$ (mod) 5
3. Originally Posted by simplependulum
Let's first recall some Pythagorean Triples :
$(3,4,5) ~,~ (5,12,13) ~,~ (8,15,17) ~,~ (9,40,41) ~,~ (11,60,61)$ etc .
Can you see a pattern ? In every triple , we must find one integer from $(x,y,z)$ , such that it is the mutiple of $5$ .
$5,5,15,40,60$ etc .
I would like you to prove this fact , the proof shouldn't be so long , i believe ...
The integers $m^2-n^2,\ 2mn,\ m^2+n^2$ form a Pythagorean triple.
Consider modulo 5.
If $m\equiv 0\ (mod\ 5)$ or $n\equiv 0\ (mod\ 5)$, then $2mn\equiv 0\ (mod\ 5)$.
If $m\equiv\pm n\ (mod\ 5)$, then $m^2-n^2\equiv 0\ (mod\ 5)$.
If $m\equiv\pm2\ (mod\ 5)$ and $n\equiv\pm1\ (mod\ 5)$ or $m\equiv\pm1\ (mod\ 5)$ and $n\equiv\pm2\ (mod\ 5)$, then $m^2+n^2\equiv 0\ (mod\ 5)$.
This concludes all the cases.
4. Hello,
The proof is trivial. Let $x \equiv a \pmod{5}$, $y \equiv b \pmod{5}$, $z \equiv c \pmod{5}$. Now consider a Pythagorean triple :
$x^2 + y^2 = z^2$
Since all terms $a$, $b$ and $c$ are modulo $5$, we may equate and say that :
$a^2 + b^2 \equiv c^2 \pmod{5}$
According to the conjecture, this congruence may only be satisfied if and only if either $a, b, c \equiv 0 \pmod{5}$ (if one of these integers is divisible by $5$). Recall the quadratic residues modulo $5$ :
$0^2 \equiv 0 \pmod{5}$
$1^2 \equiv 1 \pmod{5}$
$2^2 \equiv 4 \pmod{5}$
$3^2 \equiv 4 \pmod{5}$
$4^2 \equiv 1 \pmod{5}$
Now consider all the possibilities (I omitted those that didn't satisfy the conjecture because there's too much of them, this has been double-checked by computer) :
$0^2 + 0^2 = 0^2 \pmod{5}$
$0^2 + 1^2 = 1^2 \pmod{5}$
$0^2 + 1^2 = 4^2 \pmod{5}$
$0^2 + 2^2 = 2^2 \pmod{5}$
$0^2 + 2^2 = 3^2 \pmod{5}$
$0^2 + 3^2 = 2^2 \pmod{5}$
$0^2 + 3^2 = 3^2 \pmod{5}$
$0^2 + 4^2 = 1^2 \pmod{5}$
$0^2 + 4^2 = 4^2 \pmod{5}$
$1^2 + 0^2 = 1^2 \pmod{5}$
$1^2 + 0^2 = 4^2 \pmod{5}$
$1^2 + 2^2 = 0^2 \pmod{5}$
$1^2 + 3^2 = 0^2 \pmod{5}$
$2^2 + 0^2 = 2^2 \pmod{5}$
$2^2 + 0^2 = 3^2 \pmod{5}$
$2^2 + 4^2 = 0^2 \pmod{5}$
$3^2 + 0^2 = 2^2 \pmod{5}$
$3^2 + 0^2 = 3^2 \pmod{5}$
$3^2 + 1^2 = 0^2 \pmod{5}$
$3^2 + 4^2 = 0^2 \pmod{5}$
$4^2 + 0^2 = 1^2 \pmod{5}$
$4^2 + 0^2 = 4^2 \pmod{5}$
$4^2 + 2^2 = 0^2 \pmod{5}$
$4^2 + 3^2 = 0^2 \pmod{5}$
$2^2 + 1^2 = 0^2 \pmod{5}$
By analyzing the triples that do satisfy the conjecture, we find that all those triples have one, and only one, integer that is a multiple of $5$ ... except the first one. However, apart from $0^2 + 0^2 = 0^2$ which is not generally considered a Pythagorean triple, it is trivial to show that the sum of two squares that are divisible by $25$ cannot possibly be divisible by $25$ (the proof is left as an exercise to the reader ).
Therefore the conjecture is verified and valid.
$\mathfrak{QED}$
(okay, I admit it, that wasn't exactly beautiful maths, but it's 4 am over here, time for sleep oO)
5. Any integer is congruent to $0,\pm 1,\pm 2$ modulo 5 (complete residue system). So any square is congruent modulo 5 to $0,\pm 1$.
Assume that both $a^2$ and $b^2$ are not congruent modulo 5 to $0$, then $a^2+b^2\equiv 0,2,$ or $-2(mod\ 5)$.
Since $c^2=a^2+b^2$ and $c^2$ is a square, we must have $c^2\equiv 0(mod\ 5)$.
6. Thanks , indeed , all the solutions here are not so long . I want to include my proof in this post :
Suppose all the primitive solutions of the equation $x^2 + y^2 = z^2$ can be represented by $(m^2 - n^2 , 2mn , m^2 + n^2 )$ . $x$ is odd while $y$ is even , $m , n$ are coprime and one odd one even . (It is true but the proof not included here . )
Then take modulo $5$ , we have
$m^2 - n^2 = (m-n)(m+n)$
$m^2 + n^2 \equiv m^2 - 4n^2 = (m-2n)(m+2n)$
consider the product $xyz$ , assume $n$ is prime to $5$ , otherwise , $y$ is divisible by $5$ .
We have $xyz \equiv 2n(m-2n)(m-n)(m)(m+n)(m+2n) \equiv 2n^5 (s-2)(s-1)(s)(s+1)(s+2) \bmod{5}$ where $s \equiv m(n^{-1}) \bmod{5}$ . We can see that the product $(s-2)(s-1)(s)(s+1)(s+2)$ is divisible by $5$ so is $xyz$ , this finishes the proof . | 1,995 | 4,614 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 98, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2016-40 | longest | en | 0.667933 |
null | null | null | null | null | null | Tom Turpin
Professor of
Purdue University
Ladybug, Ladybug Get Out of My Home
Back in the 1920's, a cockroach poet named archie wrote a poem called "the froward ladybug." As it turned out, the ladybug that had become a problem for archie was not the insect most of us know as a ladybug. The ladybug of archie was a female insect, a lady bug.
Ladybugs are some of the most recognizable insects. They are beetles with hard shells or wing covers. Most are colored bright orange or red. Many have dots on their wing covers, and most are considered beneficial insects because they are well-known predators. They feed on aphids, plant-eating insects hated by most farmers and gardeners. And in spite of their name not all ladybugs are ladies. Many are gentleman bugs!
But even an insect with the name of ladybug that gets rid of insects we don't like can sometimes wear out its welcome. In the case of ladybugs, that occurs when they move into our homes during the winter months. When they do, most homeowners take a dismal view of the activity and like the cockroach archie, they acquire a dislike for the ladybug.
During the winter, ladybugs in the adult or beetle stage seek protected sites in which to hibernate. Such places include under rocks, piles of leaves, in window wells and under the eaves of houses. Most ladybugs, however, appear not to be content to stay in protected places outside. Instead, they manage to crawl into our attics or wall spaces. From there they find their way into the interior of our warm homes and crawl over cabinets, carpets and light fixtures.
Wintertime ladybug invaders vary in color, size and in number of black spots. These differences lead many folks to conclude that several ladybug species are in their homes. That may or may not be true.
There are several hundred species of ladybugs, many of which are found in homes at one time or another. However, home-invading ladybugs usually are the species known as the Asian ladybug. Asian ladybugs vary widely in color and markings even though they are of the same species.
Ladybugs do not cause damage in our homes so their presence is a nuisance rather than a real problem. They will not eat our houseplants or damage our clothing. They do not often bite, although they might if held tightly.
They do have a rather nasty characteristic however. They sometimes smell bad. Not all the time, just sometimes. The stink comes from the scientific habit known as reflex bleeding. When the insect is disturbed, it can expel a smelly, yellowish liquid from the area where the leg joins the body. The insect uses the liquid to ward off predators because it tastes as bad as it smells. So, when we touch a ladybug it sometimes responds as if we were a predator and squirts us with the ladybug equivalent of mace.
When that happens, we, like archie the cockroach, could describe the insect as froward, or--as Webster defines the word--obstinately willful!
Writer: Tom Turpin
Editor: Andrea McCann | null | null | null | null | null | null | null | null | null |
null | null | null | null | null | null | Kyle Whittingham: '(UCLA) really has no weaknesses right now' (+video/audio)
Full transcript of Utah football's Monday press conference
Published: Monday, Sept. 30 2013 3:10 p.m. MDT
Kyle Whittingham press conference, 9/30/13, prior to UCLA
SALT LAKE CITY — Utah football head coach Kyle Whittingham addressed the media prior to Thursday night's game against the No. 12 UCLA Bruins.
On mobile? Watch the press conference on YouTube.
Don't have time to watch the video? Here's the audio.
This is a full transcription of today's press conference:
Kyle Whittingham (KW): OK. I thought we made the most of our bye week. We got a lot done, got the players some rest, got some bumps and bruises hopefully healed up, so we're back today. Obviously a short week this week with the Thursday game.
UCLA, as I mentioned last week, is a tremendous football team, ranked 12th and 13th in the country in the respective polls. Really, they have no weaknesses right now. Offensively, they're leading the conference in total offense. They may be leading the nation in third downs — I haven't checked. They're leading the Pac-12 for certain at 68 percent and might be leading the nation in that category.
Bottom line is that it's a big challenge for us, but we'll be ready to play and we're looking forward to the game on Thursday night.
Reporter: Coach, looking at UCLA's offense and its ability to make explosive plays on offense, how important will ball control be for your team?
KW: That's always important for our team. Our offense has done a nice job of protecting the football and not turning it over this year. We've been very good with ball security, and that will be a premium this week, obviously. And then making first downs. …
In order to make first downs, we've got to be much better than we were on third down in the previous game. Third down was the biggest negative to come out of the rivalry game — our lack of third-down production. That's got to improve, and if that improves, then we'll move the ball better by making first downs and controlling the clock a little bit.
Reporter: Coach, it was just a few years ago that UCLA was pretty much an also-ran and they were kind of going nowhere as a program. Then they hired coach Mora, and he turned things around really quickly. Is this kind of a testament to how the right hire can come in and make a team that hasn't been so good really good in a short amount of time?
KW: I think that's part of it. Coach Mora's a heck of a football coach. He's got a great background, and he's done a lot of good things at a lot of different places, but my no means was the cupboard bare when he got there. Coach Neuheisel had done a great job recruiting there, and they had talented players.
I think it's a combination of a great deal of talent that he inherited but taking that talent and maximizing it. I guess the answer to your question is, "Yes," on one front, but there's more to it than just that.
Reporter: What is it that their quarterback, Hundley, does that makes him such a great weapon?
KW: Well, he does everything well. He's a lot like our guy, Travis (Wilson). Their numbers are almost exact, when you compare the two. They're within a point or two of each other in pass efficiency. Hundley runs the ball well like Travis does. There's a really good comparison between those two guys, and it's going to be an intriguing matchup. The quarterbacks are going against each other head-to-head Thursday night. What Travis brings to the table and makes him difficult to defend, Hundley does the same thing for the Bruins.
Reporter: In general, what are your thoughts on these very early week Thursday games, and specifically with this one, how fortuitous is it that you're coming off a bye to prepare for it?
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# Find the derivative of $\frac{x^3+x-1}{x^4+6x^2+9}$
Go!
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## Final answer to the problem
$\frac{\left(3x^{2}+1\right)\left(x^4+6x^2+9\right)+\left(-x^3-x+1\right)\left(4x^{3}+12x\right)}{\left(x^4+6x^2+9\right)^2}$
Got another answer? Verify it here!
## Step-by-step Solution
How should I solve this problem?
• Find the derivative
• Find the derivative using the definition
• Find the derivative using the product rule
• Find the derivative using the quotient rule
• Find the derivative using logarithmic differentiation
• Find the derivative
• Integrate by partial fractions
• Product of Binomials with Common Term
• FOIL Method
• Integrate by substitution
Can't find a method? Tell us so we can add it.
1
Apply the quotient rule for differentiation, which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$
$\frac{\frac{d}{dx}\left(x^3+x-1\right)\left(x^4+6x^2+9\right)-\left(x^3+x-1\right)\frac{d}{dx}\left(x^4+6x^2+9\right)}{\left(x^4+6x^2+9\right)^2}$
Learn how to solve differential calculus problems step by step online.
$\frac{\frac{d}{dx}\left(x^3+x-1\right)\left(x^4+6x^2+9\right)-\left(x^3+x-1\right)\frac{d}{dx}\left(x^4+6x^2+9\right)}{\left(x^4+6x^2+9\right)^2}$
Learn how to solve differential calculus problems step by step online. Find the derivative of (x^3+x+-1)/(x^4+6x^2+9). Apply the quotient rule for differentiation, which states that if f(x) and g(x) are functions and h(x) is the function defined by {\displaystyle h(x) = \frac{f(x)}{g(x)}}, where {g(x) \neq 0}, then {\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}. Simplify the product -(x^3+x-1). Simplify the product -(x-1). The derivative of a sum of two or more functions is the sum of the derivatives of each function.
## Final answer to the problem
$\frac{\left(3x^{2}+1\right)\left(x^4+6x^2+9\right)+\left(-x^3-x+1\right)\left(4x^{3}+12x\right)}{\left(x^4+6x^2+9\right)^2}$
## Explore different ways to solve this problem
Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more
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### Main Topic: Differential Calculus
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• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) • ### Free GMAT Strategy Webinar November 17, 2018 November 17, 2018 07:00 AM PST 09:00 AM PST Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT. # Two points are on the opposite sides of a square who sides new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Intern Joined: 24 May 2014 Posts: 13 Location: Brazil Two points are on the opposite sides of a square who sides [#permalink] ### Show Tags 26 Jul 2014, 10:24 3 00:00 Difficulty: 25% (medium) Question Stats: 73% (01:20) correct 27% (01:20) wrong based on 113 sessions ### HideShow timer Statistics Two points are on the opposite sides of a square who side measures 4. Which of the following could be the distance between these two points? (A) 3.8 (B) 5 (C) 6 (D) 7 (E) 9 Math Expert Joined: 02 Sep 2009 Posts: 50583 Re: Two points are on the opposite sides of a square who sides [#permalink] ### Show Tags 26 Jul 2014, 10:46 2 1 Reni wrote: Two points are on the opposite sides of a square who side measures 4. Which of the following could be the distance between these two points? (A)3.8 (B)5 (C)6 (D)7 (E)9 The shortest distance between any two points which are on opposite sides of a square is the length of the side of the square. The longest distance between any two points which are on a square is the length of the diagonal of the square. Attachment: Untitled.png [ 3.63 KiB | Viewed 1713 times ] So, the distance between these two points must be more than or equal to 4 and less than or equal to $$\sqrt{4^2+4^2}<6$$. The only option which fall into this range is B. Answer: B. _________________ EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12853 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Two points are on the opposite sides of a square who sides [#permalink] ### Show Tags 03 Feb 2016, 23:00 1 Hi All, With a bit of logic, and using the answer choices as reference, you can answer this question without having to do ANY calculations. We're told that two points are on the opposite sides of a square whose sides are 4. We're asked which of the following COULD be the distance between these two points. Since we don't know exactly where the two points are, we should start with what would be the SHORTEST distance between them, then we can work 'up' from there. If the points were at the exact some 'place' on opposite sides, then they would be the same distance as the length of a side: 4. That length is the SHORTEST distance between possible points, so Answer A is impossible. As we slide either (or both) points along the sides of the square, the distance will become GREATER than 4. So we could move one (or both) of the points so that they were 4.1 apart (or 4.2 or 4.325, etc.). Logically, the answer CANNOT be 6, because if 6 were a possibility, then 5 would ALSO be a possibility. Since there's only one correct answer to this question, it MUST be the smallest number that is greater than 4... Final Answer: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: [email protected] # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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# Two points are on the opposite sides of a square who sides
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,487 | 5,711 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-47 | latest | en | 0.851509 |
http://physicsinventions.com/crate-sliding-down-a-ramp/ | 1,558,670,380,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257497.4/warc/CC-MAIN-20190524024253-20190524050253-00048.warc.gz | 152,393,154 | 9,538 | Crate sliding down a ramp
"A 4.00-kg crate slides down a ramp. The ramp is 1.80 m in length and inclined at an angle of 28.0° as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 4.50 N, and continues to move a short distance on the horizontal floor after it leaves the ramp.
(A) Use energy methods to determine the speed of the crate at the bottom of the ramp.
(B) How far does the crate slide on the horizontal floor if it continues to experience a friction force of magnitude 4.50 N?"
so i solved part A for the speed of the crate at the bottom of the ramp and got 3.54m/s.
Using the speed, i then solved part B for the distance it traveled on the floor and got 5.57m using 1/2*mv2 = ffd
———————————————————————————————
Then the problem asks, what will be the final speed at the bottom of the ramp if it has an initial speed of .6m/s?
and this is where I get stuck, im not sure how to calculate that into the equation.
Thanks in advanced for the help.
http://ift.tt/1eB3QmY | 268 | 1,046 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2019-22 | latest | en | 0.870367 |
https://physics.stackexchange.com/questions/613340/does-voltage-depend-on-the-surface-area-of-a-conductor | 1,653,361,791,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00519.warc.gz | 534,235,622 | 71,933 | Does voltage depend on the surface area of a conductor?
I am beginning to study the properties of electricity and have a question. I was told that electrical voltage is basically like pressure in a pipe. However, pressure in a pipe is a measurement of force per unit (pound per square inch) of surface area, which means if I take the same force and spread it out to double the surface area I should get a pressure of half of what it was for the same surface area.
Now from what I understand, voltage isn’t measured in volts per square inch or anything like that, so can someone explain why electrical pressure and water pressure don't share the same type of pressure measurement and where the confusion comes into play? If I connect a conductor with 10x the surface area to both sides of a battery terminal will I get a voltage drop of 1/10th of the voltage since the electric pressure (voltage) is being spread out on a surface 10x the size?
The analogy is good, but you appear to misunderstand it. Water pressure is not a function of the surface area of the piping, any more so than voltage is a function of the surface area of the wiring.
The water pressure is a function of the strength of the pump which is charging the pipework (assuming the quantity of water available to charge the pipework in the first place, exceeds the capacity of the pipework). And in a situation where the water is flowing freely out after the pump, it is also a function of the resistance of the pipework that comes after the pump.
A static pressure of 1bar in a small diameter pipe is the same as a static pressure of 1bar in a large diameter pipe. Drill a small hole in either pipe, as if applying a multimeter to a wire, and you will measure the same 1bar of pressure in both.
The difference is when the water is flowing. If the water is at a pressure of 1bar leaving the pump at the pump's outlet, then there will be a pressure drop along the pipe according to the cross-section (in the ideal case of a circular pipe, the diameter) of the pipe and the distance from the pump outlet. The pressure at the open end of the pipe will not be 1bar, but something less depending on the diameter and the length from the pump.
The drop will be larger for smaller pipes than larger ones.
However, if the open end of the pipe is then closed, the pressure will charge up to the same pressure as at the outlet of the pump, and the final static pressure will be the same regardless of the diameter of the pipe.
This is why "voltage drop" has to be measured with the relevant load switched on, just as pressure drop has to be measured with the outlet open.
As to units, the electrical analogue unit for water volume - often measured in litres - is the coulomb. The electrical analogue unit for water flow - not given a first class unit, but often measured in litres/second - is amps.
Pounds per square inch - psi - is slightly confusing in this context. It does not mean the pressure is a given number of pounds distributed over the available square-inchage of the pipework (so that increasing the square-inchage of the pipework leads to a reduction in the number of pounds applied to each square inch). The psi is the amount of pressure being applied to every square inch of the pipework, howsoever many square inches there may in fact be.
Increasing the internal surface area of a solid container does not reduce the pressure of the contents. But increasing the volume of a container requires more litres to be added to charge the container to a given pressure in the first place.
Similarly, the surface area of a wire does not change the voltage - the voltage is the pressure that is apparent across all surface of the wire, and increasing the surface area of the wire will simply mean the exact same voltage level being apparent across a larger surface than before.
What a wire of larger volume does do, is require more coulombs of electricity to charge the wire to that voltage in the first place. The eletrical analogue unit of volumetric capacity - also often measured in litres, confusingly (mainly because liquids are conventionally treated as incompressible) - is the farad.
Edit: I thought I'd add further to this as it provoked my own interest about the water analogy. I hope I haven't spoiled a good start.
I was wondering how to analogise watts and joules. I've already said that coulombs are the analog unit of water quantity in litres. We would usually measure water transfer or consumption in terms of such litres. But why do we very rarely talk of electrical quantities in terms of coulombs?
We more often talk of everyday electricity consumption in terms of watt-hours - which is an equivalent to joules (1 watt-hour = 3,600 joules). What then is the water analog unit for joules?
The answer is that it is actually joules for both. Joules (and by implication watt-hours) is a unit of energy or work quantity.
"Work" here is not a description of a physical quantity. It is a description of a purely abstract mechanical quantity, based on the ability of the system in question to bring about change in the physical properties or spatial arrangements of physical things. When a quantity of work is transferred in a system, we are describing what quantity of change has occurred in that system.
The word "change" here is intentionally elusive. Specific kinds of physical change can be measured in different units - a litre of water is raised off the ground by 1 meter, or a litre of water is heated by 100 degrees. But each of these specific changes can be reduced (by applying the laws of physics) to a general description of the quantity of work transferred.
Note also that I talk of "work transferred", not "work done". Because this quantity of work is neither created nor destroyed, nor produced or consumed, only transferred. And it can be transferred on (so as to cause consequential change), or transferred back (so as to undo the change), and the laws of physics describe the terms on which this can be done. And there are always two sides to the equation - if work is transferred to some change, there must be another change elsewhere of opposite work quantity.
Nowadays we rarely use transfers of water through pipework to do work in this mechanical sense. We pipe water into the home as a solvent for washing, as a carrier for sewage, or for human consumption, and it is the water itself (i.e. it's intrinsic physical or chemical properties) which is seen as the main feature for this purpose, not the work which the piping and the water transfer system is capable of performing.
Transferring water in and out of the home does perform work, but it is not of everyday interest to measure exactly how much work is performed in doing so.
The contexts in which water, or steam, are used specifically to do mechanical work, and where that work would be subject to measurement, are in the realm of mechanical engineering. A main application is in turbines (of which a simple waterwheel is but a simple example). Now we see an application in which the quantity of water in litres is no longer intrinsically relevant, but instead it's ability to do work by driving the turbine is the quantity of interest. An unpressured quantity of water is not capable of doing any work in driving the turbine. But a pressured quantity is. And the amount of work is a function of the pressure of the transfer, not just the volume.
And that brings us back around to electricity. We do not move eletrical charges around wires for the intrinsic properties conveyed by charge, or so that we can hold the charge itself in our hands. We do it so that the movement of electrical charge performs work - usually in the end, work of a non-electrical kind, such as heating things or driving mechanical appliances. And that is why we measure electrical consumption in terms of joules, or watt-hours, because the ability to transfer work (rather than to transfer electrical charge) is the main purpose of electrical systems.
And finally, what of the watt unit? That is a measure of power, or the rate of transfer. But crucially, it is not a measure of the rate of transfer of litres of water, or of coulombs of electrical charge. It is a measure of the rate of transfer of work (measured in joules), the abstract quantity which I've described above.
It is not, as might first be thought, like the transfer of litres of water. A transfer of a very small amount of water under very great pressure, is capable of doing the same amount of work, as the transfer of a very large amount of water under more modest pressure. And watts measures the rate of work transferred electrically, regardless of the voltage or amperage used to do it (and thus regardless of the amount of coulombs actually moved, which in AC circuits is a net of zero anyway - it's like the same water volume being pumped and sucked back and forth in rapid alternation).
@Steve's answer is really very very good, and I can't improve upon it. But I want to add my thoughts anyway.
I find this analogy very useful, not just for helping understand why the currents in various legs of a circuit are what they are, but also for building intuition about what voltage actually is and where it comes from.
Water under high pressure really wants to escape--if there's any path to get to somewhere where the pressure is lower, it will take it. It's the same with electric charge--if it's somewhere where the voltage is high, it really wants to escape. That's why it's useful to think about water pipes to figure out where current will flow.
But $$why$$ does the water want to escape? What is physically different about it than if it wasn't at high pressure? The reason is that it's squeezed. Water molecules like to be so close together and no closer, and when they are a bit closer than they like, they push apart. Molecules of water under high pressure are a bit closer together than usual, and when you see pressurized water come spraying out of an opening, the force that's causing that is the molecules trying to spread back out to their desired separation. It's just the same with charge and voltage: the charges in wire at high voltage are bunched ever so slightly closer together than usual (or further apart--there's both positive and negative charge but let's not worry about that). Just like the water molecules, they want to separate back out, and that's what it is about them that makes them different from charges in another wire that's not at high voltage.
Every analogy has its limits, but if you find water pressure and flow easy to think about, this one can take you a long way, in my opinion.
As for your question about why the units of pressure are force per area while the units of voltage are just volts: it is true that pressure can be expressed as force per unit area, but maybe more aptly for this analogy, it can also be expressed as energy per unit volume: joules per meter cubed is the same thing as newtons per meter squared. So you can see that the pressure of a given amount of water tells you how much energy you can get from it by letting it flow to where it wants to go. In exactly the same way, voltage is the amount of energy you can get from a given amount of charge by letting it go where it wants. The amount of charge is not measured by volume (otherwise voltage would actually be pressure) but both voltage and pressure can be thought of as "energy per amount of stuff" (where stuff is either water or charge).
• You say not to worry about positive and negative charge, but it actually very easy to extend the water analogy to a pipework circuit, where a pump induces not only pressure at its outlet, but also suction at its inlet at the opposite end of the circuit. Even in the absence of an explicit pipework circuit, pumps do in fact function in this way - if you pump water in one direction from an open tank, the air pressure above the water in the source tank drops, and this eventually sucks air all the way back from wherever the water is being deposited. Feb 9, 2021 at 5:35
• (or further apart--there's both positive and negative charge but let's not worry about that) I just inserted so many mental caveats¹²³ Feb 9, 2021 at 21:06
That's a really bad analogy in my opinion. It's much better to understand what electric potential difference, colloquially called voltage, actually is.
From electrostatics, you might know that the electric potential at a point in an electric field is the electric potential energy per unit positive charge at that point. If there is a difference in electric potentials and thus the E.P.E of a unit positive charge between 2 points, the electric force must have done some work on the charge. A little mathematics shows you that the electric field strength (for per unit positive charge) = -dV/dx where V is the electric potential. Electric field strength is the negative derivative of electric potential and ALWAYS points in the direction of decreasing potential.
This means that wherever there is a potential difference between 2 points in a circuit, there is an electric field between them pointing in the direction of decreasing potential and positive charges from the higher potential to the lower potential. It is here where the greatly misleading analogy of "electric" pressure comes from. The only thing even remotely similar is that both a potential difference and pressure result in a force that pushes charges/ water molecules.
This analogy is just to give beginners some intuition that voltage causes charges to move. If you're comfortable with electrostatics, however, its much better to know that the potential difference creates an electric field and it is this electric field that exerts a force on charges and causes them to move.
• Steve's answer excellently elaborates why the pipe analogy is referred to in the first place, and realistically discusses its limitations. My answer hinted at how to actually calculate current and voltage for the most basic geometries, as the OP desired. I personally think your explanation is a confusing to beginner, and inadequate to guide the intuition of the OP. Feb 9, 2021 at 3:38
• Your answer is ill-informed and doesn't even answer the question. All you have done is stated a bunch have well known formulas without explaining how voltage varies and what is kept constant. Op clearly asks if connecting a wider conductor between the terminals will decrease the voltage. Voltage here is not a function of resistivity and your answer misleads gravely Feb 9, 2021 at 3:55
• The only reason I did not explain any more is that I have done so in a detailed fashion multiple times on recent questions, and indicated so in my answer. And, to be clear (to you), of course voltage drop across circuit elements depends on their resistance, once you start putting circuit elements together in series and applying a voltage across them all. Feb 9, 2021 at 4:15
• Voltage depends on resistance in the case of multiple components in series, whereas op clearly mentioned connecting a conductor to both sides of a battery, so it appears you didn't even read the question. Moreover, your formulas without any context imply that the voltage drop also depends on the current, which it obviously doesn't Feb 9, 2021 at 4:44 | 3,202 | 15,295 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2022-21 | latest | en | 0.952275 |
https://cs.stackexchange.com/questions/96960/divide-and-conquer-determining-the-top-two-candidates-and-whether-these-two-can | 1,716,813,558,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059040.32/warc/CC-MAIN-20240527113621-20240527143621-00074.warc.gz | 159,589,915 | 45,042 | Divide-and-conquer: Determining the top two candidates and whether these two candidates received more than n/2 votes
Suppose that each person in a group of n people votes for exactly two people from a set of candidates to fill two positions on a committee. The top two finishers both win positions as long as each receives more than n/2 votes.
Describe a divide-and-conquer algorithm that determines the top two candidates and whether these two candidates received more than n/2 votes.
I have the answer (below), but there's one part I don't understand:
Our algorithm will take a sequence of 2n names (two different names provided by each of n voters) and determine whether the two top vote-getters occur on our list more than n/2 times, and if so, who they are. Actually for technical reasons we will need the top 3 vote-getters.** The votes of each voter are adjacent. Note that we can have at most 3 people (but not 4) with more than half of the votes. Divide the list into two parts, the first half and the second half. (No one could have gotten more than n/2 votes on this list without having more than half votes in one half or the other, since if a candidate got less than or equal to half the votes in each half, then he got less than or equal to half the votes in all.) Thus apply the algorithm recursively to each half to come up with at most six names (three from each half). Then run through the entire list to count the number of occurrences of each of these names to decide which, if any, are the winners. This requires at most 12n additional comparisons for a list of length 2n.
I don't understand why we are coming up with at most six names. I also don't get where the 12n comes from. To be more clear, I sort of understand why the 12n is a thing given the 6 names, but I don't understand why it's 6, and if it means 6 names for each step of recursion, or 6 names after all of the recursion (and if the latter, what happens during each step of recursion?).
• I think a scenario when 3 will have $> n/2$ votes would be 5 votes for candidates (A,B,C) as follows: (A,B), (A,B), (A,C), (B,C), (B,C). A has 3 votes ($3 > 2.5$), B has 4 votes ($4 > 2.5$), C has 3 votes ($3 > 2.5$).
– ryan
Sep 4, 2018 at 17:07
• @ryan I was just able to come up with a similar list myself so that part I understand now. What I'm stuck on now is the recursion part and why we come up with 6 names and do 12n comparisons. Sep 4, 2018 at 17:12
• You said "I have the answer". Can you credit the source so that we can take a look, too? I just want to check the background and the context to gain more understanding and to check if there is something missing or unconventional. By the way, there is an algorithm that needs at most $16n$ comparisons. Sep 4, 2018 at 17:50
• @Apass.Jack The answer is the text that starts with "our algorithm". I will update the post to make that more clear. Sep 4, 2018 at 17:51
• No, the answer I provided is part of an answer key. Everything in the post is everything in the answer key so there's nothing missing. Does it sound like there's an error somewhere? Sep 4, 2018 at 18:01
The original problem is not quoted/rephrased correctly
Here is one correct version as it appears as exercise 18 in section 8.3 of the book Discrete Mathematics and Its Applications, 7th Edition by Kenneth H. Rosen, with slight modification.
Suppose that each person in a group of $$n$$ people votes for exactly two people from a set of candidates to fill two positions on a committee. The top two finishers both win positions as long as each receives more than $$n/2$$ votes.
Devise a divide-and-conquer algorithm that determines whether the two candidates who received the most votes each received more than $$n/2$$ votes. If so, determine who these two candidates are.
What is difference? As I have emphasized, only when it has been determined that each of the two candidates who received the most votes (the top two candidates) has received more than $$n/2$$ votes, the desired algorithm is required to determine these two candidates . In other word, if it has been determined that at least one of the top two candidates received at most $$n/2$$ votes, the algorithm is not required to determine these candidates.
The catch here is that it is possible to determine whether the top candidate received more than $$n/2$$ vote without identifying who is the top candidate. It is also possible to determine whether the second top candidate received more than $$n/2$$ vote without identifying who is the second candidate. For example, if we are able to ascertain the most votes received by any single candidate is at most $$n/2$$, which can happen when we have checked enough but not all of the votes (so we might not be able to identify who are the top candidates), then we are sure that none of the top two candidates received more than $$n/2$$ votes.
On the other hand, by its very nature, any (usual) divide-and-conquer algorithm cannot find the top (two) candidates unconditionally. The idea is that it can happen that the top candidates overall is not among the top candidates in either half of the list. In fact, it can happen that the top candidate overall is the bottom candidate in each half of the list. For example, 10 people can vote in the following way, $$(A,B), (A,B), (A,B), (C,D), C,D), (E,F), (E,F), (E,F), (C,D), (C,D)$$. While $$C$$ and $$D$$ are the top candidates overall, they are the bottom candidates voted by the first 5 people and they are the bottom candidates voted by the other 5 people.
The "answer" given by the OP, the fourth paragraph in the question, which starts with "our algorithm" and ends before the separator line, is in fact a part of an answer key. For brevity, I will refer to it as "the answer key".
How to understand the answer key
You can check the other cool answer written by D.W., which includes a very detailed analysis of the answer key.
D.W.'s answer describes in great detail, according to the answer key, an algorithm that returns a list of three candidates which contains all the candidates who got more than $$n/2$$ votes and possibly other candidates. More specifically, if a candidate got more than $$n/2$$ votes, then he/she must be in the returned list. However, the list may contain people who did not get more than $$n/2$$ votes. It can also happen that none of the people in the returned list is the top candidate, or the second top candidate, or the third top candidate.
D.W.'s answer also explains clearly "why we are coming up with at most six names" as well as "where the $$12n$$ comes from".
I have confirmed that D.W.'s algorithm is correct (since he, in his modest way, claim he does not "know whether this algorithm is actually correct") in that it conforms to the answer key and it is a full answer to the original problem except it omits the easy steps such as "if so, determines who these two candidates are".
A better written problem
It turns out, as you must have noticed by now, somewhat subtle to understand exactly what is the requirement of the original problem. Here is my version of the problem that should be much clearer, at least for me.
Suppose that each person in a group of $$n$$ people votes for exactly two people from a set of candidates. Devise a divide-and-conquer algorithm that determines all candidates who received more than $$n/2$$ votes.
If you are careful, you may point out the my version missed "... to fill two positions on a committee. The top two finishers both win positions as long as each receives more than n/2 votes". Well, although it is interesting to know that election goal and election rule, it has nothing to do with the specification of the desired algorithm. You may also point out that my version does not require "if so, determine who these two candidates are". Well, if we have determined all candidates who received more than n/2 votes, the number of whom is at most 3, then by just counting the number of votes received by each of those candidates, we can, easily, "if so, determine who these two candidates are".
An interesting algorithm as an exercise
There is an algorithm using linear time and constant space that determines all candidates who received more than $$n/2$$ votes. The construction of such an algorithm is left as an intriguing and challenging exercise for the readers who has read thus far.
Thanks to Tom van der Zanden, who pointed out a typo in my previous formulation of the exercise.
I can understand why you are finding this unclear. The sketch of a solution that you quote is pretty sparse, and it leaves it to you to fill in details. Here is how I interpret the proposed algorithm:
Algorithm $F[1..n][1..2]$):
1. If $n=1$, return $\{A[1][1], A[1][2]\}$. Otherwise:
2. Set $S_L := F(A[1..n/2][1..2])$ and $S_R := F(A[n/2+1..n][1..2])$ and $S := S_L \cup S_R$.
3. For each $x \in S$, count the number of times that $x$ appears in $A[1..n][1..2]$ and sort the $x$'s by that count.
4. Return the three values that occur the most often in $S$ (breaking ties arbitrarily).
where $A[i][1]$ denotes the first vote from the $i$th voter and $A[i][2]$ the second vote from the $i$th voter. At the end, after running $F(A[1..n][1..2])$, we take the three values returned by this algorithm and check which (if any of them) have received at least $n/2$ votes.
The running time of this algorithm satisfies the recurrence relation
$$T(n) = 2 T(n/2) + O(n).$$
(Why? It's easy to see that $F$ returns at most 3 elements, so $|S| \le 6$. Each iteration of step 3 can be done in $O(n)$ time, and we do only $6=O(1)$ iterations of the loop, so the total running time for steps 3 and 4 is $O(n)$.) This recurrence solves to $T(n) = O(n \log n)$.
So, this gives an $O(n \log n)$ time algorithm for the problem.
I don't know whether this algorithm is actually correct. I'm just trying to explain what algorithm I believe they are proposing, and to explain their running time analysis. I haven't tried to prove the algorithm correct. The solution you listed appears to sketch the key ideas behind such a proof (or, at least that's the claim). I haven't tried to fill in the details and verify carefully that the proof works out, so you should do that if you care, but it looks plausible to me -- all the ideas make sense to me.
Now, to answer your questions: The reason we come up with only 6 names is that each recursive call returns at most 3 names, and we perform two recursive calls, so we end up with at most $3+3=6$ different names in $S$. Why $12n$? Because each iteration of the loop in step 3 takes $2n$ comparisons (you compare $x$ to each element of $A[1..n][1..2]$, and there are $2n$ such elements), and you do at most 6 iterations of the loop, so the total number of comparisons in step is at most $6 \times 2n = 12n$.
I hope this answers your questions and makes the proposed solution clearer.
• "3. For each x∈S, count the number of times that x appears in A[1..n][1..2] and sort the x's by that count" Where did you use $S_L$ and $S_R$? Sep 7, 2018 at 3:51
• @Apass.Jack, $S_L$ and $S_R$ are used in step 2 where we set $S = S_L \cup S_R$. We could take it to this chat room if that's better!
– D.W.
Sep 7, 2018 at 4:12 | 2,805 | 11,199 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-22 | latest | en | 0.963258 |
https://www.physicsforums.com/threads/questions-about-a-normal-distribution-discrete-to-continuous.976040/ | 1,708,798,876,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474544.15/warc/CC-MAIN-20240224180245-20240224210245-00865.warc.gz | 957,993,808 | 38,472 | # Questions about a normal distribution (discrete to continuous)
• B
• PainterGuy
In summary, the probability distribution can be used to calculate the probability that the histogram of data has various shapes.
#### PainterGuy
TL;DR Summary
I was trying to understand normal distribution and had a couple of questions.
Hi,
https://youtu.(be/mtH1fmUVkfE?t=215)
While watching it, a question came to my mind. In the picture, you can easily calculate the total number of customers. It's 1000.
For my question, I'm going to use the same picture. I hope you get the point.
What if the data was continuous something as shown below. How do you calculate the total number of customers? Wouldn't the number be infinite because between any two numbers, e.g. between '2' and '3', there is an infinite set of values? I'm not sure if the question really makes sense.
In case of normal distribution the curve also represents continuous data but I believe, practically, it's discrete data made up of very thin slices as shown below and later curve fitting is used to get a continuous curve.
Thank you!
PainterGuy said:
What if the data was continuous
That's a bit difficult with 'number of ice creams'
Think of an intermediate step: at the grocery store: money spent per customer. With a sample of 1000 customers there will likely be 1000 customers in the bin 0 - 1000 $But for the bin 0-100$ the first bin will be close to 1000 and the second the remainder (more or less).
Binning from 0-10, 10-20 etc will give a wider distribution with less customers per bin. Still the total number will be 1000 if you add all the bins.
etc.
It's a matter of dimension: number per bin width times bin width = number. If the bin width is taken smaller, bins will contain proportionally less, but the sum of all bin contents will remain the same.
For a continuous distribution the vertical dimension is usually the probability density : the probability is found by multiplying with the horizontal 'bin width'. by the way, your question applies to all distributions, not just normal distributions.
PainterGuy
PainterGuy said:
In case of normal distribution the curve also represents continuous data but I believe, practically, it's discrete data made up of very thin slices as shown below and later curve fitting is used to get a continuous curve.
No. You are confusing the concept of a probability distribution with the concept of a histogram of data. A continuous probability distribution is a mathematical formula, not a histogram of data. A probability distribution gives information about the probable values of one sample of a random variable. If you take 100 independent samples of the same random variable and histogram the results then the histogram can have various shapes. The distribution of the random variable does not determine a unique shape for the histogram of the data. The probability distribution can be used to calculate the probability that the histogram of data has various shapes.
PainterGuy
Thanks a lot!
BvU said:
Think of an intermediate step: at the grocery store: money spent per customer. With a sample of 1000 customers there will likely be 1000 customers in the bin 0 - 1000 $But for the bin 0-100$ the first bin will be close to 1000 and the second the remainder (more or less).
In my humble opinion, it would depend upon the customers. Generally, as you are suggesting, most customers don't spend more than $100 in a grocery store. But It's possible that the second bin$100-200 could capture a big chunk of 1000 customers compared to $0-100 bin, at least on certain days. I just wanted to point this out. Stephen Tashi said: If you take 100 independent samples of the same random variable and histogram the results then the histogram can have various shapes. The distribution of the random variable does not determine a unique shape for the histogram of the data. The probability distribution can be used to calculate the probability that the histogram of data has various shapes. How can the probability distribution be used to calculate the probability that the histogram of data has various shapes? The distribution gives the most probable numeric guess about something. Yes, if one knows the probability, p, of something might happen, one can also calculate of probability, 1-p, that it might not happen. I think that the problem could be stated as follows. We know the probability that something will happen. If a trial or experiment is done, say, 30 times, how its probability will vary in each experiment. Should Binomial formula be used? Thank you!Note to self: Check Statistics by Levin, Rubin 7th ed, graphing frequency distribution, page 38-41 kindle version Helpful link(s): 1: https://www.statisticshowto.datasciencecentral.com/choose-bin-sizes-statistics/ 2: https://www.quora.com/What-is-the-difference-between-probability-and-probability-distribution 3: https://help.plot.ly/histogram/ Note that what is usually given and plotted is a probability density function ##P##. With the property (for a single stochastic variable)$$\int_{-\infty}^{+\infty}P(x)\,dx = 1$$and one can calculate the probability to find ##x## between ##x_1## and ##x_2## from $$\int_{x_1}^{x_2}P(x)\,dx$$ e.g. in the normal distribution lemma they clearly say ##f## is a density PainterGuy said: How can the probability distribution be used to calculate the probability that the histogram of data has various shapes? It can't (the question is a bit vague, too) The probability density function (pdf) can be used to calculate the probability that an observed value falls in a given bin. Example: between ##\mu+\sigma## and ##\mu + 2 \sigma## the probability is 0.1359 for the ##f## in the link. How many of ##N## (##N## big) observations end up in a bin for which the pdf gives a probability ##p## is again a stochastic variable that is Poisson distributed with mean ##pN##. I.e. if you repeat your set of ##N## observations a (big) number of times and then make a histogram of the number of observations in this particular ##p## bin per set, you should get a Poisson distribution with mean ##pN##. Phew... Klystron and PainterGuy PainterGuy said: I think that the problem could be stated as follows. We know the probability that something will happen. If a trial or experiment is done, say, 30 times, how its probability will vary in each experiment. Should Binomial formula be used? If we know the probability that something will happen then it's probability does not vary. I think you are confusing the concepts of frequency (how often something actually happens in a series of experiments) with the concept of probability. As mentioned before, there can be no guarantee that an event with probability ##p## will happen with a frequency of ##p##. Such a definite connection would contradict the concept of probability. Likewise if an event happened with a frequency of f in 100 independent trials then there is no mathematical law that says the probability of it happening in one trial is f. The distribution gives the most probable numeric guess about something. In general, that's not correct. I think what you mean is that if an event happens M times in N independent trials then M/N is, in an inuitive sense, the "best" guess for the probability of the event happening in a single trial. The mathematical justification for guessing M/N is that assuming the probability of the event is p= M/N is the numerical value that makes the observed frequency M/N the most probable outcome. This says nothing about the probability that p =M/N. Without an assumed probability distribution for various values of p, you can't say anything about the probability that p = M/N. To illustrate a scenario where the value of p can have an associated probabiliy, consider the following scenario: A box contains 3 coins. One coin has probability p = 1/2 of landing heads, the other two coins have probability p = 2/3 of landing heads. A coin is picked at random from the box and filipped 5 times. It lands heads 3 of those times. What is the probability that p for the selected coin is 3/5? The probability that p = 3/5 is zero because we know that the only possible values of p are 1/2 and 2/3.Guessing M/N is a specific example of the procedure of using "maximum likihood estimation". Like many statistical techniques, maximum liklihood estimation is a procedure, not an axiom or law that is absolutely guaranteed to produce the "best" result for all specific ways to define "best". How can the probability distribution be used to calculate the probability that the histogram of data has various shapes? First you must define the shape you are interested in. Then pose the problem as a specific question about probability. For example, suppose the distribution is a normal distributiion with mean 0 and variance 1. Suppose the histogram of interest had 4 bins defined by [-2,-1], [-1,0], [0,1],[0,2] and represents the result of 5 independent realizations of the random variable with results : 1 outcome in bin [-2,-1], two outcomes in bin [-1,0], two outcomes in bin [0,1]. Calculating the probability of such a histogram is a specific problem in probability. The calculation would involve "binomial coefficients", but not the "binomial distribution". As @BvU indicates, for similar problems involving a large number of bins and a large number of outcomes, it gets too complicated to compute the numerical answer. Last edited: sysprog and PainterGuy Thank you very much! I would like to expand on the previous two replies but first I need to clarify something. Suppose an event is capable of being repeated sufficiently large number of times “N”, and the frequency of the desired outcome is “f ”. Then relative frequency of the outcome is “ f ⁄ N”. The limiting value of the relative frequency can be used to define probability of the outcome. [Levin, Richard I.. Statistics for Management (p. 160). Kindle Edition. ] To illustrate several of these distributions, consider a situation at Kerr Pharmacy, where employees are often late. Five workers are in the pharmacy. The owner has studied the situation over a period of time and has determined that there is a 0.4 chance of anyone employee being late and that they arrive independently of one another. [Levin, Richard I.. Statistics for Management (p. 227). Kindle Edition.] How would the owner determine the probability of 0.4? Suppose on day #1, one employee is late, i.e. 1/6, on day #2 two employees are late, i.e. 2/6, and so on. The owner could carry on this observation for 60 days, i.e. N=60. Would the owner proceed as follows to calculate the probability of 0.4? Note to self: BvU said: Note that what is usually given and plotted is a probability density function ##P##. With the property (for a single stochastic variable)$$\int_{-\infty}^{+\infty}P(x)\,dx = 1$$and one can calculate the probability to find ##x## between ##x_1## and ##x_2## from $$\int_{x_1}^{x_2}P(x)\,dx$$ e.g. in the normal distribution lemma they clearly say ##f## is a density P(x) is probability density function where probability density, PD=probability/bin_width. Please refer to 3rd link under "Helpful link(s)" and also check page #232 of Levin, Ruben book, 7th ed, Kindle, to see how binomial could change into normal distribution as n becomes infinite. So, when you are integrating, you are summing up infinitesimal sections, {probability/dx}*dx where "x" could be molecules speeds or anything else. The main thing is that the result of integration is a scalar without any units; it gives total probability between two values of 'x'. BvU said: It can't (the question is a bit vague, too) The probability density function (pdf) can be used to calculate the probability that an observed value falls in a given bin. Example: between ##\mu+\sigma## and ##\mu + 2 \sigma## the probability is 0.1359 for the ##f## in the link. Have a look here. Helpful link(s): 1: https://www.britannica.com/topic/Poisson-distribution 2: https://stattrek.com/probability-distributions/poisson.aspx #### Attachments • normal_distri1.png 19.3 KB · Views: 576 PainterGuy said: I would like to expand on the previous two replies but first I need to clarify something. Suppose an event is capable of being repeated sufficiently large number of times “N”, and the frequency of the desired outcome is “f ”. Then relative frequency of the outcome is “ f ⁄ N”. The limiting value of the relative frequency can be used to define probability of the outcome. [Levin, Richard I.. Statistics for Management (p. 160). Kindle Edition. ] That is not the mathematical definition of probability. You can find books on statistics that give intuitive definitions for concepts used in statistics. Intuitive definitions are useful for thinking intuitively, but taking them seriously and literally is a handicap in doing mathematics. A definition based on the concept of a llimiting fequency doesn't tell us how to analyze data from a finite number of experiments. The intuitive definition you quoted is using the phrase "an event is capable of being repeated sufficiently large number of times “N”, to deal with the mathematical concepts of independent and identically distributed random variables. However, to define the mathematical concept of "independent" events requires making statements about their probabilities, so trying to define the probability of an event by using the concept of independent events would produce a circular definition. How would the owner determine the probability of 0.4? Would the owner proceed as follows to calculate the probability of 0.4? The passage you quoted simply assumes p = 0.4, so I think you are considering a problem that is beyond the scope of that section of your text. If you are asking the above questions in the sense of "What is the unique answer based on the given data?" you don't yet understand the conceptual framework for statistical estimation. Perhaps your text deals with statistical estimation in later chapters. A person who uses a phrase like "how do we calculate..." suggests that the given information plus logic and mathematics is sufficient to deduce a specific answer. A person who uses the phrase "how can we estimate..." leaves open the possibility that some additional assumptions must be made in order to get a definite result. The two major categories of statistical methods are 1) Estimation and 2) Hypothesis Testing. A set of data, by itself, does not tell us how to estimate the parameters of a probability distribution. Various procedures of statistical estimation make different assumptions, set different objectives, and can get different answers. It is the nature of probabilistic phenomena, that no method of estimation will always produce the true value of a parameter. A result from a statistical method is only "correct' in the sense that it is the correct answer to a problem that is stated using the data plus additional assumptions and objectives. One method of statistical estimation is Moment Matching. In the example, there are (5)(60) = 300 opportunities for an absence. In those opportunities there occurred a total of: (1)(24)+2(20)+3(11)+4(3)+5(2)=119 absences. Model the distribution as a binomial with N = 300 and probability of success ##p##. The mean of the distribution is ##Np = 300p##. Set our objective as matching the observed number of absences in the data equal to the mean number of absences for the distribution. This defines the problem as solving ##300p = 119## for ##p##. Instead of matching the mean (first moment) of a binomial distribution to the mean of the data, one could take the approach of matching the variance of the data to the variance of the distribution. Which method would be best? That depends on the conceptually complicated ways of defining "best"! Last edited: sysprog, BvU and PainterGuy Thank you! Stephen Tashi said: That is not the mathematical definition of probability. You can find books on statistics that give intuitive definitions for concepts used in statistics. Intuitive definitions are useful for thinking intuitively, but taking them seriously and literally is a handicap in doing mathematics. I understand your point. But, in my humble opinion, when someone starts learning a concept or subject, it's okay to start with simple, a little bit informal, and intuitive definitions which are not very rigorous and precise. At the same time, one should always bear in mind that as the conceptual progress is made, one needs to evolve the defining framework, and work out and accept more rigorous framework of the subject. Actually, academically, this is what is done in every subject and it seems to work. For example, a simple framework of atomic models and atomic particles are later refined and formalized using quantum theory models, concepts of numbers such as natural, real, complex are gradually formalized, concepts of limit which is later rigorously defined using delta epsilon definition, etc. Pedagogically, refining a concept in small steps is a best to learn something, in my view. I believe that you are saying the same thing. Stephen Tashi said: One method of statistical estimation is Moment Matching. In the example, there are (5)(60) = 300 opportunities for an absence. In those opportunities there occurred a total of: (1)(24)+2(20)+3(11)+4(3)+5(2)=119 absences. Model the distribution as a binomial with N = 300 and probability of success p. Well, how do I graph such a binomial distribution with N=(5)(60)=300 where "5" is number of employees, "60" is number of observed days. Initially, I assumed that N=60 and I could picture it where x-axis represented number of late arrivals from '0' to '5', and y-axis represented probability of those late arrivals. In case of "300", I think it represents each opportunity of being late but how do I relate the probability of an employee being late to it? For the proper context I'm quoting some sections on the same example from the same book. In Picture #2, the author also uses n=10 and n=30 cases to show that how the binomial starts looking like normal distribution once the sample size, i.e. number of employees, increases. Thank you! Picture #1: https://imagizer.imageshack.com/img921/7509/xJaZOg.jpg Picture #2: https://imagizer.imageshack.com/img921/1979/64jLcB.jpg Note to self: BvU said: How many of ##N## (##N## big) observations end up in a bin for which the pdf gives a probability ##p## is again a stochastic variable that is Poisson distributed with mean ##pN##. I.e. if you repeat your set of ##N## observations a (big) number of times and then make a histogram of the number of observations in this particular ##p## bin per set, you should get a Poisson distribution with mean ##pN##. To apply Poisson distribution, average or mean, is required, and to calculate it you need to know the probability, mean = λ = np, where n is number of trials and p is probability. Check this link: https://www.britannica.com/topic/Poisson-distribution and check Levin, Richard I.. Statistics for Management (p. 239). Kindle Edition. Check post #7 and #8 to get an idea how to calculate probability. The Poisson distribution will tell you the probability of each observation ending up into a bin with probability p. Last edited: PainterGuy said: Pedagogically, refining a concept in small steps is a best to learn something, in my view. I believe that you are saying the same thing. In a college level statistics course, I think it's best to follow the standard mathematical presentation of probability theory. This includes the concept of a probability space, which is a probability measure, a set of outcomes, and an algebra of subsets of outcomes. Without much harm those concepts can be presented as the special cases of "a sample space for an experiment" consisting of a "probability distribution" over a "range of values", and limiting the subsets considered to sets consisting of intervals and single points. Presenting the concept of probability as a limiting frequency is something I'd expect in course designed for people who only need to do statistical calculations by rote - perhaps lab technicians, inspectors in factories - maybe managers! You are asking questions that go beyond the material in your textbook, so I assume you want to deal with statistics at a college level. Well, how do I graph such a binomial distribution with N=(5)(60)=300 where "5" is number of employees, "60" is number of observed days. The binomial distribution I used has the parameters N = 300 and p. So it doesn't have any parameter representing the 5. If we take seriously the idea that each lateness by each employee is an independent event, the fact that data is collected in groups of 5 doesn't matter. It's like flipping a coin 300 times and recording the data. Grouping the data in sets of 5 records doesn't add anything to the information it contains. I agree that using a binomial distribution with N=300 does not model the fact that at most 5 employees can be late on a given day. However, if you grouped the coin flip data in sets of 5, that would automatically impose such a constraint. To repeat an earlier point, a histogram of data (or a table of data) and a probability distribution are different things. So if you pick a probability distribution for the data, you don't necessarily get a probability distribution whose graph involves all the parameters involved in a table or histogram of the data. Initially, I assumed that N=60 and I could picture it where x-axis represented number of late arrivals from '0' to '5' That's reasonable way to represent a binomial distribution where N = 5. The 60 days of data are not represented on the graph and they don't need to be. A distribution for a random variable gives the probabilities for the outcomes of a single realization of a random variable. Students taking introductory statistics are tempted to think that a statistics problem will involve exactly one random variable and its distribution. Perhaps that will be true in Chapter 1, but even moderately sophisticated problems involve several different random variables. In the example at hand, we can consider the random variable that answers the question "How many employees are late today"? We can consider a different random variable that answers the question "How many latenesses are there in 300 opportunities for a lateness to occur?". In the example at hand, we might choose to model the number of employee latenesses on a given day by some non-binomial distribution. For example, we might think the latenesses are not independent events. Perhaps some employees depend on others for a ride to work. Let the random variable ##X## answer the question "How many employees are late today?". Suppose we consider what happens each day to represent an independent realization of ##X## (i.e. we ignore things like hangovers on Monday mornings) We can define another random variable ##Y## that answers the question "How many latenesses occur in 60 days?". If we have specific distribution for ##X## we can compute the distribution for ##Y## by the procedure of "convolution". Probability distributions can be defined on outcomes in two-dimensional spaces. If we want to consider things like hangovers on Mondays, we can pick a model where the distribution is a function of two variables ##f(k,d)## where ##k## is the number of employees absent and ##d## is the day of the week: 1,2,..7. sysprog and PainterGuy Thank you very much! You can find the full example of employees here (ignore the highlights): https://imagizer.imageshack.com/img922/6692/uuUrm0.jpg Stephen Tashi said: The binomial distribution I used has the parameters N = 300 and p. So it doesn't have any parameter representing the 5. If we take seriously the idea that each lateness by each employee is an independent event, the fact that data is collected in groups of 5 doesn't matter. It's like flipping a coin 300 times and recording the data. Grouping the data in sets of 5 records doesn't add anything to the information it contains. I agree that using a binomial distribution with N=300 does not model the fact that at most 5 employees can be late on a given day. However, if you grouped the coin flip data in sets of 5, that would automatically impose such a constraint. I will focus on the case of n=10 from the lateness example, plot is shown below. Initially it was said that n=5 is total number of employees and r is number of late arrivals. If n=10 and r=10, how do these relate to the original number of employees and late arrivals? Let's try to find an answer. Originally, the maximum possible number of late arrivals was 5, as is shown below. Assume that the data set was collected over a period of two weeks. Suppose that the owner collects another independent data set over the period of two more weeks. Now there are two independent data sets conveying the same information. We notice that for n=5, the probability for two late arrivals, i.e. r=2, is 0.3456. But when n=10, the probability for r=4 is 0.2508; 0.3456 - 0.2508 = 0.0948. What does this difference tell us? I'd say that it mean that the probability of two employees being late in both of the data sets, i.e. r=2+2, is 0.2508. We can see below that when n=30, which according to my understanding means 6 independent data sets, the probability for r=2+2+2+2+2+2=12 is 0.14738. Do I make any sense at all? Thanks a lot your time and help! PainterGuy said: Originally, the maximum possible number of late arrivals was 5, as is shown below. Assume that the data set was collected over a period of two weeks. Suppose that the owner collects another independent data set over the period of two more weeks. Now there are two independent data sets conveying the same information. I'd say that it mean that the probability of two employees being late in both of the data sets, i.e. r=2+2, is 0.2508. If ##q## is the probability of something happening in one experiment, then ##q^2## is the probability of it happening in each of two given independent experiments. Check your idea against that fact. PainterGuy Thank you! For n=5, the probability for two late arrivals, i.e. r=2, is 0.3456, and when n=10, the probability for r=4 is 0.2508. 0.3456² = 0.1194 ≠ 0.2508 I'm wrong so I'm back to the original question as quoted below. Could you please guide me with this? PainterGuy said: We notice that for n=5, the probability for two late arrivals, i.e. r=2, is 0.3456. But when n=10, the probability for r=4 is 0.2508; 0.3456 - 0.2508 = 0.0948. What does this difference tell us? Think precisely about the outcomes in "sample spaces" involved. A detailed way to define the outcomes of an experiment that can answer "How many employees are late today" in the case of 5 total employees is to use probability space ##\Omega_a## , each of whose outcomes is a sequences of 5 zeroes or ones. For example: (1,1,1,1,1), (0,1,1,1,1), (1,0,1,1,1), ... (0,0,0,0,0) where a "0" in the 3rd place indicates employee 3 is late, etc. For outcomes in the case of 10 total employees, we can use probability space ##\Omega_b## whose outcomes are sequences of 10 zeroes and ones like: (1,1,1,1,1,1,1,1,1,1) or (0,1,1,1,1,1,1,1,1,1), etc. For the experiment to answer "How many employees are late on day 1 and how many employees are late on day 2", the we can use probability space ##\Omega_c##, whose outcomes are a pair of sequences like: ( (1,1,0,0,1),(0,1,1,0,0)) or ((1,1,0,0,1),(1,1,1,1,1)) etc. The set "two employees are late on both days" in ##\Omega_c## does not contain all pairs of sequences with 4 total latenesses For example it does not include ( (0,0,0,1,1)(1,1,1,0,1) ). However, in the sample space ##\Omega_b## we do include sequences like (0,0,0,1,1,1,1,0,1). So the set of outcomes defined by "4 employees are late out of 10 total employees" in ##\Omega_b## is a larger set than the set defined by "On both days, 2 out of 5 employees are late" in ##\Omega_c##. ##\Omega_b## and ##\Omega_c## have the same total number of outcomes and we are assuming each outcome is equiprobable. So comparing the probability of sets between them only involves counting the elements in the sets.Edit: Instead of say that each outcome is "equiprobable", I should say that each outcome on one sample space has a corresponding outcome in the other sample space that has the same probability. Last edited: sysprog and PainterGuy Thank you! I'm sorry that I'm going to ask about the same problem again. But I believe that you are approaching the problem differently. In post #11 above, this example, https://imagizer.imageshack.com/img922/6692/uuUrm0.jpg, was discussed. I was mainly focused on the plots at the bottom where n=10 and n=30. The following was my interpretation for n=10 case. As there were 5 employees, n=5, therefore the maximum possible number of late arrivals was also 5, i.e. r=5. Assume that the data set was collected over a period of two weeks. Suppose that the owner collects another independent data set over the period of two more weeks. Now there are two independent data sets conveying the same information. The combined elements of both data sets are 10. I was assuming that the actual number of employees does not change and I was trying to interpret what n=10 actually means. Well, it looks like that my underlying interpretation is wrong. When, n=10, it might mean that the actual number of employees is really 10 instead of 5. Perhaps, the author was trying to show that how probabilities are affected when number of employees change from 5 to 10 and 30. Thanks a lot! Last edited: PainterGuy said: I'm sorry that I'm going to ask about the same problem again. But I believe that you are approaching the problem differently. How would you define "the problem"? I haven't seen a specific problem stated. Perhaps, the author was trying to show that how probabilities are affected when number of employees change from 5 to 10 and 30. It looks like the author is trying to illustrate the idea that a binomial distribution B(n,p) for large N can be approximated by a normal distribution. What precisely does "can be approximated" mean? How would you, or the textbook author do the approximation? Thank you! Stephen Tashi said: How would you define "the problem"? I haven't seen a specific problem stated. Actually, I was trying to understand it for myself that if the assumption of 5 employees could hold even when n=10, or n=30? If you watch this video until 4:33, it shows how the distribution starts looks like a bell curve. https://youtu.(be/mtH1fmUVkfE?t=197) (remove the parentheses) I was under the impression somewhere in my mind that such a procedure, like the one in video, is also done on employees example to make it look like a bell curve. I had assumed that n=2(5)=10 represents two independent data sets collected by the owner over different periods. The same goes for n=6(5)=30. But original number of employees which is 5 remains the same. So, I'd say that the question is if in case of n=10 or n=30, the employees number is same as the original, which is 5. Well, it looks like I was wrong because the way author presents it, the number of employees would be 10 when n=10 and 30 when n=30. Stephen Tashi said: It looks like the author is trying to illustrate the idea that a binomial distribution B(n,p) for large N can be approximated by a normal distribution. What precisely does "can be approximated" mean? How would you, or the textbook author do the approximation? The author only says this, "Let us examine graphically what happens to the binomial distribution when p stays constant but n is increased. Figure 5-6 illustrates the general shape of a family of binomial distributions with a constant p of 0.4 and n’s from 5 to 30. As n increases, the vertical lines not only become more numerous but also tend to bunch up together to form a bell shape. We shall have more to say about this bell shape shortly." PainterGuy said: The author only says this, "Let us examine graphically what happens to the binomial distribution when p stays constant but n is increased. Figure 5-6 illustrates the general shape of a family of binomial distributions with a constant p of 0.4 and n’s from 5 to 30. As n increases, the vertical lines not only become more numerous but also tend to bunch up together to form a bell shape. We shall have more to say about this bell shape shortly." We'll have to see what the author says later in the text. I don't know if there is an intuitive way to understand the binomial distribution ##B_{2N,p}## in terms of the binomial distribution ##B_{N,p}##. Resorting to algebra we have ##B_{2N,p}(k) = \binom{2N}{k} p^k (1-p)^{2N-k}## Can we write ##B_{2N,p}(k)## as an expression that is a function of the values of ##B_{N,p}##? sysprog and PainterGuy Thank you! Stephen Tashi said: We'll have to see what the author says later in the text. I don't think the author is going to say much about it. Perhaps, I was reading too much into it. Stephen Tashi said: I don't know if there is an intuitive way to understand the binomial distribution ##B_{2N,p}## in terms of the binomial distribution ##B_{N,p}##. Resorting to algebra we have ##B_{2N,p}(k) = \binom{2N}{k} p^k (1-p)^{2N-k}## Can we write ##B_{2N,p}(k)## as an expression that is a function of the values of ##B_{N,p}##? I don't see a way. The general way to look at things is to consider sums of large number of independent identically distributed random variables. The distribution of such a sum is roughly a bell shaped curve. The "Central Limit Theorem" says this and more. The binomial distribution B(n,p) is special case of summing independent identically distributed random variables. The binomial distribution B(n,p) can be regarded as the sum of n independent Bernoulli random variables. A Bernoulli random variable ##X## is defined as have the probability density ##Pr(X=1) = p ,\ Pr(X=0) = 1-p ##. In your example, the binomial distribution B(5,p) that gives data for 5 employee latenesses can be regarded as summing 5 Bernoulli random variables, one for each of 5 employees. We add 1 to the sum if the employee is late and add zero otherwise. The general way to look at distribution of the sum of independent random variables is to understand the concept of taking the "convolution" of two distributions. It is correct that a random variable ##Y## that has distribution ##B(10,p)## can expressed in terms of two independent random variables ##X_1, X_2## each of which has distribution ##B(5,p)##. The way to do this is ##Y = X_1 + X_2##. In your example, this says the data ##Y## from two days of business with 5 employees is computed by adding the lateness data from each of the two days. We get ##Y = 4## in more ways than getting ##X_1 =2## and ##X_2= 2##. We also get ##Y = 4## from ##X_1 = 0,\ X_2 =4## and ##X_1=1,\ X_2 = 3##, etc. So to compute the probability that ##Y = 4## we must add up the probabilities of all these possibilities. Performing that calculation is called taking the convolution of the distributions of ##X_1## and ##X_2##. sysprog and PainterGuy Thank you very much! I'm reading on central limit theorem and other topics to get a big picture. Stephen Tashi said: The general way to look at distribution of the sum of independent random variables is to understand the concept of taking the "convolution" of two distributions. It is correct that a random variable ##Y## that has distribution ##B(10,p)## can expressed in terms of two independent random variables ##X_1, X_2## each of which has distribution ##B(5,p)##. The way to do this is ##Y = X_1 + X_2##. In your example, this says the data ##Y## from two days of business with 5 employees is computed by adding the lateness data from each of the two days. We get ##Y = 4## in more ways than getting ##X_1 =2## and ##X_2= 2##. We also get ##Y = 4## from ##X_1 = 0,\ X_2 =4## and ##X_1=1,\ X_2 = 3##, etc. So to compute the probability that ##Y = 4## we must add up the probabilities of all these possibilities. Performing that calculation is called taking the convolution of the distributions of ##X_1## and ##X_2##. I understand your comments about convolution and I will try to do it as I get some time. Can we add up the values individually to get an idea? Let me try. When n=10, the probability for Y=r=4 is 0.2508 . If I understand you correctly, it means that the following values (r=0, r=4), (r=1, r=3), (r=2, r=2) should be used from the case n=5. I tried to do it three different ways but it didn't work for me. The answer should have been 0.2508. If it's complicated, then no worries. We can get back to it later. (0.0778*0.0768)+(0.2592*0.2304)+(0.3456*0.3456)= 0.18513 (0.0778+0.0768)+(0.2592+0.2304)+(0.3456+0.3456)= 1. 3354⋅2 (0.0778+0.0768)*(0.2592+0.2304)*(0.3456+0.3456)= 5. 2318×10⁻² Code: n = 5; p = 0.4 B(5,p) r Probability 0 0.0778 1 0.2592 2 0.3456 3 0.2304 4 0.0768 5 0.0102 n = 10; p = 0.4 B(10,p) r Probability 0 0.0060 1 0.0403 2 0.1209 3 0.2150 4 0.2508 5 0.2007 6 0.1115 7 0.0425 8 0.0106 9 0.0016 10 0.0001 This question is rather about how 'this' would be done practically. Please check the green highlights; the arrows point those highlights. Link: https://imagizer.imageshack.com/img923/3776/q6Og35.jpg "Because only five people are involved, the population is too small to be approximated by a normal distribution. We’ll take all of the possible samples of the owners in groups of three, compute the sample means (x_bar), list them, and compute the mean of the sampling distribution (μ_x_bar)." You can see in Table 6-10, how 10 samples are created each with n=3 from a population with N=5. "This distribution has a mean of$19,000 and a standard deviation of $2,000. If we draw a random sample of 30 tellers, what is the probability that their earnings will average more than$19,750 annually?"
In Figure 6-8, you see a plot of actual population distribution of bank tellers' earnings and also sampling distribution. My question is how such plots are created in real life.
To keep it simple, I would say that the actual population plot is based on actual database of bank tellers' earnings.
The sampling distribution plot could be created, say, using data of only 30 tellers, and then 'mathematically' a lot of samples could be created with n=25, total samples would be = 30(nCr)25 = 142506. This would give bell shaped distribution which could be approximated using normal distribution.
Do I make sense?
Actually I was reading about the molecular speed distribution when I started this thread. We can see below that population distribution is asymmetric - negatively skewed; it's not normal. I believe that they should have used a label "probability" instead of "number of molecules". I'm focusing on the curve with label "low T".
Total number of molecules = Nm
Number of molecules having certain speed =
Do I have it correct? And is Cp the modal speed? Thanks a lot!
1: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map:_Physical_Chemistry_(McQuarrie_and_Simon)/27:_The_Kinetic_Theory_of_Gases/27.3:_The_Distribution_of_Molecular_Speeds_is_Given_by_the_Maxwell-Boltzmann_Distribution
2: https://chem.libretexts.org/Bookshe...ing_Gas_Properties_to_Kinetic_Theory_of_Gases
Last edited:
sysprog
PainterGuy said:
When n=10, the probability for Y=r=4 is 0.2508 . If I understand you correctly, it means that the following values (r=0, r=4), (r=1, r=3), (r=2, r=2) should be used from the case n=5.
I don't know what you are denoting by "r". In my notation, one must not only consider the case ##X_1=1,\ X_2 = 3## but also the case ##X_1=3,\ X_2 = 1##. So there are more than three probabilities to be added up.
This question is rather about how 'this' would be done practically. Please check the green highlights; the arrows point those highlights. Link: https://imagizer.imageshack.com/img923/3776/q6Og35.jpg
I don't know what you mean by "this".
"Because only five people are involved, the population is too small to be approximated by a normal distribution. We’ll take all of the possible samples of the owners in groups of three, compute the sample means (x_bar), list them, and compute the mean of the sampling distribution (μ_x_bar)."
I don't know what that example is supposed to show. It's a poor choice for showing anything related to the Central Limit Theorem because the central limit theorem deals with samples of identically distributed random variables. To get such a sample of three from a population of 5, you would give each member of the population the same probability of being chosen each time. That allows the possibility that the same member is selected on each of those three times.
The example illustrates "random sampling without replacement". The sampling method relevant to the Central Limit Theorem is "random sampling with replacement".
In Figure 6-8, you see a plot of actual population distribution of bank tellers' earnings and also sampling distribution. My question is how such plots are created in real life.
If there is data for an entire finite population then, in theory, a sampling distribution for the mean value of a sample of size K can be computed by computing the probability for each possible value of that mean. This can be done for random sampling with replacement or for random sampling without replacement.
If the data itself is only a sample from a larger population, then there is no unique answer to your question. People use various ways of estimating the sample distribution.
The sampling distribution plot could be created, say, using data of only 30 tellers, and then 'mathematically' a lot of samples could be created with n=25, total samples would be = 30(nCr)25 = 142506. This would give bell shaped distribution which could be approximated using normal distribution.
You are describing how to create a sampling distribution (for what?) where the sampling method is random sampling without replacement. The resulting curve might be approximately bell shaped. For large populations and small sample sizes, random sampling without replacment gives similar results to random sampling with replacement.
You'll find it difficult or impossible to interpret statistical physics in the language of mathematical statistics and probability theory.
The develoment of statistical physics preceded modern mathematical probability theory. Statistical physics has its own jargon. Expositions of statistical physics use terminology developed before the modern concept of a probability space. When it comes to defining the set of outcomes, expositions of statistical physics are vague. For example, if a graph shows "The distribution of molecular velocities", mathematical probability theory asks what random variable the distribution describes. Does it refer to selecting a molecule at random from a container of gas? The physical picture is that molecules are colliding and changing velocities, so we don't get a unique velocity by selecting one molecule. We have to pick both a molecule and a specific time. In what finite time interval are we to make the choice?
Instead of thinking of "the distribution of molecular velocities" as a probability distribution, statistical physics tends to view such a graph as the histogram of data for the entire population of molecules. However, there is the difficulty that a given molecule has a changing velocity. So you must pretend that the distribution is a snapshot taken at one instant of time if you want to think of it as being analogous to data about the income of bank tellers.
And is Cp the modal speed?
I think it is, just from looking at the graph.
PainterGuy
I'm sorry that I couldn't get back to you earlier.
Stephen Tashi said:
I don't know what that example is supposed to show. It's a poor choice for showing anything related to the Central Limit Theorem because the central limit theorem deals with samples of identically distributed random variables. To get such a sample of three from a population of 5, you would give each member of the population the same probability of being chosen each time. That allows the possibility that the same member is selected on each of those three times.
The example illustrates "random sampling without replacement". The sampling method relevant to the Central Limit Theorem is "random sampling with replacement".
Thank you for pointing this out.
I was thinking of a scenario and not sure if the answer is simple, and if it's not then no need to help me with it! :) Suppose, I'm trying to sample a population of 50 with a sample size 5. I'm trying to sample the population along with another guy; two persons are involved in sampling simultaneously.
If it was only me doing the sampling alone, it would be easy and I could use either 'sampling with replacement' or 'sampling without replacement'. But when sampling is done by both of us, a member from the population is chosen at the same time, wouldn't it create a conflict? How is such a situation is handled in such scenarios and which one of the two, 'sampling with replacement' or 'sampling without replacement', should be chosen as a more practical and sensible methodology?
Note to self:
The combination formula nCr gives you a number of unique combinations without repetition. Use the following formula to find number of combinations with repetition.
1: https://www.statisticshowto.datasciencecentral.com/sampling-with-replacement-without/
2: https://sites.math.northwestern.edu/~mlerma/courses/cs310-05s/notes/dm-gcomb
3: https://keisan.casio.com/exec/system/1223622559
Stephen Tashi said:
Instead of thinking of "the distribution of molecular velocities" as a probability distribution, statistical physics tends to view such a graph as the histogram of data for the entire population of molecules. However, there is the difficulty that a given molecule has a changing velocity. So you must pretend that the distribution is a snapshot taken at one instant of time
Thank you. Actually, I had a question and also wanted to add more content to complete this topic since it's related to the probability but then I thought it'd better to start a new thread.
Stephen Tashi said:
I don't know what you are denoting by "r". In my notation, one must not only consider the case ##X_1=1,\ X_2 = 3## but also the case ##X_1=3,\ X_2 = 1##. So there are more than three probabilities to be added up.
I'm still struggling with this part. I understand that it's quite frustrating but if it's possible, help me out with this. I'm just curious. In case, you wanted to have a look on the original problem: https://imagizer.imageshack.com/img922/6692/uuUrm0.jpg
"r" is the number of late employees.
You are using Y as a random variable to expresses lateness of of employees from a data set of 10 employees. If Y=4, it'd give you probability of 4 employees being late.
X1 and X2 are both random variables expressing the probability of employees being late out of total 5 employees from each data set.
Below I'm going to repeat some parts from earlier posts.
Stephen Tashi said:
It is correct that a random variable ##Y## that has distribution ##B(10,p)## can expressed in terms of two independent random variables ##X_1, X_2## each of which has distribution ##B(5,p)##. The way to do this is ##Y = X_1 + X_2##. In your example, this says the data ##Y## from two days of business with 5 employees is computed by adding the lateness data from each of the two days. We get ##Y = 4## in more ways than getting ##X_1 =2## and ##X_2= 2##. We also get ##Y = 4## from ##X_1 = 0,\ X_2 =4## and ##X_1=1,\ X_2 = 3##, etc. So to compute the probability that ##Y = 4## we must add up the probabilities of all these possibilities. Performing that calculation is called taking the convolution of the distributions of ##X_1## and ##X_2##.
Can we add up the values individually to get an idea? Let me try.
When n=10, the probability for Y=4 is 0.2508 . If I understand you correctly, it means that the following values (X1=0, X2=4), (X1=1, X2=3), (X1=2, X2=2) should be used from the case n=5. I tried to do it but it didn't work for me. The answer should have been 0.2508 instead of 1.9796.
(X1=0 + X2=4) + (X1=4 + X2=0) + (X1=1 + X2=3) + (X1=3 + X2=1) + (X1=2 + X2=2) = (0.0778 + 0.0768) + (0.0768 + 0.0778) + (0.2592 + 0.2304) + (0.2304 + 0.2592) + (0.3456 + 0.3456) = 1. 9796
Code:
n = 5; p = 0.4
B(5,p)
r Probability
0 0.0778
1 0.2592
2 0.3456
3 0.2304
4 0.0768
5 0.0102 n = 10; p = 0.4
B(10,p)
r Probability
0 0.0060
1 0.0403
2 0.1209
3 0.2150
4 0.2508
5 0.2007
6 0.1115
7 0.0425
8 0.0106
9 0.0016
10 0.0001
PainterGuy said:
. But when sampling is done by both of us, a member from the population is chosen at the same time, wouldn't it create a conflict?
I don't detect a mathematical question in that. It sounds like a question about psychology or sociology.
(X1=0 + X2=4) + (X1=4 + X2=0) + (X1=1 + X2=3) + (X1=3 + X2=1) + (X1=2 + X2=2) = (0.0778 + 0.0768) + (0.0768 + 0.0778) + (0.2592 + 0.2304) + (0.2304 + 0.2592) + (0.3456 + 0.3456) = 1. 9796
Use notation that distinguishes between an outcome versus the probability of an outcome.
"##X_1 = 0##" is an outcome, not a probability.
Don't use the symbol "+" to mean the word "and". You can use "##\land##" to abbreviate "and".
##Pr( X_1 = 0 \land X_2 = 4) = Pr(X_1 = 0) Pr(X_2 = 4)##, not ## Pr(X_1 = 0) + Pr(X_2 = 4)##
sysprog and PainterGuy
Thank you! I really appreciate your help.
Stephen Tashi said:
I don't detect a mathematical question in that. It sounds like a question about psychology or sociology.
I didn't put forward my problem properly so let me try again.
Suppose, I'm trying to sample a population of 50 with a sample size 5. I'm trying to sample the population along with another guy; two persons, A and B, are involved in sampling simultaneously.
If it was only me doing the sampling alone, it would be easy and I could use either 'sampling with replacement' or 'sampling without replacement'. But when sampling is done by both of us, a member from the population is chosen at the same time.
Suppose there is only one "Michael" and "John" in the population of 50, and sampling with replacement is being done.
When John is selected by Person A, the Person B wouldn't have a chance of selected John. The same goes for Michael. This situation would happen for each member of the population.
How is such a situation is handled in such scenarios and which one of the two, 'sampling with replacement' or 'sampling without replacement', should be chosen as a more practical and sensible methodology?
Stephen Tashi said:
Use notation that distinguishes between an outcome versus the probability of an outcome.
"##X_1 = 0##" is an outcome, not a probability.
Don't use the symbol "+" to mean the word "and". You can use "##\land##" to abbreviate "and".
##Pr( X_1 = 0 \land X_2 = 4) = Pr(X_1 = 0) Pr(X_2 = 4)##, not ## Pr(X_1 = 0) + Pr(X_2 = 4)##
I got the correct number now. But how do interpret the result.
In post #11 the following was said.
Initially it was said that n=5 is total number of employees and r is number of late arrivals. If n=10 and r=10, how do these relate to the original number of employees and late arrivals?
Originally, the maximum possible number of late arrivals was 5, as is shown. Assume that the data set was collected over a period of two weeks. Suppose that the owner collects another independent data set over the period of two more weeks. Now there are two independent data sets, each collected separately over the period of 2 weeks, but conveying the same information.
We notice that for n=5, the probability for four late arrivals, i.e. r=4, is 0.0768. But when n=10, the probability for r=4 is 0.2508. The difference is 0.2508 - 0.0768 = 0.174. For n=10, the probability of 4 late arrivals is more.
Complete problem: https://imagizer.imageshack.com/img922/6692/uuUrm0.jpg
In our calculation of Y=4, the probability was found to be 0.25083. Obviously,it provides the probability of 4 employees being late over the period of 4 weeks. What does the case
mean? It'd mean that if you look at the clock-in data of first 2 weeks, and also at the clock-in data of another 2 weeks, the probability of finding total of 7 instances of being late; we can say that n=10 gives us maximum number of instances of being late although the number of employees is still 5. I hope it makes sense.
sysprog
PainterGuy said:
How is such a situation is handled in such scenarios
I don't know what you mean specifically by "handled". Are you asking how to compute a sampling distribution?
To talk about a sampling distribution, you must first specified what statistic is being sampled. One can compute statistics from a sample in various ways. For example, from a sample of 5 persons, one could record the height of the tallest of those 5 persons ( an example of "order statistics"). Or one could compute the average height of those 5 persons, or one could compute how many of those 5 persons wear suspenders.
As I understand your example, we have two investigators, I1, I2. They draw a sample of 5 persons from a population of 30 as follows. I1 selects 1 person from the 30. Then I2 selects 1 person from the 29 persons remaining. The the process is repeated on the original population of 30 until each investigator has selected 5 people.
Intuitively, I see no reason why this changes the probability of I2 getting a particular sample. For example assume the persons are designated by numbers 1,2,...30. When I2 does random sampling with replacement by himself there is a certain probability that I2 selects a multi-set such as {1,1,15, 18, 25, }. Does the probability of getting that multi-set change when I1 is picking a person before I2 picks a person?
and which one of the two, 'sampling with replacement' or 'sampling without replacement', should be chosen as a more practical and sensible methodology?
There is no general rule.
I got the correct number now. But how do interpret the result.
The result is ##Pr(X_1 + X_2 = 4)##.
In post #11 the following was said.
...
The way you are forming questions in post #11 is incoherent because it fails to distinguish between the concept of data and the concept of a probability distribution.
In post #11 you are referring to the collection of data. The calculation you are asking about is done from the values in a probability distribution. To repeat advice from previous posts, a set of data representing samples from a probability distribution is not the same thing as the probability distribution.
A given binomial distribution such as ##B(70,p)## can be used to model different situations. ##B(70,p)## might model the probability that there are ##k## latenesses among 70 employees on 1 day. It might also model the lateness of 1 employee over a period of 70 days. Or it might model the total latenesses of 5 employees over 14 days.
One interpretation of the probability distribution ##B(5,p)## is that it gives the probability that ##k## employees will be late on 1 given day. The possible values of ##k## are 0,1,2,3,4,5.
If you are talking about 2 weeks of data about lateness , then we consider both ##k## , the number of employees late on a given day and also ##d##, the number of days where ##k## employees were late. The possible values of ##d## are 0,1,2,...14. The data concerns the frequency of latenesses. Frequencies are not the same concept as probabilities.
Apparently you are imagining that 2 weeks worth of data are taken and that the data is used to estimate a probability distribution. Ignoring the order of the 14 days, two weeks worth of data can be represented by a set of pairs of numbers of the form ##(k,d)##. It might turn out that the data consists of one pair of numbers such as (2,14). Or it might turn out that we need as many as 6 pairs of numbers such as { (0,7),(1,1),(2,1),(3,1)(4,1)(5,3)}. That data could be used to estimate the probability for ##k## employees being late on 1 given day. The procedure for estimation can be done in different ways.If we take ##B(5,p)## as the probability distribution for ##k## employees being late on 1 given day and assume each day is an independent experiment then we can compute the probability distribution for x total lateness in D total days. This done by taking the "D-fold convolution" of ##B(5,p)##. Because of the special nature of the binomial distribution, the D-fold convolution of ##B(5,p)## is ##B(5D,p)##. Those calculations have nothing to do with specific data.
The binomial distribution is special. Let ##f(K)## be a non-binomial distribution for the probability that ##k = 0,1,2,..K## employees are late on a 1 given day. It is not, in general, true that the D-fold convolution of ##f(K)## is ##f(KD)##. Constraints about the total number of employees and the total number of days might be critical if we use a non-binomial distribution to model what happens on 1 day.
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sysprog and PainterGuy | 13,847 | 57,079 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-10 | latest | en | 0.935656 |
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BBC News
watch One-Minute World News
Last Updated: Saturday, 9 September 2006, 05:07 GMT 06:07 UK
Indian town buries bomb victims
A victim of a bomb blast in India is buried
Victims were buried where they had prayed for the dead
Muslims in India have begun burying their relatives after bomb attacks near a mosque killed at least 37 people.
Police patrolled the city of Malegaon overnight, but a curfew was lifted on Saturday morning with no reports of any violent reaction to the attacks.
The city, north-east of Mumbai, has a history of riots between its Hindu minority and the Muslim majority.
Three bombs exploded near the mosque and a graveyard on a feast day when Muslims pray for the dead.
More than 125 others were hurt in the explosions. The hospital in the western Indian town said 20 of them were in a serious condition.
No group has said it planted the bombs, which were reported to have been attached to bicycles, and police have not named any suspects.
"There's a high alert across the state," said PS Pasricha, director-general of police in Maharashtra state.
"The motive appears to be to create panic and make Hindus and Muslims fight with each other."
A message from the office of Prime Minister Manmohan Singh said he "appealed for peace and communal harmony".
Within 24 hours of the blasts, Muslims returned to the scene of the carnage to bury the victims in the mosque graveyard where they had earlier prayed for the dead.
Beggars killed
Survivors recalled the horror that followed the bomb attacks.
"People were stunned. There was blood all over," said Nadeem Shaikh.
"Children tried to run out and many were trampled. Many beggars waiting outside for alms were also killed," he told the Associated Press news agency
Wrecked bicycle after a bomb in India
There were reports that the bombs were attached to bicycles
At a city hospital Dr Saeed Farami said he had treated more than 40 patients through the night.
"Most of those who died had head injuries and others had abdomen, chest and face wounds," he said.
With many people critically injured there are fears that the death toll could rise.
Sectarian tension
The blasts came days after Prime Minister Singh told the country there were intelligence warnings of attacks.
Only two months ago a series of explosions killed more than 180 people on the train networks of Mumbai, the capital of Maharashtra state.
Muslim-majority Malegaon, which has a population of about 500,000 and a large community of weavers, is a town with a history of communal rioting.
Earlier this year, the police recovered a large cache of arms and explosives from the area.
In October 2001, violence in protest at the US attacks on Afghanistan left a total of 12 people dead in the town.
Mourners attend the victims' funerals
In pictures: India blasts
08 Sep 06 | In Pictures
Eyewitness: Blast carnage in Malegaon
08 Sep 06 | South Asia
Indian town tense after protest
29 Oct 01 | South Asia
Anti-US protesters killed in India
27 Oct 01 | South Asia
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http://www.ecmag.com/section/your-business/perception-percent?qt-issues_block=0 | 1,508,701,826,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825436.78/warc/CC-MAIN-20171022184824-20171022204824-00185.warc.gz | 424,230,494 | 15,209 | In an attempt to simplify complex mathematical equations, the summary value for harmonics called total harmonic distortion (THD) is derived from the individual harmonic values. The individual harmonic values are a magnitude (and phase angle) at the frequencies that are integer multiples of the fundamental frequency. In North America, these are 120 Hz, 180 Hz, 240 Hz, 300 Hz, and so on. By taking each magnitude, squaring, summing, taking the square root, and dividing by the magnitude at the fundamental frequencies, we end up with THD.
Whereas the math works for both voltage and current, it is only the voltage THD that should be normally used as a measure of the quality of electrical supply. A voltage harmonic distortion level of less than 3 percent at the service entrance (point of common coupling) is generally considered acceptable, though individual systems may vary. Measured at an individual piece of equipment, such as an adjustable-speed drive, this value may be a couple of percentage points higher. On the neutral conductor of a wye circuit, it is possible to have the VTHD greater than 100 percent, as there may be larger third or other triplen harmonics magnitudes than the fundamental itself, since the fundamental waveforms of the three-phase conductors should cancel out if balanced. (See Figures 1 and 2.)
Concerning the current, the use of the THD can be very misleading. Current is like the water flowing through a pipe. If you have a 5-inch pipe that has the capacity to flow 1,000 gpm at 60 psi, what would be the effect if you had just 75 gpm leaking out of it? The people in the neighborhood would still have plenty of water available to them, despite the system capacity’s 3 percent loss. But a 75-gpm leak in a 1-inch pipe with a 100 gpm at 60 psi is quite significant. Water flow on the upper floors of houses could be reduced to trickles.
The same concept is true for harmonic current, which causes unwanted losses in transformers, motors and other electromagnetic equipment. These losses also go up as the square of the harmonic number increases. The effect of the 11th harmonic (11 x 11 = 121) would be 12 times worse than the third harmonic (3 x 3 = 9).
What matters is the amount of harmonic current relative to the circuit’s current capacity, usually measured as the short-circuit current (SCC), which is how much current could be drawn from the system if the load were a short circuit. If the circuit has SCC of 30A, the total current is 1.5A and the harmonic current is 0.5A, the resulting 50 percent THD is basically meaningless. We are hardly tickling the systems capacity. However, take that same circuit with 24A total current of which 8A is harmonic current, then this 50 percent THD level is quite significant. If it is made up of higher harmonics, such as 13th, 19th or even 25th, then the eddy current and other losses in transformers and motors may result in significant derating of their capacity to do useful work.
IEEE Std 519-1992, IEEE Recommended Practices and Requirements for Harmonic Control in Electrical Power Systems has a more detailed set of recommended limits for VTHD as well as individual harmonic current magnitudes based on the short-circuit current capacity, as measured at the point of common coupling to the utility service. It is also important to monitor the trending of the harmonic levels, as well as which harmonics are contributing to the change. If the seventh harmonic suddenly gets much larger in steady state levels, and no six-pole converters (such as ASDs) have been added to the system, look for a system resonance condition that is magnifying such, perhaps from a change in power-factor capacitors on the system. Such resonance conditions can create overvoltages that may damage equipment.
So the next time someone starts getting all excited about a current THD of 53 percent, remember which percents are meaningful, and which are just a misperception. EC
BINGHAM, a contributing editor for power quality, can be reached at 732.287.3680. | 884 | 4,031 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-43 | latest | en | 0.941373 |
https://www.calculushowto.com/types-of-functions/continuous-function-check-continuity/#Cvar | 1,660,056,116,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570977.50/warc/CC-MAIN-20220809124724-20220809154724-00792.warc.gz | 624,537,916 | 29,163 | # Continuous Function / Check the Continuity of a Function
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## 1. What is a Continuous Function?
Watch the video for an overview and examples, or read on below:
Continuous Functions
This function is continuous.
In layman’s terms, a continuous function is one that you can draw without taking your pencil from the paper. If you have holes, jumps, or vertical asymptotes, you will have to lift your pencil up and so do not have a continuous function.
If your function jumps like this, it isn’t continuous.
More formally, a function (f) is continuous if, for every point x = a:
1. The function is defined at a. In other words, point a is in the domain of f,
2. The limit of the function exists at that point, and is equal as x approaches a from both sides,
3. The limit of the function, as x approaches a, is the same as the function output (i.e. the y-value) at a.
## Order of Continuity: C0, C1, C2 Functions
Order of continuity, or “smoothness” of a function, is determined by how that function behaves on an interval as well as the behavior of derivatives.
## C0 Function
A C0 function is a continuous function. More specifically, it is a real-valued function that is continuous on a defined closed interval .
This simple definition forms a building block for higher orders of continuity.
## C1 Function
A C1 function is continuous and has a first derivative that is also continuous.
## C2 Function
A C2 function has both a continuous first derivative and a continuous second derivative.
Note here that the superscript equals the number of derivatives that are continuous, so the order of continuity is sometimes described as “the number of derivatives that must match.” This is a simple way to look at the order of continuity, but care must be taken if you use that definition as the derivatives must also match in order (first, second, third…) with no gaps. For example, let’s say you have a continuous first derivative and third derivative with a discontinuous second derivative. There are two “matching” continuous derivatives (first and third), but this wouldn’t be a C2 function—it would be a C1 function because of the missing continuity of the second derivative.
## Modeling with Order of Continuity
Continuity in engineering and physics are also defined a little more specifically than just simple “continuity.” For example, this EU report of PDE-based geometric modeling techniques describes mathematical models where the C0 surfaces is position, C1 is positional and tangential, and C3 is positional, tangential, and curvature. Which continuity is required depends on the application. For example, modeling a high speed vehicle (i.e. an airplane) needs a high order of continuity compared to a slow vehicle.
## 2. Different Types of Continuous
If a function is simply “continuous” without any further information given, then you would generally assume that the function is continuous everywhere (i.e. the set of all real numbers from -∞ to + ∞). However, sometimes a particular piece of a function can be continuous, while the rest may not be.
• Continuous on an interval: A function f is continuous on an interval if it is continuous at every point in the interval. For example, you could define your interval to be from -1 to +1. As long as the function is continuous in that little area, then you can say it’s continuous on that specific interval.
• Continuity at an endpoint: There are two possible endpoints for a function: all the way to the left (in the far negative direction), and all the way to the right (in the far positive direction). Assuming the limit exists, “continuity at an endpoint” means that the function is continuous from the right (for the left endpoint) or continuous from the left (for the right endpoint).
## Absolutely Continuous
Absolutely continuous real-numbered functions are those functions for which the Fundamental Theorem of Calculus (FTC) holds [1]. In other words, absolute continuity identifies which functions can be antiderivatives: a function on a closed, bounded interval is absolutely continuous on that interval if it is also an antiderivative over that same interval [2].
These functions have the “smoothest” type of continuity, followed by uniform continuity and then plain old continuity. All absolutely continuous functions are continuous, but the converse is not true.
Absolutely continuous functions and random variables are related to each other in the following way: A real-valued random variable X is absolutely continuous if its distribution function FX is absolutely continuous [3]
## Formal Definition
The formal definition is frequently used in real analysis, particularly for proving the Fundamental Theorem of Calculus for the Lebesgue Integral [4].
An absolutely continuous function, defined on a closed interval, has the following property. The property is based on a positive number ε and its counterpart, another positive number δ.
1. Take the interval for which we want to define absolute continuity, then break it into a set of finite, nonoverlapping intervals. The lengths of these intervals have a sum less than δ,
2. Next, consider the absolute values of the differences in the function values at the ends of the intervals; The sum over these intervals is less than ε [5].
In summation notation, we can state the above as: when a finite sequence of non-overlapping intervals satisfies:
then
Where (xk, yk) are the non-overlapping subintervals.
## Examples and Properties of Absolutely Continuous Functions
Every convex function and every continuously differentiable function is absolutely continuous [3].
Given a real-valued absolutely continuous function, the following properties hold [6]:
• cf, where c ∈ ℝ
• f + g
• fg
• 1/f, if f(x) ≠ 0 for every x ∈ [a, b]
• |f|.
A few specific examples: The Lipschitz function is absolutely continuous; The Cantor function, is not (although it is continuous everywhere) [7]. The function tan(x) is neither uniformly continuous nor absolutely continuous on the interval [0, π/2].
## Absolutely Continuous: References
[1] Heil, C. Real Analysis Lecture notes. 3.5. Abs. cont. and singular functions. Retrieved May 4, 2021 from: https://people.math.gatech.edu/~heil/handouts/ac.pdf
[2] 7.4 Abs. Cont. & Singular Functions. Retrieved May 4, 2021 from: https://www.math.lsu.edu/~rich/Absolute_Continuity
[3] Hill, T. & Berger, A. (2015). An Introduction to Benford’s Law. Princeton University Press.
[4] Pouso, R. (2012). A simple proof of the Fundamental Theorem of Calculus for the Lebesgue integral. Retrieved May 5, 2021 from: https://arxiv.org/pdf/1203.1462.pdf
[5] McGraw-Hill Dictionary of Mathematics, 2/E. (2002). McGraw-Hill.
[6] World Scientific. (2014). Problems and proofs in real analysis. pp. 314-352. Retrieved May 5, 2021 from: https://www.worldscientific.com/doi/10.1142/9789814578516_0013
[7] Real Analysis January 9, 2016 Chapter 6. Differentiation and Integration. Retrieved May 4, 2021 from: https://faculty.etsu.edu/gardnerr/5210/Beamer-Proofs/Proofs-6-5.pdf
## Left Continuous Function
A left-continuous function is continuous for all points from only one direction (when approached from the left).
It is a function defined up to a certain point, c, where:
1. The function is defined on an closed interval [d, c], lying to the left of c,
2. The limit at that point, c, equals the function’s value at that point.
The following image shows a left continuous function up to the point x = 4:
Note how the function value, at x = 4, is equal to the function’s limit as the function approaches the point from the left.
## Formal Definition of a Left-Continuous Function
Formally, a left-continuous function f is left-continuous at point c if
limxc--f(x) = f(c).
In other words, f(x) approaches c from below, or from the left, or for x < c (Morris, 1992). The right-continuous function is defined in the same way (replacing the left hand limit c- with the right hand limit c+ in the subscript).
## Right Continuous Function
A right continuous function is defined up to a certain point. The following image shows a right continuous function up to point, x = 4:
This function is right continuous at point x = 4. The limit at x = 4 is equal to the function value at that point (y = 6). This is equal to the limit of the function as it approaches x = 4.
Note that the point in the above image is filled in. On a graph, this tells you that the point is included in the domain of the function. If the point was represented by a hollow circle, then the point is not included in the domain (just every point to the right of it, in this graph) and the function would not be right continuous.
The point doesn’t exist at x = 4, so the function isn’t right continuous at that point.
Note that this type of continuity is defined for a point, not for an entire function.
## More Formal Definition of a Right Continuous Function
The reason why the function isn’t considered right continuous is because of how these functions are formally defined. Two conditions must be true about the behavior of the function as it leads up to the point:
1. exists: The limit of functions must exist at the point. The + sign above the “a” means that the point is being approached from the positive end of the number line; In other words, it’s approaching from the right.
2. The right hand limit leading up to the point a, must equal the limit of the function at that point.
In the second example above, the circle was hollowed out, indicating that the point isn’t included in the domain of the function. As the point doesn’t exist, the limit at that point doesn’t exist either.
## These Aren’t Really Continuous Functions!
The label “right continuous function” is a little bit of a misnomer, because these are not continuous functions. In order for a function to be continuous, the right hand limit must equal f(a) and the left hand limit must also equal f(a). The definition for a right continuous function mentions nothing about what’s happening on the left side of the point. The function may be continuous there, or it may not be. The only way to know for sure is to also consider the definition of a left continuous function.
## Uniformly Continuous Function
A uniformly continuous function on a given set A is continuous at every point on A. The way this is checked is by checking the neighborhoods around every point, defining a small region where the function has to stay inside.
More formally, this is stated as:
A function f : A → ℝ is uniformly continuous on A if, for every number ε > 0, there is a δ > 0; whenever x, y ∈ A and |x − y| < δ it follows that |f(x) − f(y)| < ε.
What that formal definition is basically saying is choose some values for ε, then find a δ that works for all of the x-values in the set. If the same values work, the function meets the definition.
## Graph of a Uniformly Continuous Function
Graphically, look for points where a function suddenly increases or decreases curvature. The definition doesn’t allow for these large changes; It’s very unlikely you’ll be able to create a “box” of uniform size that will contain the graph. The function might be continuous, but it isn’t uniformly continuous.
The uniformly continuous function g(x) = √(x) stays within the edges of the red box. The function f(x) = 1/x escapes through the top and bottom, so is not uniformly continuous. Image: Eskil Simon Kanne Wadsholt | Wikimedia Commons.
The game here is to try and find a uniform box of height x width = 2ε x 2 δ that, when moved, will keep the graph contained within the edges of the box. So, in a way, the “uniform” part of the definition refers to a “box of uniform size”.
## Properties of a Uniformly Continuous Function
These functions share some common properties.
• All of these functions are bounded on a closed interval [a, b] and will achieve a maximum in the set (a, b).
• Every uniformly continuous function is also a continuous function. However, not all continuous functions are uniformly continuous. Therefore, you can think of a these function as ones that are “more” continuous.
• They may or may not be differentiable. Uniform continuity doesn’t necessarily imply differentiability.
## 3. List of Continuous Functions
All of the following functions are generally continuous:
## 5. How to check for the continuity of a function
There are a few general rules you can refer to when trying to determine if your function is continuous. For some functions, you need to do a little detective work. (For a more rigorous approach, see: Continuity Test).
Sin(x) is an example of a continuous function.
Step 1: Draw the graph with a pencil to check for the continuity of a function. If your pencil stays on the paper from the left to right of the entire graph, without lifting the pencil, your function is continuous. In other words, if your graph has gaps, holes or is a split graph, your graph isn’t continuous.
Step 2: Figure out if your function is listed in the List of Continuous Functions. If it is, then there’s no need to go further; your function is continuous.
Step 3: Check if your function is the sum (addition), difference (subtraction), or product (multiplication) of one of the continuous functions listed in Step 2. If it is, your function is continuous. For example, sin(x) * cos(x) is the product of two continuous functions and so is continuous.
Step 4: Check your function for the possibility of zero as a denominator. The ratio f(x)/g(x) is continuous at all points x where the denominator isn’t zero. In other words, there’s going to be a gap at x = 0, which means your function is not continuous.
That’s it!
## Definition of a Continuous Variable
A continuous variable has an infinite number of potential values.
It’s the opposite of a discrete variable, which can only take on a finite (fixed) number of values.
Watch the video or read on below:
Discrete vs continuous variables
## Continuous Variable Range
A continuous variable doesn’t have to include every possible number from negative infinity to positive infinity. In most cases, it’s defined over a range. For example, the range might be between 9 and 10 or 0 to 100. Even though these ranges differ by a factor of 100, they have an infinite number of possible values.
As an example, let’s take the range of 9 to 10. Possible continuous variables include:
• 9, 9.01, 9.001, 9.051, 9.000301, 9.000000801
.
## Examples of Continuous Data
A few examples:
• Time it takes you to run the 100 m dash. Watch any timed sporting even and you’ll see times reported to the tiniest hair width. At the time of writing, the world record is held by Usain Bolt, at 9.572 seconds.
• Your weight. Your bathroom scale probably rounds to one tenth of a pound, so you might weight 151.1 lbs. But do you really? You could weight 151.14, or 151.143. The possibilities for weight are endless.
• Money. There was a time in the distant past when money was finite (it was the number of bills and coins in circulation). With the advent of the banking system and cryptocurrency, “money” is now uncountable.
• Car prices. Sure, that Ford might cost you \$30k right now. But what about in 100 years? 1,000 years? The price of a car is going to go up and up. You could argue that cars will become obsolete in the future, and therefore there’s a point where car prices will max out. But someone, somewhere is going to have an ancient car worth millions or billions).
Heights and weights are both examples of quantities that are continuous variables.
## What is a Discrete Variable?
A discrete variable can only take on a certain number of values. In other words, they don’t have an infinite number of values. If you can count a set of items, then the variables in that set are discrete variables.
The opposite of a discrete variable is a continuous variable. Continuous variables can take on an infinite number of possibilities.
## What is a Discrete Variable? Examples
Some examples of discrete variables:
• Number of quarters in a purse, jar, or bank. Discrete because there can only be a certain number of coins (1, 2, 3, 4, 5…). Coins don’t come in amounts of 2.3 coins or 10 ½ coins, so it isn’t possible for there to be an infinite number of possibilities. In addition, a purse or even a bank is restricted by size so there can only be so many coins.
• The number of cars in a parking lot. A parking lot can only hold a certain number of cars.
• Points on a 10-point rating scale. If you’re graded on a 10-point scale, the only possible values are 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10.
• Ages on birthday cards. Birthday cards only come in years—they don’t come in fractions. So there are a finite amount of possibilities (presumably, about one hundred).
## Discrete random variables
Discrete random variables are variables that are a result of a random event. For example, the roll of a die. Discrete random variables are represented by the letter X and have a probability distribution P(X). If you flipped a coin two times and counted the number of tails, that’s a discrete random variable. It’s represented by the letter X. X in this case can only take on one of three possible variables: 0, 1 or 2 [tails].
## Continuous Variable Subtype: The Interval Variable & Scale
An interval variable is simply any variable on an interval scale. An interval scale has meaningful intervals between values. The intervals between points on the interval scale are the same. For example, the difference between 10°C and 20°C is the same as the difference between 40°F and 50° F.
An interval variable is a type of continuous variable. Technically (and this is really splitting hairs), the scale is the interval variable, not the variable itself. For example, the variable 102°F is in the interval scale; you wouldn’t actually define “102 degrees” as being an interval variable. That’s because on its own, it’s pretty meaningless.
## Problems with Zero
Dates are interval scale variables. For example, a century is 100 years long no matter which time period you’re measuring: 100 years between the 29th and 20th century is the same as 100 years between the 5th and 6th centuries. Although this seems intuitive, dates highlight a significant problem with interval scales: the zero is arbitrary. This means you have to be very careful when interpreting intervals. For example, in the A.D. system, the 0 year doesn’t exist (A.D. starts at year 1). However, some calendars include zero, like the Buddhist and Hindu calendars.
Arbitrary zeros mean that you can’t say that “the 1st millennium is the same length as the 2nd millennium.”
This leads to another issue with zeros in the interval scale: Zero doesn’t mean that something doesn’t exist. For example, just because there isn’t a year zero in the A.D. calendar doesn’t mean that time didn’t exist at that point. To the contrary, it must have, because there are years before 1 A.D. (B.C.!). Similarly, a temperature of zero doesn’t mean that temperature doesn’t exist at that point (it must do, because temperatures drop below freezing).
Arbitrary zeros also means that you can’t calculate ratios. Ratio scales (which have meaningful zeros) don’t have these problems, so that scale is sometimes preferred.
## Ratio Variable & Scale
Weight is measured on the ratio scale (no pun intended!).
A ratio scale has all the properties of an interval scale, plus a fixed natural zero.
Ratio data this scale has measurable intervals. For example, the difference between a height of six feet and five feet is the same as the interval between two feet and three feet. Where the ratio scale differs from the interval scale is that it also has a meaningful zero. Zero means that something doesn’t exist, or lacks the property being measured. For example, the zero in the Kelvin temperature scale means that the property of temperature does not exist at zero.
Other examples of the ratio scale:
• Age. The clock starts ticking when you are born, but an age of “0” technically means you don’t exist.
• Weight. At 0 pounds, you would weight nothing and therefore wouldn’t exist.
• Height. If you were 0″, you would have no height.
• Sales figures. Sales of zero means that you sold nothing and so sales didn’t exist.
• Quantity purchased. If you bought 0 items, then there were no quantities purchased.
• Time measured from the “Big Bang.”
## The Ratio Scale and Negative Numbers
As the “0” in the ratio scale means the complete absence of anything, there are no negative numbers on this scale.
## Invariant Under a Similarity Transformation
Data on a ratio scale is invariant under a similarity transformation, y= ax, a >0. For example, you could convert pounds to kilograms with the similarity transformation K = 2.2 P. The ratio stays the same whether you use pounds or kilograms. Although the ratio scale is described as having a “meaningful” zero, it would be more accurate to say that it has a meaningful absence of a property; Zero isn’t actually a measurement of anything—it’s an indication that something doesn’t have the property being measured. For example, 0 pounds means that the item being measured doesn’t have the property of “weight in pounds.”
## The Ratio Scale and Ratios
As the name suggests, we can create meaningful ratios between numbers on a ratio scale.
For example, a count of how many tests you took last semester could be zero if you didn’t take any tests. However, if you took two exams this semester and four the last semester, you could say that the frequency of your test taking this semester was half what it was last semester.
## Use in Calculus
Scales of measurement, like the ratio scale, are infrequently mentioned in calculus classes. But in applied calculus (a.k.a. in the real world), you likely be using them a lot. For example, economic research using vector calculus is often limited by a measurement scale; only those values forming a ratio scale can form a field (Nermend, 2009).
## References
Image: By Eskil Simon Kanne Wadsholt – Own work, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=50614728
Bogachev, V. (2006). Measure Theory Volume 1. Springer.
Carothers, N. L. Real Analysis. New York: Cambridge University Press, 2000.
Dartmouth University (2005). Continuity. Retrieved December 14, 2018 from: https://math.dartmouth.edu//archive/m3f05/public_html/ionescuslides/Lecture8.pdf
Guha, S. (2018). Computer Graphics Through OpenGL®: From Theory to Experiments. CRC Press.
Kaplan, W. “Limits and Continuity.” §2.4 in Advanced Calculus, 4th ed. Reading, MA: Addison-Wesley, pp. 82-86, 1992.
Larsen, R. Brief Calculus: An Applied Approach.
Morris, C. (1992). Academic Press Dictionary of Science and Technology. Elsevier Science.
Nermend, K. (2009). Vector Calculus in Regional Development Analysis. Comparative Regional Analysis Using the Example of Poland.
Ross, K. (2013). Elementary Analysis: The Theory of Calculus (Undergraduate Texts in Mathematics) 2nd ed. Springer.
Titchmarsh, E. (1964). The theory of functions, 2nd Edition. Oxford University Press.
Tseng, Z. (n.d.). Continuity. Article posted on PennState website. Retrieved December 14, 2018 from: http://www.math.psu.edu/tseng/class/Math140A/Notes-Continuity.pdf
CITE THIS AS:
Stephanie Glen. "Continuous Function / Check the Continuity of a Function" From CalculusHowTo.com: Calculus for the rest of us! https://www.calculushowto.com/types-of-functions/continuous-function-check-continuity/
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Need help with a homework or test question? With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Your first 30 minutes with a Chegg tutor is free! | 5,417 | 23,744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2022-33 | latest | en | 0.951696 |
null | null | null | null | null | null | Discussion: Name that genre! And…does it matter?
I don’t know about you, but I sometimes get genre-burned. I’ll pick up a book, thinking it’s one thing, and then be disappointed when it turns out to be something else. It’s not that the something else isn’t good, or even that I didn’t like the book. It’s that it wasn’t what I was expecting. And while I sometimes welcome the unexpected, like with a juicy plot twist, I find myself wishing sometimes that the book world as a whole — bookstores, bloggers, even authors sometimes — would try to be a tad more accurate with genre labeling.
Also, let me just throw this out there: Young Adult is not a genre. Nor is Middle Grade, Adult, or New Adult. Those are audiences. They encompass the age demographic a book is targeting. But they don’t tell you a thing about what the book is about, other than the relative age of the characters (give or take a few decades, in the case of Adult).
So let’s talk just a minute about genres, what defines them, and which ones tend to have an identity crisis.
These are highlights from the Goodreads definitions. Sometimes it’s just easier than trying to type it all out myself.
Science FictionScience fiction is a broad genre of fiction that often involves speculations based on current or future science or technology. Science fiction differs from fantasy in that, within the context of the story, its imaginary elements are largely possible within scientifically established or scientifically postulated laws of nature (though some elements in a story might still be pure imaginative speculation).
RomanceAccording to the Romance Writers of America, “Two basic elements comprise every romance novel: a central love story and an emotionally-satisfying and optimistic ending.” Both the conflict and the climax of the novel should be directly related to that core theme of developing a romantic relationship, although the novel can also contain subplots that do not specifically relate to the main characters’ romantic love.
Historical FictionHistorical fiction presents a story set in the past, often during a significant time period. In historical fiction, the time period is an important part of the setting and often of the story itself. Historical fiction may include fictional characters, well-known historical figures or a mixture of the two.
HorrorHorror fiction is fiction in any medium intended to scare, unsettle, or horrify the audience. Historically, the cause of the “horror” experience has often been the intrusion of a supernatural element into everyday human experience. Since the 1960s, any work of fiction with a morbid, gruesome, surreal, or exceptionally suspenseful or frightening theme has come to be called “horror”.
ContemporaryContemporary literature is literature with its setting generally after World War II.
These are just some of the biggies. There’s tons of genres and subgenres out there, and often authors like to mash them up. For example, romance can be incorporated into nearly all of these genres, which gives you Historical Romance, Paranormal Romance, etc. I think one of the reasons that genres get so muddled is that they’re not mutually exclusive at all. Contemporary is anything that takes place after World War II? Well, that could encompass pretty much everything (except Historical), couldn’t it? And obviously there’s tons of crossover between Science Fiction/Fantasy/Horror/Paranormal.
I think the problem happens when we get these main categories confused. I think the problem is twofold:
1) Certain genres are really popular, and everyone wants their book (or their client’s book, or their friend’s book) to be the next Big Thing. So they say it fits the genre, when in reality, it doesn’t. (I’m looking at you, Dystopian Fiction.)
2) Lots of books are really hard to classify, because the authors have mixed a bunch of genres together in a delicious cocktail of imagination. It’s a bit more understandable how these get confused.
So what’s the trick in figuring out how to classify what you’re reading? Just ask yourself a few questions:
1) What’s the setting? Is it past, present, future, or a made-up world? Is it based in reality, or could it plausibly happen in our reality, or is it in no way related to our reality? Does magic factor into it? Science? Is it based on a historical event that actually happened, or a historical event that might have happened if things were different?
2) If it is the future, what shaped the world? Was it a cataclysmic event? Government conspiracy? Aliens? Magic? Technological advancement? Just because it’s the future doesn’t automatically make it sci-fi or dystopian or post-apocalyptic. Look at why the world is the way it is, and that’s a big clue.
3) What’s the conflict? Is it about whether or not Jim and Sally will get together, or is it about whether or not Jim will save Sally’s a ghost, or is it about whether or not Jim will discover that he’s really a prince and the only one who can free Sally from the dragon? Granted, Jim and Sally may get together in all of these scenarios, but it’s only the main conflict in one of them.
Am I alone in caring about this? I’m not sure. Maybe you don’t care how something’s labeled; a good book is a good book. So what if you were expecting dystopian and got sci-fi instead? Or you wanted steampunk but wound up reading historical fiction? What’s the big deal?
But if you’re like me, it’s kind of like ice cream flavors. If I’m in the mood for chocolate and I get strawberry, I’m going to be disappointed. I like strawberry. Sometimes, all I want in the whole world is strawberry. But if I’m in the mood for chocolate, strawberry won’t cut it.
Here’s some examples of books I’ve seen miscategorized (a lot):
The Dark Unwinding by Sharon Cameron. I’ve heard this book described as Steampunk and Paranormal, but really it’s just Historical Fiction. The automatons in the story are things that actually existed during that time period (you can ask Sharon. It’s fascinating), and there’s no supernatural elements that defy scientific explanation.
What’s Left of Me by Kat Zhang. This one always gets called Dystopian or Sci-Fi. But really, if you look close, it’s neither. It’s a modern alternate reality. So really, it doesn’t fit into any of the above categories. Broadly, it can go under the Speculative Fiction umbrella, but none of the other terms really fit. So there’s really little wonder why bookstores want to label it as something else.
Defiance by C.J. Redwine. This book is a cornucopia of so many genres, it’s easy to see why people can’t seem to label it. I’ve actually had a few discussions with C.J. about what to call this book, and even she is at a bit of a loss. I’ve heard it called Steampunk, Dystopian, Fantasy, and Sci-Fi. It’s marketed as Fantasy Adventure, but there’s no magic (although there is a blind wingless subterranean dragon). What it actually is, I believe, is a Post-Apocalyptic Adventure. I think.
Miss Peregrine’s Home for Peculiar Children by Ransom Riggs. I always see this book on the Horror shelf, and it’s just not. It’s not designed to scare or horrify. It’s about magical powers and adventure. It’s Fantasy.
How about you? Do you long to sneak into bookstores and reshelve the books to more accurately reflect what’s in them? Or do you figure, hey, I don’t care why someone picked up the book, as long as they’re reading it? What books do you see commonly misclassified, and do you care?
The Plot (from Goodreads)
My Thoughts
The Plot
My Thoughts
Content Guide: Contains violence, sexual situations
Review: Fire by Kristin Cashore (@kristincashore)
The Plot
My Thoughts
My review of Graceling
Here’s how it works:
My Throwback this week is…
Ship of Magic by Robin Hobb
If you are a fan of fantasy, pirates, adventure, magic, and — oh yeah – dragons, this is a must-read.
| null | null | null | null | null | null | null | null | null |
https://fersch.de/beispiel?nr=Raute&nrform=GeoRautFlaecheUf&datei=1005RauteGeoRautFlaecheUf | 1,627,178,535,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151563.91/warc/CC-MAIN-20210725014052-20210725044052-00165.warc.gz | 262,455,144 | 4,616 | Geometrie-Viereck-Raute
Beispiel Nr: 05
$\text{Gegeben:}\\\text{Fläche} \qquad A \qquad [m^{2}] \\ \text{Diagonale e} \qquad e \qquad [m] \\ \\ \text{Gesucht:} \\\text{Diagonale f} \qquad f \qquad [m] \\ \\ f = \frac{2\cdot A}{ e}\\ \textbf{Gegeben:} \\ A=\frac{1}{3}m^{2} \qquad e=\frac{3}{4}m \qquad \\ \\ \textbf{Rechnung:} \\ f = \frac{2\cdot A}{ e} \\ A=\frac{1}{3}m^{2}\\ e=\frac{3}{4}m\\ f = \frac{2\cdot \frac{1}{3}m^{2}}{ \frac{3}{4}m}\\\\f=\frac{8}{9}m \\\\\\ \small \begin{array}{|l|} \hline A=\\ \hline \frac{1}{3} m^2 \\ \hline 33\frac{1}{3} dm^2 \\ \hline 3333\frac{1}{3} cm^2 \\ \hline 333333\frac{1}{3} mm^2 \\ \hline 0,00333 a \\ \hline 3,33\cdot 10^{-5} ha \\ \hline \end{array} \small \begin{array}{|l|} \hline e=\\ \hline \frac{3}{4} m \\ \hline 7\frac{1}{2} dm \\ \hline 75 cm \\ \hline 750 mm \\ \hline 7,5\cdot 10^{5} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline f=\\ \hline \frac{8}{9} m \\ \hline 8\frac{8}{9} dm \\ \hline 88\frac{8}{9} cm \\ \hline 888\frac{8}{9} mm \\ \hline 888888\frac{8}{9} \mu m \\ \hline \end{array}$ | 524 | 1,066 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2021-31 | longest | en | 0.109922 |
http://ibelitetutor.com/blog/tag/online-ib-tutors/ | 1,524,309,088,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945143.71/warc/CC-MAIN-20180421110245-20180421130245-00261.warc.gz | 154,833,758 | 14,958 | ## Applications of Integration
In my previous posts, we discussed Definite and Indefinite Integrations. Now we shall learn about Applications of Derivatives. Initially, we shall discuss “Area Under Curves”.
Area Under Curve-: If we want to calculate the area between the curves y=f(x) and y=g(x) then there are actually two cases-
First Case when Below is the figure showing this case
here area under these two curves
The second Case When Below figure shows this case
## Types of Functions-
IB Maths Tutors should give twenty-two hours for teaching functions and equations as per IBO recommendations. This is my third article on functions in the series of ib mathematics
IB Maths Tutors should give twenty hours in teaching functions and equations.This is my third article on functions in the series of ib mathematics
For the sake of a comprehensive discussion, some standard functions and their graphs are discussed here
For the sake of a comprehensive discussion, some standard functions and their graphs are discussed here.
1.Greatest Integer Function– This is an interesting function. It is defined as the largest integer less than or equal to x
y = [x].
For all real numbers, x, this function gives the largest integer less
than or equal to x.
For example: [1] = 1 [2.5] = 2 [4.7] = 4 [5.3] = 5
Beware! [-2] = -2 [-2.6] = -3 [-4.1] = -5 [-6.5] = -7
domain=R
range=Z
greatest integer function
## IB Tutors-useful pdfs
I am sharing few documents prepared by experienced IB Tutors
Below is a past year paper for ib mathematics
ib_elite_tutor.com
Below is an important worksheet for Differentiability of Functions
Worksheets_on_Differentiability_(22-09-15)
Below is an important worksheet for logarithms
Worksheets_on_Log
Below is an important worksheet for maxima and minima value of Functions
Worksheets_on_Maxima_&_Minima_(22-09-15)
worksheet of INDUCTION (PMI)
You can request PDF of any IB BOOK, we shall try to provide that to you for free
## How to Become a Good Learner in Classroom or in Online sessions
Every moment of our life offers something to learn.We should always be ready to grab the opportunity.In this post, I have tried to identify characteristics of a good learner in classroom as well as with online tutors
Positive Attitude- A positive attitude plays a very important part in improving our learning capacities.Actually, its the backbone of our learning capacities.If we believe that we can do it, we will surely be able to accomplish the work Read more
## Online IB Tuitions
In Live Online IB Tuitions, an online tutor and student are not in the same location they use the Internet to teach and learn remotely. The tutor communicates real-time with the student using Skype or Zoom (Voice over the Internet), a digital pen tablet, enabling tutor and student to interact through a shared whiteboard or paint page (on which both teacher and student can write using different paint tools).Online Tutors and students can interact in a fun and dynamic learning environment, choosing individually customized set of the online tools which best help the student achieve his/her academic target. It is like having a tutor sitting right in front of you and is extremely popular with today’s tech savvy younger generation.
IB Elite Tutor offers LIVE ONLINE TUTION CLASSES for INTERNATIONAL BACCALAUREATE (all Groups HL, SL), IGCSE, Cambridge A.I.C.E Diploma, International A-S, A, O- levels, SAT and SAT 2nd (for all subjects). WE have 7 years’ of ONLINE TUTORING experience. The service is ‘‘very effective’’ and helps the students from anywhere in India and the world to access the leading and finest of the tutors. Our online classes have stretched to countries like UK, UAE, USA, Singapore, Australia, Canada, India, Mexico, Spain, China Ecuador, Argentina, Hong Kong, Germany Switzerland, Sweden, Poland, Turkey, Oman, Bahrain, Netherlands, Egypt, Qatar and Japan. By every passing year, the number of our online students is increasing.
IB Elite Tutor’s team of Online tutors is highly experienced and well trained to give online tutoring. Our Live Online tutors have been selected from the highest performing tutors across the world. They have great track records of tutoring students online and bring the best of the International education system.
## Why join Live Online IB Tuitions?
• Best Quality Teachers. Our Online Tutors are highly experienced & well qualified IB well Trained teachers. The quality of IB online tutors is far more than IB Home Tutors and avail all the supporting material viz, past ten-year papers chapter wise notes, Chapter wise assignment sheets, chapterwise PPTs and solved past papers to all the students.
• It’s Effective. Way more effective than face-to-face home tuition.
• It’s Convenient. A student can access their tutor and have lessons anywhere and anytime.
• It’s Flexible. A student can fix the time for an online session according to his/her’s comfort and need and it can be done wherever the student happens to be – home, school, travel, university etc.
• It’s Simple. All you need is a broadband connection and a Skype account.
• It’s Safe. Online tutoring avoids the busy and hectic traffic travelling for both tutor and pupil
• It’s Fun. Many young people like online IB tuitions (to Face-to-Face) due to their life-long familiarity with computers. | 1,199 | 5,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-17 | longest | en | 0.873421 |
http://math.stackexchange.com/questions/27717/convolution-a-continuous-function/27725 | 1,469,485,247,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824395.52/warc/CC-MAIN-20160723071024-00248-ip-10-185-27-174.ec2.internal.warc.gz | 153,426,797 | 18,491 | # convolution a continuous function?
define $$h(x)=\int_0^{2\pi}f(x-y)g(y)dy=f*g(x)$$ if $f,g \in L^2$ are $2\pi$ periodic, show that h is continuous on $[0,2\pi)$
so let $x_n \to x$, then
$$|h(x)-h(x_n)|=|\int f(x-y)g(y)-\int f(x_n-y)g(y)|$$
i tried to manipulate this but couldnt get anywhere. I was also thinking of using Holder's inequality somehow but im not sure...
-
Note first that $|(f \ast g)(x)| \leq \Vert f \Vert_2 \, \Vert g\Vert_2$ by Hölder's inequality. Therefore convolution is a continuous bilinear map from $L^2([0,2\pi])$ to the space of bounded functions.
Since the space of continuous functions is dense in $L^2$, we may find a sequence of continuous $2\pi$-periodic functions such that $f_{n} \to f$ in $L^2$. Then the above inequality gives $f_{n} \ast g \to f \ast g$ uniformly on $[0,2\pi]$. Since a uniform limit of continuous functions is continuous, it suffices to prove that $f \ast g$ is continuous if $f$ is continuous.
If $f$ is continuous then $f$ is uniformly continuous (because $[0,2\pi]$ is compact), that is: for each $\varepsilon \gt 0$ there is $\delta \gt 0$ such that for all $|z - z'| \lt \delta$ we have $|f(z) - f(z')| \lt \varepsilon$. In other words, $\|f(\cdot - x) - f(\cdot - x_{n})\|_{\infty} \lt \varepsilon$.
Let $\varepsilon \gt 0$. For $n$ so large that $|x_{n} - x| \lt \delta$ we then have $|(f \ast g)(x_{n}) - (f \ast g)(x)| \leq \Vert f(\cdot - x) - f(\cdot - x_{n})\Vert_{\infty} \Vert g \Vert_{1} \lt \varepsilon \cdot \Vert g \Vert_1$ again by Hölder (note that $L^2([0,2\pi)] \subset L^1([0,2\pi])$ so that $\|g\|_{1}$ is finite). This shows that $f \ast g$ is continuous if $f$ is continuous and as argued in the second paragraph this shows that $f \ast g$ is continuous for all $f,g \in L^2([0,2\pi])$.
Added: To avoid confusion, let me mention that this argument is not circular. In order to show that the space of continuous functions is dense in $L^2$ one often uses convolution with a mollifying sequence. Of course, we should avoid that and this can be done without too much pain (the crux of the proof lies in showing that the step functions are dense in $L^p$ for $1 \leq p \lt \infty$), see e.g. Royden, Real Analysis, Proposition 8 of Chapter 4, on p.128 of the third edition.
-
I think the second part of the proof can be slightly simplified if we approximate continuous functions by smooth functions, and use the fact that a convolution with a smooth function is smooth, therefore continuous a fortiori. – timur Mar 23 '14 at 17:46
I believe I've found an alternative proof for this fact (using some arguably much-less-elementary facts).
Given $f,g\in L^2(S^1)$, we can write their Fourier series $(\hat{f_n}),(\hat{g_n})$, defined by $$f(x)=\sum_{n\in \mathbb{Z}} \hat{f_n} e^{inx}$$ and same for $g$.
Then by the convolution theorem, $h\equiv f\star g$ has Fourier coefficients $\hat{h_n}=2\pi\hat{f_n}\hat{g_n}$. It follows that $\{\hat{h_n}\}$ is absolutely summable, since
$$\sum_{n\in\mathbb{Z}} |\hat{h_n}|=\sum_{n\in\mathbb{Z}} 2 \pi |\hat{f_n}\hat{g_n}|\le 2\pi\langle f,g\rangle <\infty$$ by Parseval's identity.
Finally, the Lebesgue dominated convergence theorem gives
\begin{align} \lim_{x\to y} h(x) &= \lim_{x\to y} \sum_{n\in\mathbb{Z}} \hat{h_n}e^{inx} \\ &= \sum_{n\in\mathbb{Z}} \lim_{x\to y} \hat{h_n}e^{inx} \\ &= \sum_{n\in\mathbb{Z}} \hat{h_n}e^{iny} = h(y) \end{align}
since for all $x$, the sequence $\{\hat{h_n}e^{inx}\}$ is dominated in absolute value by the summable sequence $\{|\hat{h_n}|\}$. Hence, $h=f\star g$ is continuous.
- | 1,208 | 3,557 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2016-30 | latest | en | 0.849095 |
https://www.physicsforums.com/threads/line-integral-evaluation.789101/ | 1,674,773,159,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494826.88/warc/CC-MAIN-20230126210844-20230127000844-00139.warc.gz | 956,053,760 | 21,997 | # Line Integral Evaluation
NewtonApple
## Homework Statement
Evaluate the line integral $∫^{(1,0)}_{(0,1)} (x^2-y)dx + (y^2+x)dy$ along
(a) a straight line from (0,1) to (1,2);
(b) the parabola x=t, y=t2 + 1;
(c) straight lines from (0,1) to (1,1) and then from (1,1) to (1,2).
## Homework Equations
Equation of line: y = mx + c
## The Attempt at a Solution
(a) [/B]To convert given integral into one variable I used Equation of line
y=mx + c
where slope m = $\frac {y_2 - y_1}{x_2 - x_1} = \frac{2-1}{1-0}=1$
and y intercept c = 1
which gives us y = x+1
dy = dx
Thus given integral
$∫^{(1,0)}_{(0,1)} (x^2-y)dx + (y^2+x)dy$
becomes one variable x dependent
$= ∫^{1}_{0} (x^2-x-1)dx + (({x+1})^2+x)dx$
$= ∫^{1}_{0} (x^2-x-1)dx + (x^2+2x+1+x)dx$
$= ∫^{1}_{0} (x^2-x-1)dx + (x^2+3x+1)dx$
$= ∫^{1}_{0} (x^2-x-1+x^2+3x+1)dx$
$= ∫^{1}_{0} (2x^2+2x)dx$ ------------------ (1)
solving
$\frac{2x^3}{3} +\frac{2x^2}{2}$
applying limits gives
$\frac{2}{3} + 1 = \frac{2+3}{3} = \frac{5}{3}$
(b) need some hints for this part
(c) straight lines from (0,1) to (1,1) and then from (1,1) to (1,2).
from equation (1) above
$= ∫^{1}_{0} (2x^2+2x)dx$
solving
$\frac{2x^3}{3} +\frac{2x^2}{2}$
from (0,1) to (1,1)
we use only x: 0 to 1
we get $\frac{2}{3} + 1 = \frac{2+3}{3} = \frac{5}{3}$
from (1,1) to (1,2)
we use only x: 1 to 1
we get $\frac{2}{3} - \frac{2}{3} + 1 - 1 = 0$
hence adding both parts gives $\frac{5}{3}$
Please tell me am I on the right track?
Last edited:
Homework Helper
## Homework Statement
Evaluate the line integral $∫^{(1,0)}_{(0,1)} (x^2-y)dx + (y^2+x)dy$ along
(a) a straight line from (0,1) to (1,2);
Check the upper boundary, it must be (1,2). With that boundary, your answer to (a) is correct.
(b) the parabola x=t, y=t2 + 1;
What is t at the lower boundary x=0, y=1, and what is it at x=1, y=2?
Write x, dx, and y, dy in the integrand in terms of t, and integrate with respect to t.
(c) straight lines from (0,1) to (1,1) and then from (1,1) to (1,2).
from equation (1) above
$= ∫^{1}_{0} (2x^2+2x)dx$
You can not use eq (1). Separate the integral from(0,1) to (1,1) along the straight line connecting these points (a horizontal path) and then from (1,1) to (1,2) (a vertical path).
NewtonApple
ARaslan
a)
A vector parallel to the line segment from (0,1) to (1,2) is <1,1>. Therefore the vector equation for the line segment is r(t) = <0,1> + t<1,1> = <t, t +1>. So the parameter of the line integral is :
x =t
y = t+1
0 <= t <= 1.
dy/dt = 1
dx/dt = 1
so the integral becomes: ∫(t^2 - t - 1) + (t+1)^2 +t dt from 0 to 1
= ∫2t^2 + 2t dt from 0 to to 1 which evaluates to 5/3.
follow the same procedure for the rest of the section.
1)parameter
2)find dy/dt and dx/dt
3)plug in the x's and y's in term of t
4)plug in dy = dy/dt * dt and dx = dx/dt *dt
5) integrate from 0 to 1
note: You will have to divide C into to problems identical to A.
NewtonApple
Check the upper boundary, it must be (1,2). With that boundary, your answer to (a) is correct.
yes, you are right. It's a typo, upper bound should be (1,2).
ARaslan
The line integral is evaluated along a line segment C. It makes no sense to have limits in the integral.
NewtonApple
The line integral is evaluated along a line segment C. It makes no sense to have limits in the integral.
The problem is from Mathematical Methods for Physicists by Tai L. Chow, see the attached file.
#### Attachments
• image026.gif
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ARaslan
The problem is from Mathematical Methods for Physicists by Tai L. Chow, see the attached file.
Yes, those limits given are the exact same limits given by the line segments. My guess is that those limits are given for problem b. For problems A and C, ignore the given limits and focus on the limits given by the line segment
Homework Helper
Gold Member
2022 Award
The line integral is evaluated along a line segment C. It makes no sense to have limits in the integral.
No, it makes sense, but it in (a) and (c) it's redundant because the bounds are also given separately. You could equally well complain, instead, that the curve specifications should be:
(a) a straight line segment;
(b) a segment of the parabola x=t, y=t2 + 1;
(c) two straight line segments meeting at (1,1)
Does it really matter?
NewtonApple
So for part (b)
Evaluate the line integral $∫^{(1,2)}_{(0,1)} (x^2-y)dx + (y^2+x)dy$ along
(b) the parabola x=t, y=t2 + 1
x= t
dx = dt
$y= t^2 +1$
dy = 2tdt
when x=0 then t=0
when x=1 then t=1
when y=1 then t = imaginary
when y=2 then t= 1
Hence given integral
$∫^{(1,2)}_{(0,1)} (x^2-y)dx + (y^2+x)dy$
in terms of t is
$(t^2 -t^2 -1)dt + [(t+1)^2 + t]2tdt$
$(-1)dt + [(t^2 + 2t +1) + t]2tdt$
$∫^{1}_{0} [-1 + 2t^3 + 6t^2 +2t]dt$
$-t + \frac{2}{4}t^4 + \frac{6}{3}t^3 + \frac{2}{2}t$
after applying lower and upper bound we get
$-1 +\frac{1}{2} + 2 + 1$
$\frac{1}{2} + 2 = \frac{5}{2}$
Which doesn't seem right.
Homework Helper
Gold Member
2022 Award
when y=1 then t = imaginary
No. Try that again.
##(t^2 -t^2 -1)dt + [(t+1)^2 + t]2tdt##
You dropped an exponent.
NewtonApple
NewtonApple
When y=1 then t = 0
No. Try that again.
You dropped an exponent.
thx! yes and it should be
$(t^2 -t^2 -1)dt + [(t^2 + 1)^2 +1]2dt$
After solving and applying bounds we get $\frac{5}{3}$
Homework Helper
Gold Member
2022 Award
When y=1 then t = 0
thx! yes and it should be
$(t^2 -t^2 -1)dt + [(t^2 + 1)^2 +1]2dt$
After solving and applying bounds we get $\frac{5}{3}$
ARaslan
No, it makes sense, but it in (a) and (c) it's redundant because the bounds are also given separately. You could equally well complain, instead, that the curve specifications should be:
(a) a straight line segment;
(b) a segment of the parabola x=t, y=t2 + 1;
(c) two straight line segments meeting at (1,1)
Does it really matter?
Well technically it does not; however a line integral is defined as an integral which is taken around a curve. By definition line integrals must be calculated along a certain path, so it does not make sense to give coordinates in the integral itself.
NewtonApple
Ok! solving again
Evaluate the line integral $∫^{(1,2)}_{(0,1)} (x^2-y)dx + (y^2+x)dy$ along
(b) the parabola x=t, y=t2 + 1
x= t
dx = dt
$y= t^2 +1$
dy = 2tdt
when x=0 then t=0
when x=1 then t=1
when y=1 then t = 0
when y=2 then t= 1
Hence given integral
$∫^{(1,2)}_{(0,1)} (x^2-y)dx + (y^2+x)dy$
in terms of t is
$(t^2 -t^2 -1)dt + [(t^2+1)^2 + t]2tdt$
$(-1)dt + [(t^4 + 2t^2 +1) + t]2tdt$
$[-1 + t^4 + 2t^2 +1 + t]2tdt$
$∫^{1}_{0} [2t^5 + 4t^3 +2t^2]dt$
$\frac{2}{6}t^6 + \frac{4}{4}t^4 + \frac{2}{3}t^3$
after applying lower and upper bound we get
$\frac{1}{3} + 1 + \frac{2}{3}$
$\frac{1+3+2}{3} = \frac{5}{3}$
Homework Helper
after applying lower and upper bound we get
$\frac{1}{3} + 1 + \frac{2}{3}$
$\frac{1+3+2}{3} = \frac{5}{3}$
1+3+2=6 :D
NewtonApple
NewtonApple
1+3+2=6 :D
now I know my problem, I don't know simple arithmetic :).
Jokes apart. Integral suppose to be path independent so why two different answers for part (a) and (b)?
Homework Helper
Gold Member
2022 Award
You also went wrong in this step
$$(-1)dt + [(t^4 + 2t^2 +1) + t]2tdt$$
$$[-1 + t^4 + 2t^2 +1 + t]2tdt$$
but it turned out to make no numerical difference.
Homework Helper
Gold Member
2022 Award
Integral suppose to be path independent
Why should it be?
Homework Helper
now I know my problem, I don't know simple arithmetic :).
Jokes apart. Integral suppose to be path independent so why two different answers for part (a) and (b)?
No, it is only path-independent if you integrate an exact differential. http://mathworld.wolfram.com/ExactDifferential.html
NewtonApple
You also went wrong in this step
but it turned out to make no numerical difference.
Part(b) Correction
$=\left(-1+2t^{5}+4t^{3}+2t^{2}+2t\right)dt$
$=-t+\frac{2t^{6}}{6}+\frac{4t^{4}}{4}+\frac{2t^{3}}{3}+\frac{2t^{2}}{2}$
$=-t+\frac{t^{6}}{3}+t^{4}+\frac{2t^{3}}{3}+t^{2}$
applying bounds from 0 to 1
$=-1+\frac{1}{3}+1+\frac{2}{3}+1$
$=\frac{1}{3}+\frac{2}{3}+1=\frac{1+2+3}{3}=\frac{6}{3}=2$
NewtonApple
Part (c)
Evaluate the line integral $∫^{(1,2)}_{(0,1)} (x^2-y)dx + (y^2+x)dy$ along
(c) straight lines from (0,1) to (1,1) and then from (1,1) to (1,2).
Solution:
Two paths. Let's call it Path 1 and Path 2.
(i) Path 1: (x1=0, x2=1, y1=1, y2=1)
Equation of line is y=1; dy=0
Integral becomes
$\intop_{0}^{1}(x^{2}-1)dx=\frac{x^{3}}{3}-x=\frac{1}{3}-1=\frac{1-3}{3}=\frac{-2}{3}$
(ii) Path 2: (x1=1, x2=1, y1=1, y2=2)
Equation of line is x=1; dx=0
Integral becomes
$\intop_{1}^{2}(y^{2}+1)dy=\frac{y^{3}}{3}-y=\frac{2^{3}}{3}-\frac{1^{3}}{3}+2-1=\frac{8}{3}-\frac{1}{3}+1=\frac{8-1+3}{3}=\frac{11-1}{3}=\frac{10}{3}$
$\frac{-2}{3}+\frac{10}{3}=\frac{10-2}{3}=\frac{8}{3}$
NewtonApple
How do I insert LaTeX? I don't see any tool bar. Right now I'm doing it manually.
Homework Helper
(ii) Path 2: (x1=1, x2=1, y1=1, y2=2)
Equation of line is x=1; dx=0
Integral becomes
$\ \intop_{1}^{2}(y^{2}+1)dy=\frac{y^{3}}{3}-y=\frac{2^{3}}{3}-\frac{1^{3}}{3}+2-1=\frac{8}{3}-\frac{1}{3}+1=\frac{8-1+3}{3}=\frac{11-1}{3}=\frac{10}{3}$
$\frac{-2}{3}+\frac{10}{3}=\frac{10-2}{3}=\frac{8}{3}$
It is correct, but you made a sign mistake in the integral, it should be ##\frac{y^{3}}{3}+y## | 3,653 | 9,204 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2023-06 | latest | en | 0.690602 |
http://vkundeti.blogspot.com/2008/04/tech-computing-all-derivatives-of-n.html | 1,527,251,140,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867092.48/warc/CC-MAIN-20180525121739-20180525141739-00549.warc.gz | 303,048,090 | 12,852 | ## Wednesday, April 23, 2008
### [TECH] Computing all the derivatives of a 'n' degree polynomial at point 'x=a' in O(nlog(n)) time.
I came across this problem recently, basically you are given a polynomial f(x) = anxn+an-1xn-1+....a0 and you need to evaluate all its derivatives at a given point say x=a. I guess one of the suggested algorithms for this problem runs in O(nlog2(n)). But I found an idea to solve this in just O(nlog(n)). The following is the brief description of my algorithm
The idea is that we can post this as a simple multiplication of two n degree polynomials, the coefficients resulting from the multiplication will be the evaluated derivatives at a given point x=k. And as we might know that we can use FFT to multiply two n-degree polynomials in O(nlog(n))
• STEP1: Compute the values of k2,k3....,kn in O(n) time. Also compute the values of n!,(n-1)!,....2! incrementally. The runtime of this step is O(n)
• STEP2: Create polynomial p(x) = (n!an)xn-1+((n-1)!an-1)xn-2+.....a1.
• STEP 3: Create polynomial g(x) = (kn-1/(n-1)!)x+(kn-2/(n-2)!)x2+(kn-3/(n-3)!)x3+.... +kxn-1 .
• STEP 4: Multiply p(x) and g(x) using FFT in O(nlog(n) time.
• STEP 5: Clearly coefficient of xn-i+1 gives the value of ith derivative. | 372 | 1,238 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-22 | latest | en | 0.868203 |
https://leanprover-community.github.io/archive/stream/116395-maths/topic/product.20with.20variable.20number.20of.20factors.html | 1,621,023,962,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991207.44/warc/CC-MAIN-20210514183414-20210514213414-00072.warc.gz | 397,932,072 | 4,487 | ## Stream: maths
### Topic: product with variable number of factors
#### Nicholas McConnell (Feb 20 2020 at 16:18):
To prove that if p is prime and p = 4k + 1, then -1 is a square modulo p.
I propose to use Wilson's Theorem, which states (p-1)! == -1 (mod p).
In Lean, what's the best way to tackle a product of a variable number of factors?
#### Kevin Buzzard (Feb 20 2020 at 16:19):
finset.prod
#### Nicholas McConnell (Feb 20 2020 at 16:19):
Alright, thanks, I'll look into that
#### Kevin Buzzard (Feb 20 2020 at 16:20):
Things like this are often more challenging than they look for beginners, but go for it and get stuck and ask here and people will be happy to help. This is a great little project.
#### Nicholas McConnell (Feb 29 2020 at 23:33):
I just imported finset today, and apparently, finset.prod actually takes the Cartesian product of two sets
#### Nicholas McConnell (Feb 29 2020 at 23:35):
But I'm willing to believe they relate.
#### Reid Barton (Feb 29 2020 at 23:35):
Are you sure? I think finset.product does that
#### Nicholas McConnell (Feb 29 2020 at 23:35):
"product" is an unknown identifier
#### Reid Barton (Feb 29 2020 at 23:36):
I'm doubly confused now, but finset.prod is in algebra.big_operators.
Oh, okay...
#### Kevin Buzzard (Feb 29 2020 at 23:38):
Another confusing thing about finset.prod is that it doesn't take the product of the elements of the finset, it also takes a map from the underlying type to a comm_monoid and takes the product of the images. In particular you can have repeated factors in your product.
#### Kevin Buzzard (Feb 29 2020 at 23:39):
Another confusing thing about it is that there's also finset.sum but it's hard to find the definition, because it's generated automatically by the @[to_additive] tag on finset.prod.
#### Kevin Buzzard (Feb 29 2020 at 23:42):
#eval finset.prod (finset.range 3) (λ n, n^2+1) -- 1*2*5=10
#### Kevin Buzzard (Feb 29 2020 at 23:46):
It might be worth adding that factorials are already in mathlib (as is the fact that -1 is a square mod p iff p isn't 3 mod 4, and probably Wilson's theorem too, but this definitely shouldn't stop you trying them yourself)
#### Nicholas McConnell (Feb 29 2020 at 23:57):
Out of curiosity, where are factorials and -1 being a square mod p in mathlib?
#### Bryan Gin-ge Chen (Mar 01 2020 at 00:03):
Here are factorials (I searched for "factorial"). The other fact that Kevin mentioned is here.
#### Nicholas McConnell (Mar 01 2020 at 00:06):
Alright, thanks Bryan
#### Nicholas McConnell (Mar 01 2020 at 18:55):
I'm just going to use the builtin ones now because my file is already really big and I have little time left to complete the project.
When I imported data.zmod.quadratic_reciprocity, the file broke down with 1000 problems, and they even persisted when I deleted that import and made the file have the same text it started with, until I restarted VSC. Is that supposedly the wrong import?
#### Bryan Gin-ge Chen (Mar 01 2020 at 19:11):
I don't think it's the wrong import, but I can't reproduce your issue either. Sometimes things like this happen if Lean runs out of memory. If it goes away after you restart Lean, I wouldn't worry too much about it. (Note that you can restart Lean without starting VS Code by hitting ctrl+shift+p and typing 'Lean: restart')
Last updated: May 14 2021 at 20:13 UTC | 908 | 3,376 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2021-21 | latest | en | 0.931708 |
http://scienceblogs.com/dotphysics/2008/10/07/fake-vs-real-forces/ | 1,490,365,334,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188132.48/warc/CC-MAIN-20170322212948-00093-ip-10-233-31-227.ec2.internal.warc.gz | 327,853,355 | 14,998 | # Fake vs. Real Forces
In this post, I am going to talk about real and not real forces as well as the fake centrifugal force (if you don’t like the word “fake” you could replace that with “fictitious”)
First, an example: suppose you are in a car at rest and press the gas pedal all the way down causing the car to accelerate. What does this feel like? If I weren’t skilled in the art of physics, I might draw a diagram something like this:
Yes, maybe someone would add gravity and the chair pushing up, but this shows the important points. What is this force of acceleration? What causes this? This is EXACTLY the same thing as centrifugal force. If you think centrifugal force is real, this also should be real. I think this is enough discussion to show that this force (and centrifugal) is not real, but I will continue. There is another mystery: why does it feel like there is a force pushing you back when you accelerate? (if you have read all my blog posts, you may have a hint to the answer).
Let me replace the person with a model of a person. Here is my model (very simplistic)
In this model of a person, there are 4 masses each connect to the adjacent “atoms” with a spring (I represent the springs as rectangles because of my laziness). Now suppose I push on this model from both sides with equal forces.
I put these big bars on the side to make it clear the force was applied to both “atoms” on that side. So, when these two forces are applied, 1) the object stays at rest and 2) the horizontal springs are compressed.
Now what if I just apply 1 of these forces:
Notice that the compression is EXACTLY the same before (Eye-dentical). Hey wait! How do I know that this one force would compress this exactly the same? Well, you or I could easily model this and in fact I have done so for a [previous article (weightlessness and gravity)](http://scienceblogs.com/dotphysics/2008/09/gravity-weightlessness-and-apparent-weight/)
If the above model looks the same, it means a person would feel the same. The only difference is that this person would be accelerating. The point of this story is that when a person accelerates, it FEELS like a force is pushing on you in the opposite way. One note: when you accelerate, it doesn’t feel exactly the same as if someone was pushing on you. When someone pushes on you, they are exerting a force on just part of you. When you accelerate, it feels like something is pulling on ALL of you.
Ok, now on to circular motion and centrifugal force. In the above case, what if I took a “picture” of the velocity vector after 1 second? The two vectors would look like this:
And using the definition of acceleration:
I can find the direction of the acceleration by finding the change in the two velocity vectors:
Ok, so maybe we are all happy with this? (I am happy) Let me move to circular motion. I will once again “take a picture” of the velocity vectors for an object moving in a circle.
Now, I can do the same thing as before to find the direction of the acceleration. (it is ok to move a vector as long as you don’t change its direction or length)
Key points: 1) the velocity did change (although only in direction and not in magnitude). 2) This change in velocity means the object accelerated. 3) in this case, the acceleration is towards the center of the circle.
This would make it “feel” like a force is pushing outwards. It is this force that people call centrifugal force.
Whenever one is thinking about forces, it is important to realize that forces are an interaction between two objects and there are only a few real forces. They are:
• Gravity – an interaction between objects with mass
• Electromagnetic – an interaction between objects with electrical charge
• Strong nuclear – an interaction between hadrons (protons and neutrons are two examples of hadrons)
• Weak nuclear – an interaction between quarks and leptons
Anything that is a real force should be one of these. Gravity is an easy one to pick out. What about me pushing on a book? That would be the electromagnetic force because the atoms in my hand are interacting with the atoms in the book (and that is what prevents my hand from going through the book).
What about centrifugal force? What are the objects that are interacting? (hmmmm…..) Which of the fundamental forces is it? (hmmmm…..). Well, it must not be a real force.
Don’t get me wrong, sometimes the idea of a centrifugal force is useful, but that does not make it real.
1. #1 Alex R
October 8, 2008
In your models, assuming that the four masses are equal, the springs will compress half as much in the second case as in your first case.
To correctly model the fictitious force of acceleration, it must act on each mass in an amount proportional to that mass — in your case, if the blue “seat force” is F, then the red “fictitious acceleration force” will be F/4 on each mass to balance the “seat force” and keep things static. Then the force compressing each of the two horizontal springs will be F/4, rather than F/2 as it would be if you model the force as just pushing on the left-hand masses…
2. #2 rhett
October 8, 2008
True, that is why I put that big “plate” on the side so that I could push with just one force.
3. #3 Aatish
October 30, 2008
Hi,
Great post! The thought experiment with the springs was a really nice way of looking at the problem.
I wanted to point out that in addition to these 4 forces, there is also the degeneracy force that you get when you try to squeeze a bunch of identical fermions together. It’s different from the other four forces in that it’s doesn’t have any particle to mediate it – it’s purely a quantum effect. But it’s still quite real.
It would be interesting to try and figure out how much of matter not collapsing is due to electromagnetic repulsion and how much is due to degeneracy pressure.
4. #4 mefistofeles
September 26, 2010
This is incorrect.
When you exert a force to one end of the system with spring and masses (so it’s accelerating) the compression is NOT the same as if you exert the same force to both ends of the system.
You should make a correct free body diagram to notice this. | 1,420 | 6,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-13 | longest | en | 0.953676 |
http://www.ck12.org/book/CK-12-Middle-School-Math-Concepts---Grade-7/r4/section/12.14/ | 1,493,083,131,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917120001.0/warc/CC-MAIN-20170423031200-00093-ip-10-145-167-34.ec2.internal.warc.gz | 500,170,737 | 40,034 | # 12.14: Combinations
Difficulty Level: Basic Created by: CK-12
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Have you ever been on a decorating committee?
The decorating committee is getting the stage ready for the Talent Show. There was a bunch of different decorating supplies ordered, and the students on the committee are working on figuring out the best way to decorate the stage.
They have four different colors of streamers to use to decorate.
Red
Blue
Green
Yellow
“I think four is too many colors. How about if we choose three of the four colors to decorate with?” Keith asks the group.
“I like that idea,” Sara chimes in. “How many ways can we decorate the stage if we do that?”
The group begins to figure this out on a piece of paper.
Is this a combination or a permutation?
Use what you will learn in this Concept to answer this question.
### Guidance
In the last Concept, you saw that order is important for some groups of items but not important for others. Consider a list of three words: HOPS, SHOP, and POSH.
• For the spelling of each individual word, order is important. The words HOPS, SHOP, and POSH all use the same letters, but spell out very different words.
• For the list itself, order is not important. Whether the words are presented in one order–such as HOPS, SHOP, POSH, or another order, such as SHOP, POSH, HOPS, or a third order, such as POSH, HOPS, SHOP–makes no difference. As long as the list includes all 3 words, the order of the 3 words doesn’t matter.
A combination is a collection of items in which order, or how the items are arranged, is not important. The collection of one order of the items is not functionally different than any other order.
Combinations and permutations are related. To solve problems in which order matters, you use permutations. To solve problems in which order does NOT matter, use combinations.
Take a look at this situation.
The winning 3-digit lottery numbers are drawn from a drum as 641, 224, and 806. Does order matter in the way the three winning numbers are drawn?
Step 1: Write out a single order.
641, 224, 806
Step 2: Now rearrange the order. Did you change the outcome? If so, then order matters.
\begin{align*}224, 806, 641\Longleftarrow \end{align*} different order, same 3 winning numbers
Order does NOT matter for this problem. Use combinations.
Write the difference between combinations and permutations down in your notebook.
Here is another one.
A bag has 4 marbles: red, blue, yellow, and green. In how many different ways can you reach into the bag and draw out 1 marble, then return the marble to the bag and draw out a second marble?
Step 1: Write out a single order.
red, blue
Step 2: Now rearrange the order. Did you change the outcome? If so, then order matters.
blue, red \begin{align*}\Longleftarrow \end{align*} different order, meaning is DIFFERENT
Order DOES matter for this problem. Use permutations.
Write whether you would use combinations or permutations for each example.
#### Example A
Cesar the dog-walker has 5 dogs but only 3 leashes. How many different ways can Cesar take a walk with all 3 dogs at once?
Solution: Combination
#### Example B
Five different horses entered the Kentucky Derby. How many different ways can the horses finish the race?
Solution: Combination
#### Example C
How many different 5-player teams can you choose from a total of 8 basketball players?
Solution: Combination
Here is the original problem once again.
The decorating committee is getting the stage ready for the Talent Show. There was a bunch of different decorating supplies ordered, and the students on the committee are working on figuring out the best way to decorate the stage.
They have four different colors of streamers to use to decorate.
Red
Blue
Green
Yellow
“I think four is too many colors. How about if we choose three of the four colors to decorate with?” Keith asks the group.
“I like that idea,” Sara chimes in. “How many ways can we decorate the stage if we do that?”
The group begins to figure this out on a piece of paper.
Combinations are arrangements where order does not make a difference. The decorating committee is selected three colors from the possible four options. Therefore, the order of the colors doesn’t matter. This is a combination.
### Vocabulary
Here are the vocabulary words in this Concept.
Combination
an arrangement of objects or events where order does not matter.
Permutations
an arrangement of objects or events where the order does matter.
### Guided Practice
Here is one for you to try on your own.
Is this a permutation or a combination?
Mario’s gym locker uses the numbers 14, 6, and 32. How many different arrangements of the three numbers must Mario try to be sure he opens his locker?
There is only one way that the numbers can be arranged to open Mario's locker. This is a permutation because order does matter.
### Video Review
Here is a video for review.
### Practice
Directions: Write whether you are more likely to use permutations or combinations for each of the following examples.
1. A bag has 4 marbles: red, blue, yellow, and green. In how many different ways can you reach into the bag and draw out 2 marbles at once and drop them in a cup?
2. A bag contains 5 slips of paper with letters \begin{align*}A, B, C, D\end{align*}, and \begin{align*}E\end{align*} written on them. Pull out one slip, mark down the letter and replace it in the bag. Do this 3 times so you have written 3 letters. How many different ways can you write the 3 letters?
3. Eight candidates are running for the 4-person Student Council. How many different Student Councils are possible?
4. Mark's gym locker uses the numbers 24, 36, and 2. How many different arrangements of the three numbers must Mark try to be sure he opens his locker?
5. Five horn players are running for 2 seats in a jazz band. How many different ways can the two horn players be chosen?
Directions: Identify each situation as a permutation or combination.
6. The medals awarded for a swimming meet.
7. The order that you put toppings on a pizza.
8. The numbers that open a combination lock.
9. The way that you organize your clothing in a drawer.
10. A baseball line up according to ability.
11. A group of people organized randomly.
12. Walking five dogs three at a time.
13. Walking five dogs with a specific order.
14. Nine candidates are running for the 5-person Student Council. How many different Student Councils are possible?
15. Twelve candidates are running for the 6-person Student Council. How many different Student Councils are possible?
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1
Let me start by saying you are right in a sense, and I think the article is at best being unclear, but the larger point that the equation of ellipse cuts out a sphere is also right. There are a few things going on, hence a long answer below. What one is trying to describe is the topology of a set defined by a quadratic equation $\frac{x^2}{a^2}+\frac{y^2}{... 1 Your$\Sigma$does not satisfy condition (a): if you take the subset$\emptyset\subset\Sigma$, then the union$\bigcup\emptyset=\emptyset$is not an element of$\Sigma$. That is, the union of elements of$\Sigma$consisting of no elements at all is not in$\Sigma$. 4 For the Baire category proof. Use the OP proof up to $$C^{'}=\displaystyle\bigcup_{i=1}^{\infty} \partial B_{i} .$$ Then note$C^{'}$is a complete metric space, written as a countable union of sets$\partial B_i$, which are closed (in$C'$) sets with empty interior (in$C'$). 3 To show no such cover via open balls exists is easy: any such cover$\mathcal{O}$must involve at least 2 open balls (since the radii are finite). So pick$O\in\mathcal{O}$, and let$A=O$,$B=\bigcup (\mathcal{O}\setminus\{O\})$. Then$A$is open by assumption, while$B$is open as a union of open sets; and$A\sqcup B=\mathbb{R}^n$. But this contradicts ... 0 The following answer is not rigorous. To make it rigorous you may need to know more about smooth manifold.$\mathbb S^3 = \{ (u, v) \in \mathbb C^2 : |u|^2 + |v|^2 =1\}$is itself a$3$-manifold, which means that locally it looks like an open sets in$\mathbb R^3$. Mathematically speaking, it means that for each point$p\in \mathbb S^3$, there is a ... 1 There is a map from$T^3$to$SO(3)$, called "Euler angles" -- basically, you use each angular coordinate to specify "yaw", "pitch", and "roll". This map is, however, singular, much like the map from$T^2$to$S^2$discussed in the comments. The singularities are often referred to by the generic name "gimbal lock". You can read about this in books on ... 2 Here's another one, in$\mathbb R$.$X = $the set of midpoints of the complementary intervals of the Cantor set. So every point of$X$is the midpoint of an interval with no other points of$X$; that is every point is isolated. But$X'$is the Cantor set, uncountable. 2 I'm not sure what you're doing with your approach. Let me sketch another one: For each$x\in X,$set$r_x = d(x,X\setminus \{x\}).$Because each$x\in X$is an isolated point of$X,$each$r_x>0.$Show that the collection$\{B(x,r_x/2): x \in X\}$is pairwise disjoint. Note that each$B(x,r_x/2)$contains an element in$\mathbb Q^n.$5 Unfortunately, what you’re trying to prove is false. Let $$X=\left\{\left\langle\frac{2m+1}{2^n},\frac1{2^n}\right\rangle:n\in\Bbb N\text{ and }m\in\Bbb Z\right\}\;;$$ every point of$X$is isolated in$X$, but$X'$is the whole$x$-axis. Added: For that matter, you could just as well use the simpler set $$\left\{\left\langle\frac{m}{2^n},\frac1{2^n}\... 0 Hint: all point of X are isolated i.e there exists open balls B_x(x,r_x>0) with B(x,r_x)\cap B(y,r_y) is empty if x\neq y. Take an element q_x\in Q^n\cap B(x,r_x) an define f:X\rightarrow Q^n f(x)=q_x. Use the fact that Q^n is numerable and f is injective. Remark. A point x is isolated if there is a ball B(x,r_x) such that B(x,r_x)\... 2 As you try this on points farther and farther along the way, you start approaching the point * from the other side. To be continuous also there, you'd have to drag * itself around the circle once, and then the points slightly past it once and a bit. But that's in contradiction to your rule that you only move them until you hit * the first time. So your ... 1 it is not continuous at *. You construct an homotopy H_t:S^1\rightarrow S^1 such that H_t(x)=c_t(x) where c_t is the path going towards * from x clockwise. You must have H_t(*)=*. There exists a sequence lim_nx_n=* anticlockwise such that x_n\neq * and lim_nH_t(x_n)=c_t(x_n)=* anticlockwise otherwise H_t(S^1), t<1 is homeomorphic to ... 4 It seems to me that in the usual polar coordinates on the circle, OP is talking about the homotopy$$ H: S^1 \times [0, 1] \to S^1 : (\theta, s) \mapsto (1-s) \theta. $$For s > 0, this map is not continuous at \theta = 0, because no neighborhood of 0 maps to a small neighborhood of H(0, s). Hence it's not a deformation retraction from S^1 to ... 2 You have a family (\gamma_t)_{t\in[0,1]} of continuous maps from the circle to itself such that \gamma_0 is the identity, \gamma_1 is constant, and \gamma_t\to\gamma_1 pointwise is t\to1. But contractibility requires that \gamma_t\to\gamma_1 uniformly. At least that's the way I read your description. It's sufficiently vague that it could be the ... 0 Let U=\prod_{m=1}^\infty (a_m,b_m)\cap \mathbb R^\infty be an arbitrary set of a canonical base of the box topology. Since for each n the intersection U\cap\mathbb R’^n is \prod_{m=1}^n (a_m,b_m) \times\prod_{m=n+1}^\infty \{0\} iff a_m<0<b_m for each m>n and is empty, otherwise, we see that the set U is open in \mathbb R^\infty. ... 1 I think you're working far too hard. Pulling X=\mathbb{R}^2 apart works well! Define f\colon X\to\mathbb{R}^2\setminus\{ [-1,1]\times(-\infty,\infty)\} by f(x,y)=(x,y) if |x|>1, f(x,y)=(x+4,y) if |x|\le 1. Let's call Y=\mathbb{R}^2\setminus\{ [-1,1]\times(-\infty,\infty)\}. This isn't open because the image of the infinite vertical strip ... 0 As the user Nobody pointed out, the map F is continuous as it is a composition of continuous maps: The map \iota: [0,1] \rightarrow [0,1], t \mapsto 1-t, and the Homotopy G. Then the map F = G \circ (Id_X , \iota), is continuous as both G and \iota are contiuous. 1 r(t) : t \mapsto 1-t is a continuous function [0,1] \to [0,1] (1_X, r) is thus a continuous function X \times [0,1] \to X \times [0,1] F \circ (1_X, r) is thus a continuous function X \times [0,1] \to X 1 Note that each one of the three balls is relatively open AND CLOSED in K, being the intersection of a closed ball of \mathbb{C} with K. For instance, B_1 = \overline{B_1} \cap K, where the closure is meant in \mathbb{C}. So you have found a subset of K (namely, B_1) which is relatively closed and open in K, hence K is disconnected. Your ... 0 Trigonometric functions are circular functions. Complex cosine is: cos z = (e^(iz) + e^(-iz))/2 where z = x + iy where x and y are real numbers and z is a complex number. Suppose we restrict z so modulus z = 1 = modulus (x + iy). Hence, the image of z is the unit circle in the complex plane. What if we were to look at the map X = (x, y)? Let X ... 2 The function f(z) = (z+1/2)^2 will do the job. It maps the upper half of the circle bijectively to a curve that starts in the first quadrant, moves left to the fourth quadrant, comes down and intersects the negative real axis, continues downward and rightward under 0 and then comes back up to end at 1/4. The bottom half of the circle gets matched to ... 1 Take any two points \exp(i\alpha) \ne \exp(i\beta) on the unit circle, and a nonconstant analytic function f on a neighbourhood of the circle such that f(\exp(i\alpha)) = f(\exp(i\beta)). Then the curve \gamma(t) = f(\exp(it)), 0 \le t \le 2\pi intersects itself (\gamma(\alpha) = \gamma(\beta)). If you want the curve to be regular, require the ... 0 I think$$cos(x)+i sin(2x)$$is a good, simple example. (x=\frac{lnz}{i}) 1 The essential property is that given a compact metric space X and r>0 there is a finite number of balls covering X. Let 0<\lambda<1 and make the simplifying assumption that X may be covered by two closed balls B_0 and B_1 of size 1. Then assume (again for simplicity) that each of \Lambda_0=B_0 and \Lambda_1=B_1 (both are compact ... 3 Suppose towards contradiction that \mathbb{R}=U\sqcup V, U and V nonempty and open. Then there are real numbers x\in U and y\in V - without loss of generality let's assume x<y. Now - looking at the interval [x, y] - let$$z=\inf\{a\in [x, y]: a\in V\}.$$Such a real z exists, by the completeness of \mathbb{R}. Now, which piece of \... 1 The general outline is to write your given compact metric space A as a decomposition into two sets, then four sets, then eight sets, etc., in the exact same fashion that the Cantor set is decomposed, except that whereas the decomposition elements of the Cantor set are pairwise disjoint, the decomposition elements in A need not be pairwise disjoint. Be ... 1 Let X be a compact metric space. The key point is that we can subdivide X into smaller and smaller regions, such that at each stage there are only finitely many regions total. Specifically, we want for each n a partition$$X=Y^n_1\sqcup . . . \sqcup Y^n_{k_n}$$of X such that each Y^n_i has size (that is, diameter) less than (say) 2^{-n}. If we ... 2 Unfortunately, you are not free to morph the two halves in the very restricted fashion described in your question. You simply do not have that much control over the three given sets. Those three sets can each be much wilder and off kilter than you are imagining. If your proof were correct, then the given plane P that you found, passing through the centers ... 1 A misinterpretation I think you have misunderstood the stated fact in the linked answer. When it is stated that given any basis of a topological space, you can always find a subset of that basis which itself is a basis, and of minimum possible size. what is meant is the following given any basis of a topological space, you can always find a subset ... 1 T_0 is not enough. For n\in\Bbb N let U_n=\{k\in\Bbb N:k\ge n\}, and let \tau=\{\varnothing\}\cup\{U_n:n\in\Bbb N\}; \tau is a T_0 topology on \Bbb N. Let$$f:\Bbb N\to\Bbb N:n\mapsto n+1$$and$$g:\Bbb N\to\Bbb N:n\mapsto\begin{cases} 0,&\text{if }n=0\\ 1,&\text{if }n\ge 1\;. \end{cases}$$Check that \langle\Bbb N,\tau\rangle is ... 0 Suppose this is false and take x \in S, where S is the subset of non isolated points. Let y_0 \in S and take a neighborhood U_1 of x which does not contain y_0. Now take y_1 \in S, y_1 \in U_1 (this is clearly possible because every neighborhood of x contains infinite points of S), and U_2 \subset U_1 such that y_1 \notin U_2. Take ... 0 A very general result is true; I’ll leave just part of the proof to you. We’ll need the following result, which is proved here. Let \langle X,\le\rangle be any linear order, and let \tau be the order topology on X. Then the space \langle X,\tau\rangle is T_5 (i.e., completely normal and T_1). Now let \langle X,\le\rangle be a linear order,... 1 Your conclusion is false. At least part of it. For any convergent sequence \{a_n\} with limit q in \mathbb{R}^n, if we let B=\{a_n; n\in \mathbb{N}\}, then \overline{B}=B\bigcup\{q\}, regardless of whether or not it is on the boundary of an open set (every point in \mathbb{R}^n is on the boundary of some open set). However, your second ... 1 Given the function f:\overline{B}_r(p)\to\overline{B}_r(p)\setminus\{p\}, let$$g:\overline{B}_r(0)\to\overline{B}_r(0)\setminus\{0\}:x\mapsto f(x+p)-p\;,$$and apply the original result to g. 1 The following should answer your question unless I made a mistake (it's sketchy in a couple places), and is probably a bit overkill but it clarifies some relevant concepts: Let X be path-connected, let a,b\in X, and let \Omega_{[a,b]}(X) denote the space (with the compact-open topology) of continuous paths \gamma\colon [0,1]\to X such that \gamma(0)... -1 Here's another example that uses the idea of infinite oscillation as suggested by @YCor: On \mathbb R, let f(t) be a C^\infty function that is 0 for t<0, that strictly increases to 1 on [0,1], and that is 1 for t>1. The oscillatory piece: Define$$g(t) = \begin{cases}\sin(1/(t(1-t))\exp (-1/(t(1-t)), & 0< t < 1 \\ 0, &... 3 I assume the sequences are of abelian groups. The first one splits since the right term is a free abelian group. The second one does not necessarily split, since one can take$A=\mathbb Q$and$B$the quotient. Since the middle term is torsion free, the sequence cannot split. 1 I assume the groups involved here are commutative. The first sequence splits because$Z^2$is a free abelian group. if$f:B\rightarrow Z^2$,$f(u)=e_1,f(v)=e_2,$where$e_1,e_2$are generators of$Z^2$, write$g(e_1)=u, g(e_2)=v$. 0 See Henno Brandsma’s answer for a way to express your idea much more clearly. A slightly different approach is to let$\mathscr{B}$be a countable base for$X$, and let $$\mathscr{B}_0=\{B\in\mathscr{B}:|B\cap A|\text{ is countable}\}\;,$$ the family of basic open sets that contain at most countably many points of$A$.$\mathscr{B}_0$is a subset of the ... 2 Along the same lines: For the first question,$E_i$is nowhere dense iff$\overline{E_i}$is nowhere dense (both saying that$\overline{E_i}$has empty interior). And if$X$is not a countable union of$ \overline{E_i}$then it is not a countable union of$E_i$either. For the second question:${\mathbb R} = (\mathbb R \setminus {\mathbb Q} ) \cup {\...
1
First, a comment: I think you are implicitly assuming that every set is either open or closed. This is not true: consider e.g. the rationals! On to your questions: For the first one, there is no restriction on the type of sets involved: they may be open, closed, or neither, so long as they are nowhere dense. (Although in fact it's not hard to prove the ...
1
You essentially seem to be having the right idea, but the proof is very unclear. Start by picking a countable base $\{B_n: n \in \mathbb{N}\}$ of $X$. For every $x \in A\setminus A'$ ($A' =$ the set of limit points of $A$, that you call $D$) we can pick a basic open set $B_{n(x)}$ such that $B_{n(x)} \cap A = \{x\}$. If $x, y$ in $A \setminus A'$, then if ...
0
More or less ok, but you should perhaps pay more attention as to the choice of the sets $B_a$? Let $B_n$, $n\geq 1$ be a countable basis. For $a\in A\setminus D$ there is an element $B_{n(a)}$ which does not intersect $A\setminus\{a\}$ and since $a\in B_{n(a)}$ the sets $B_{n(a)}$, $a\in A\setminus D$ must be disjoint. So the map $a\in A\setminus D \mapsto n(... 2 Your last paragraph is basically already the answer to your question. Suppose$A$is hollow. Then for any (nonempty) open set$O$,$O$must contain some$x\in \mathbb{R}\setminus A$(why? otherwise$O$would be a subset of the interior of$A$, which would make$A$not hollow). But this is exactly the statement that the complement of$A$is dense! So ... 3 Yes - assuming the axiom of choice, cardinalities are well-ordered. In fact, this statement is equivalent to the axiom of choice. So your argument does indeed work - assuming the axiom of choice. To polish it up a bit, here's how I'd write it: Let$S$be the set of cardinalities of bases of$X$. By AC, the cardinalities are well-ordered, so$S$has a ... 1 That$A_i \cap A_j \not= \emptyset$is not always true. Consider for each$n \in \mathbb{Z}$:$A_n:=\{n\}\times \Bbb{R}$as a subspace of$\Bbb{R}^2$and let$A=\mathbb{R}\times \{0\}$. Then$A_n\cap A \not= \emptyset$for each$n$, but$A_n \cap A_m = \emptyset$whenever$m\not=n$. Let$X$be a topological space. Then, show that every continuous function$...
1
Hint that should make it clear how to proceed: The connected component that contains $A$, in $A\cup\big(\bigcup_{i\in I} A_i\big)$, contains a point in each $A_i$ by your non-empty intersection condition. If some connected component contains a point in $A_i$, then can it happen that it does not contain all of $A_i$?
2
Urysohn's Lemma implies that in a normal space you can separate points and closed sets by continuous functions; in fact you can separate disjoint closed sets. But what you're trying to do is impossible in general. Not every Hausdorff space is normal, and what you're trying to do is not always possible in normal spaces. If $C$ and $f$ are as in your post ...
0
Let $f$ be a $C^\infty$ function on $\mathbf{R}$ that is 2-antiperiodic ($f(x)+2=-f(x),$ $\forall x$) and such that $f(x)=\sin(1/x)\exp(-1/x)$ for all $x\in ]0,1]$. Let $g$ be a $C^\infty$ function on $\mathbf{R}$ that is 2-periodic and such that $g(x)=x$ for all $x\in]0,1]$. Note that $x\mapsto \exp(i\pi x/2)$ is 4-periodic and injective on $[0,4[$. ...
0
It seems that I have got it: It is necessary to establish that $U = U_q \setminus {U_p}^-$ is an open set and thus a neighborhood of $x_0$. It is not said explicitly in the proof.
Top 50 recent answers are included | 4,981 | 16,242 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2016-30 | latest | en | 0.862469 |
https://www.physicsforums.com/threads/inconsistent-vs-consistent-augmented-matrices.902782/ | 1,532,264,771,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593223.90/warc/CC-MAIN-20180722120017-20180722140017-00467.warc.gz | 963,127,786 | 16,493 | # Homework Help: Inconsistent vs consistent augmented matrices
1. Feb 5, 2017
### Arnoldjavs3
1. The problem statement, all variables and given/known data
Consider the following matrix where * indicates an arbitrary number and a ■ indicates a non zero number.
http://prntscr.com/e4xqkx
[■ * * * * | *]
[0 ■ * * 0 |* ]
[0 0 ■ * * | *]
[0 0 0 0 ■ | *]
(Sorry for poorly formatted matrix. The link above contains a screenshot if you can see it)
http://prntscr.com/e4xuoj
[■ * * | * ]
[0 ■ * | * ]
[0 0 ■ | * ]
2. Relevant equations
3. The attempt at a solution
Okay, so i know that a consistent system means that there has to be a unique solution or several of them. So in this isituation, i know that a non zero number(the last pivot) is equal to an arbritrary number. I believe this means that x5 is 'free'? This means that it is not constrained to any number and can be represented as x5 = t.
So how do I use this information to determine whether this augmented matrix is consistent or inconsistent? Are there any patterns to take note of here?
In the second matrix, it is similar in the sense that x3 is free and is not constrained. I'm having a hard time putting these matrices in perspective because they aren't actual numbers. Could something of this nature occur?:
x3 = t
x2 = s - t
2. Feb 5, 2017
### Staff: Mentor
No. Assuming you're talking about the first matrix above, x5 is specified, so you could solve for x5.
In the same matrix, x4 isn't specified, so it is free; i.e., x4 = t, where t can be any real value.
The system represented by this matrix is consistent, which means in this case there will be an infinite number of solutions.
Again, no. Since the reduced matrix (on the left) is diagonal, you could solve for unique solutions for each of the three variables. This system is also consistent, but there is a unique solution in this case.
From the last row, you could solve for x3, then back-substitute to solve for x2, and back-substitute these values to solve for x1.
Not for the second matrix.
For the first matrix, your solution would look something like $\vec{x} = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \\ c_5\end{bmatrix} + t\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0\end{bmatrix}$, with t being a real parameter.
3. Feb 5, 2017
### Arnoldjavs3
So in the second matrix, since it is a diagonal with non-zero numbers and no variables being free, it is consistent with a unique solution as every variable can be pointed to a specific value(Hope that makes sense)?
And in the first matrix, there is a free variable. If there is a free variable(x4 like you said, i understand why now) then it is going to have infinite solutions?
4. Feb 5, 2017
### Staff: Mentor
Yes and yes. I wouldn't say "pointed to," though. Each variable has a unique value would be a better way to say this.
5. Feb 5, 2017
### Ray Vickson
Are all the * numbers the same, or are they possibly all different? | 785 | 2,923 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-30 | latest | en | 0.91508 |
https://www.ahbcs.org/2nd-3rd-week-of-dec-4/ | 1,708,825,794,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474573.20/warc/CC-MAIN-20240225003942-20240225033942-00657.warc.gz | 643,574,403 | 43,574 | # 2nd/3rd: Week of Dec 4
## Math –
Ms. Andrea’s math group finally officially learned to subtract with regrouping! We call it: trading down. The Betas are off to a great start – so proud of them!! On Wednesday we solved many addition and subtraction story problems – with regrouping – independently and in partners. We came together at the end to explain our thinking. At the end of the week we practiced a couple of new strategies for adding triple digit numbers: using place value to break up the numbers (much like expanded form) and adding/subtracting from one number to make it easier to work with. Afterwards we played some fluency practice games. What a week!
Ms. Kelly’s math group learned several more multiplication strategies this week. The first- arrays. A multiplication array is simply an arrangement of rows or columns that matches a multiplication equation. Arrays make multiplication visual! The next strategy, number lines, clearly shows repeated addition by repeating jumps along the numberline. We also explored patterns on a multiplication table and practiced the Commutative Property for multiplication (a rule that says that the order of factors in a multiplication sentence has no effect on the product. For example, 3 × 5 = 15 and 5 × 3 = 15).
Multiplication is the main tool for many forms of maths such as algebra, calculus, equations and more. The ability to rehearse and understand multiplication will enable your child to confidently and skilfully tackle more complex mathematical subjects later. Now that we have a full understanding of what multiplication actually is, we jump to practicing facts! Be on the lookout after Winter Break for extra multiplication facts practice! ;)
## Literacy –
Book club is beginning to wrap up their latest reads for the quarter while kiddos are rocking reading assessments! We also squeezed in some word work games, dictation and silly sentences using homonyms..
In writing we wrote our conclusions and finished the rough drafts of our opinion pieces! We worked through the editing and revising process with a checklist – first on our own and then with the helpful eyes of a peer. We finished off the week with a creative quick write in our journals on an opinion of our choosing.
## Theme –
We explored pyramids this week!
The pyramids were built as burial places and monuments to the Pharaohs. As part of their religion, the Egyptians believed that the Pharaoh needed certain things to succeed in the afterlife. Deep inside the pyramid the Pharaoh would be buried with all sorts of items and treasure that he may need to survive in the afterlife.
There are around 138 Egyptian pyramids. Some of them are huge. The largest is the Pyramid of Khufu, also called the Great Pyramid of Giza. When it was first built it was over 480 feet tall! It was the tallest man-made structure for over 3800 years and is one of the Seven Wonders of the World. It’s estimated that this pyramid was made from 2.3 million blocks of rock weighing 5.9 million tons.
How the pyramids were built has been a mystery that archeologists have been trying to solve for many years. It is believed that thousands of slaves were used to cut up the large blocks and then slowly move them up the pyramid on ramps. The pyramid would get slowly built, one block at a time. Scientists estimate it took at least 20,000 workers over 23 years to build the Great Pyramid of Giza. Because it took so long to build them, Pharaohs generally started the construction of their pyramids as soon as they became ruler.
Deep inside the pyramids lies the Pharaoh’s burial chamber which would be filled with treasure and items for the Pharaoh to use in the afterlife. The walls were often covered with carvings and paintings. Near the Pharaoh’s chamber would be other rooms where family members and servants were buried. There were often small rooms that acted as temples and larger rooms for storage. Narrow passageways led to outside.
Sometimes fake burial chambers or passages would be used to try and trick grave robbers. Because there was such valuable treasure buried within the pyramid, grave robbers would try to break in and steal the treasure. Despite the Egyptians’ efforts, nearly all of the pyramids were robbed of their treasures by 1000 B.C.
We have created petite pyramids that will house a mini mummified pharaoh in a small sarcophagus. Cute, right? We worked on the “stone walls” on the outside, and the tomb walls on the inside, adding loads of Egyptian “paintings” and hieroglyphics. Next week we will make a mini mummy in a small sarcophagus.
Winter Faire Fun! | 972 | 4,611 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-10 | longest | en | 0.929718 |
http://healthresearchfunding.org/distributive-property-definition-for-kids/ | 1,475,113,904,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661775.4/warc/CC-MAIN-20160924173741-00191-ip-10-143-35-109.ec2.internal.warc.gz | 124,380,541 | 10,758 | Distributive Property Definition for Kids
# Distributive Property Definition for Kids
446
When it comes to the distributive property in mathematics, you’re talking about something that is extremely important to math as a whole. Once you have learned more about the concept of distributive property, you will be able to apply these fundamentals to a wide variety of problems.
### 1. It Is One Of The Most Frequently Used Properties In Math
Obviously, when it comes to math, we understand that some properties are utilized more than the others. One of the most significant things to keep in mind with the distributive property is that it is one of the most frequently utilized properties in mathematics. In a general sense, the property refers to the distributive property of multiplication.
### 2. The Most Straightforward Definition
In the end, the best way to learn about the distributive property is to apply the concept to actual examples. However, if you are interested in a straightforward definition of distributive property, there are a couple of simple things you can keep in mind. Basically, this property gives you the ability to multiply sums by multiplying each component separately, before setting about the task of adding your figures.
### 3. Understanding Basic Examples
The above definition may not be the easiest thing in the world to understand. To that end, consider an example such as 4 (x+2) = 4 X x + 5 X 2. With the distributive property, you can use the 4 to multiply your number to both the x and the 2. This is a fairly simple definition of the distributive property, but it is one that is worth keeping in mind nonetheless. This is a concept that can be applied to a wide variety of mathematical problems. You can use this property to solve very simple problems. By the same token, you can use this property to solve considerably more complex problems, as well.
### 4. It Is A Very Flexible Concept
One of the neat things about the distributive property is that you are talking about a very flexible concept. Not only can you apply the particulars of this property to numbers, but you can also apply the particulars of this property to expressions, as well.
### 5. You Can Actually Use The Distributive Property Twice
When we multiply 2 binomials, we are going to utilize this property. However, when we multiply 2 binomials, we are actually using the distributive property two times. This concept is known by the name foiling, and it is particularly prevalent with quadratic equations. | 522 | 2,526 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2016-40 | latest | en | 0.925926 |
https://socratic.org/questions/what-is-the-solution-set-for-absx-1-4 | 1,623,760,086,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621273.31/warc/CC-MAIN-20210615114909-20210615144909-00107.warc.gz | 493,381,608 | 6,198 | # What is the solution set for absx - 1 < 4?
Aug 14, 2015
$- 5 < x < 5$
#### Explanation:
To solve this absolute value inequality, first isolate the modulus on one side by adding $1$ to both sides of the inequality
$| x | - \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}} < 4 + 1$
$| x | < 5$
Now, depending on the possible sign of $x$, you have two possiblities to account for
• $x > 0 \implies | x | = x$
This means that the inequality becomes
$x < 5$
• $x < 0 \implies | x | = - x$
This time, you have
$- x < 5 \implies x > - 5$
These two conditions will determine the solution set for the absolute value inequality. Since the inequality holds true for $x > - 5$, any value of $x$ that's smaller than that will be excluded.
LIkewise, since $x < 5$, any value of $x$ greater than $5$ will also be excluded. This means that the solution set to this inequality will be $- 5 < x < 5$, or $x \in \left(- 5 , 5\right)$. | 318 | 987 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 16, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2021-25 | latest | en | 0.847782 |
mrbrownlee.blog.macc.nsw.edu.au | 1,709,306,495,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475311.93/warc/CC-MAIN-20240301125520-20240301155520-00744.warc.gz | 23,405,645 | 17,441 | ## Wednesday 22 March 2017
### Times tables...easy!
Having trouble with your times tables?
Knowing your times tables is one of the most important building blocks in your Mathematical learning. I know what some of you are thinking, 'That's what calculators are for!' Yes, calculators make large computations quicker, however, as students are learning Mathematical concepts all through school they need to be able to quickly find solutions so that they can focus on the working out and not the times table fact.
In my experience as a teacher for over 12 years, those students who know their times tables spend all their brain power on learning concepts and do so more quickly and effectively.
It's also my experience that the best place to learn (by this I mean practicing) times tables is at home.
Lucky for you, I have a couple of resources that I have seen work. Also, here are some other ideas that assist in the learning of your times tables:
1. Understand that there is not that much to learn in the first place. After learning your 1's and 10's the amount you need to remember goes from 100 to 64. Then, understand that multiplication works both ways, ie. 5x7 = 7x5, so there are only 36 combinations to remember.
2. Start with skip counting. Learn the skips really, really well before you start memorising the multiplication.
3. Revise times tables you know, but focus on one times tables at a time.
4. As you are learning a new times table write it out so that their mind can associate the numbers together (3 x 5 = 15)
5. Practice for short amounts of time but often.
Some effective strategies video clip
## Tuesday 21 March 2017
### Some Week 7 activities
Spelling
After over a year of investigation into a particular spelling program based on current research that focuses on students specific skills and needs, MACC Primary has commenced a revamped spelling program this year.
Words Their Way is the program we have been using for spelling this term. Students are involved in sorting and writing words into categories based on word patterns. Students often discuss and give evidence for reasons why each word should go into each category, which builds their understanding of each word pattern.
Here are a few word sort discussions taking place:
Maths Groups
At the start of most Maths lessons students take part in some warm up questions that revise concepts they have explored. But....every now and then, students will split up into Maths groups and complete a range of activities that give them a chance to brush up on a few areas that they need work on or extension in, as well as providing a chance to use their skills in some Maths games.
Happy Birthday Brianna!
## Friday 10 March 2017
### Highlights from Week 6
Music
5B have been learning a percussion piece called Carmen. It's been lots of fun practising different parts and playing a range of percussion instruments.
History Research on the Australian Gold Rush
I have been very impressed with the level of researching skills 5B have displayed in the last few weeks. They have attentively listened and quickly applied some simple strategies that they were presented.
Clean up Australia day
A Friday afternoon walk to Jack Nash reserve with all of stage 3 and year 10 has made our local parks, walkways and waterways a much cleaner place.
The exercise was great, however, the helpful attitude and enthusiasm of our students was most commendable.
Here were two of the stars.
## Thursday 2 March 2017
### Joy in the midst of trials
'Consider it pure joy, my brothers and sisters, whenever you face trials of many kinds, because you know that the testing of your faith produces perseverance.' James 1:2-3
We completed an art work today that was started in week 1. Not that we were that slow. No, we spent a couple of devotions talking about this passage found in the Bible. We discussed that, although we can be in hard times, we can have a joy inside us that is the hope of Jesus and of his great love and ability to see us through.
Students have also been exploring the concept of tone, light and shape when drawing.
Joy in the midst of trials
### Primary Assembly
We had our first Primary Assembly today. It was a wonderful opportunity for our choir to show what they have learned in the short couple of weeks prior.
We also presented some Key Learning Area awards for Christian Discovery.
Congratulations to Odelia and Joshua. | 936 | 4,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-10 | longest | en | 0.962437 |
null | null | null | null | null | null |
Re: Openldap reliability
--On Wednesday, March 26, 2003 13:08:36 +0200 Graham Leggett <[email protected]> wrote:
Hi all,
Running openldap v2.0.27 under Redhat v7.3, we have encountered
tremendous reliability problems.
Openldap doesn't die, it just stops accepting new connections - LDAP
on the same machine, so network is not a problem.
No, but 2.1.16 is. That is why it is labeled the "stable" release. It does need a small fix to slurpd (a one character change to properly terminate a comment) for slurpd to work correctly however. Why would you be looking at 2.1.12? It was just announced to the list recently that all versions from 2.1.5 to 2.1.15 have a serious DB corruption problem.
| null | null | null | null | null | null | null | null | null |
https://www.scribd.com/document/79488867/Sylow-Theorem | 1,501,054,961,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426050.2/warc/CC-MAIN-20170726062224-20170726082224-00589.warc.gz | 825,015,551 | 24,810 | # CHAPTER 4
The Sylow Theorems
Lagrange’s Theorem says that if H is a subgroup of a finite group G then |H| divides |G|. The converse is false: if G is a finite group and m is a number which divides |G| then it is not necessarily true that G has a subgroup of order m. For example, A5 has order 60 which is divisible by 15 and yet it can be shown that A5 has no subgroups of order 15. The Sylow theorems provide a partial converse to Lagrange’s theorem. In particular, Sylow proved that if m divides |G| and m is a power of a prime number then G has a subgroup of order m. The Sylow theorems also give useful information about the number of such subgroups. For example, if G is a group of order 60 then G always has at least one subgroup of order 4 and the number of subgroups of order 4 is 1, 3, 5 or 15. Here is another example. If G has order 168 then the Sylow theorems tell us that G has subgroups of all the following orders: 1 2 3 4 7 8. These are the only prime powers which divide 168. For other numbers such as 21 which are factors of 168 it is not immediately clear whether there is or is not a subgroup of corresponding order. p-groups D EFINITION 4.1. Let p be a prime. A group G of a group is called a p-group if every element g of G has order pm for some m = m(g) ≥ 0.
L EMMA 4.2 (Corollary of Lagrange’s and Cauchy’s Theorem). Let p be a prime. A finite group G is a p-group if and only if G has order pn for some n ≥ 0.
P ROOF. Suppose that G is a finite p-group. If q is any prime dividing the order of G then G contains an element of order q by Cauchy’s theorem. Therefore q = p. This shows that p is the only prime which can divide |G|. Therefore |G| = pn for some n ≥ 0. Conversely suppose that G has order pn for some n ≥ 0. Let g be any element of G. Let r be the order of g. The cyclic subgroup g has order r and so r divides pn by Lagrange’s theorem. Therefore r is a power of p. (Theorem 4.2 1 ). Every non-trivial finite p-group has non-trivial centre. 2 Groups of order pn for n ≤ 3
T HEOREM 4.3. Let p be a prime and let Cpn denote a cyclic group of order pn . (i) Every group of order p is cyclic and so isomorphic to Cp .
33
[Hint: 3420 is divisible by 19. Therefore. E XERCISE 4.5. L EMMA 4. is isomorphic to one of the groups (iv) If G is a non-abelian group of order p3 then ζ (G) ∼ Cp and G/ζ (G) ∼ Cp ×Cp . A subgroup H of G is called a Sylow p-subgroup if and only if H is a p-group and |G : H| is not divisible by p. Then PQ is also a p-subgroup. Suppose that Q normalizes P. with p = 2 we have the expression and we see that 22 is the largest power of 2 which divides 60. Since Q normalizes P we know that PQ is a subgroup. Let G be a group and let P and Q be p-subgroups. The factor s is called the p -part of n. . = = D EFINITION 4.8. E XAMPLE 4. To find this form you need to factorize n into a product of primes.34 4. Cp ×Cp2 Cp3 . (v) 7! has the factorization 7 · 6 · 5 · 4 · 3 · 2 · 1 and from this we can see that 32 is the largest power of 3 which divides 7!. P ROOF.6. (ii) If G is a group of order 60 then every Sylow 2-subgroup has order 4. Let G be a finite group and let p be a prime which divides the order of G.4. (i) Find the prime factorization of 3420. Now |PQ| = |P| · |Q| and |P ∩ Q| 168 = 23 · 21 60 = 22 · 15 so PQ is a p-group. Note that if n is a natural number and p is a prime which divides n then there is a unique way to write n in the form pm s where m and s are natural numbers such that p | s. (iv) What is the order of a Sylow 3 subgroup of S9 ? The Statement and Proof of the Sylow Theorems We need two lemmas in order to prove the Sylow theorems. (iv) If G is a group of order 168 then every Sylow 2-subgroup of G has order 8. R EMARK 4. (vi) Every Sylow 3-subgroup of the symmetric group S7 has order 9. (i) 60 has the prime factorization 2 · 2 · 3 · 5.7. (iii) 168 has the prime factorization 23 · 3 · 7 and therefore we have the expression and we see that 23 is the largest power of 2 which divides 168. THE SYLOW THEOREMS (ii) Every group of order p2 is abelian and is isomorphic to one of the two groups (iii) Every abelian group of order Cp ×Cp ×Cp Cp ×Cp p3 Cp2 .] (ii) If G is a group of order 3420 what is the order of a Sylow 19-subgroup of G? (iii) Find the largest power of 3 which divides 9!.
First note that for all X ∈ X . p | |X |.THE STATEMENT AND PROOF OF THE SYLOW THEOREMS 35 L EMMA 4. r|s where s is the p -part of the order of G. T HEOREM 4. Every Sylow p-subgroup of G is conjugate to P. Let X be the set pm of all subsets of G with exactly pm elements. Let G be a finite group and let p be a prime which divides |G|. Choose X ∈ X such that From this together with the first displayed fact it follows that This shows that there is at least one Sylow p-subgroup. Then n p | .. (iii) Every p-subgroup of G is contained in a Sylow p-subgroup of G. For now.10 (The Sylow Theorems). Then (i) G has at least one Sylow p-subgroup. if p divides |G| for some ≥ 0 then G has a subgroup of order p . m . Let n be a natural number and let p be a prime which divides n. · · pm p − 1 p − 2 p −i 3 2 1 and observe that in each fraction the power of p in the denominator exactly matches the power of p in the numerator. Then G acts on X by left multiplication. pm P ROOF.9 (Wielandt’s Observation). Now we are in a position to state and prove the Sylow Theorems. The second displayed fact tells us that there is an orbit whose size is not divisible by p. (ii) More generally. (v) The number r of Sylow p-subgroups of G satisfies and in addition r ≡ 1 (mod p). |GX | = pm . (ii) We postpone the proof of this part. . Write n = pm s where p | s. (i) Let pm be the largest power of p which divides |G|.. (iv) Let P be a Sylow p-subgroup of G. notice that • if = 0 then p = 1 and the claim is true for the trivial subgroup of G.. G P ROOF. • if = 1 then p = p and the claim is true by Cauchy’s theorem. We also know that | stabG (X)| ≤ pm . Write out the binomial coefficient like this: pm s pm s − 1 pm s − 2 pm s − i pm s − pm + 3 pm s − pm + 2 pm s − pm + 1 · m · m . p | |OX |..
P ROOF. Let g P be an element in an orbit of size 1. Then G is a non-trivial p-group and therefore it has non-trivial centre. Show that r is equal to 1 or 20. (v) Moreover g P is clearly the only orbit of size 1.13. We proceed by induction on the order of G. Let G be a group of order 3420. A permutation representation of G of degree n is a homomorphism from G to Sn . Hence g PQ = g P and so Q ⊆ g P. E XERCISE 4. Z contains a subgroup K of order p. If G has order pn then the quotient group G/K is defined and it has order pn−1 . (i) Let r be the number of Sylow 19 subgroups of G. Let G be a group. (iii) Let P be a Sylow p-subgroup. By induction G/K has subgroups of every possible order p j with 0 ≤ j ≤ n − 1. Note that p | r. . namely {e} and G. . Let G be a finite p-group. Since Z is central. A group is said to be simple if and only if it has exactly two normal subgroups. So there is an orbit of size 1.14. Permutation Representations D EFINITION 4. Simple groups D EFINITION 4.12. If |G| = 1 there is nothing to prove. a subgroup of order p j is of the form H/K where H is some subgroup of G containing K and so H has order p j+1 . . Therefore r ≡ 1 (mod p). What about part (ii) of the Sylow theorems? It suffices to prove the following: L EMMA 4. There must be an orbit whose order is not divisible by p. Then G acts on P by conjugation. Then Q normalizes g P. Let Z denote the centre of G. . Thus every Sylow p-subgroup is conjugate to P. However the only power of p which is not divisible by p is p0 = 1. Then Q acts on P by conjugation.11. L EMMA 4. There is a bijective correspondence between permutation representations of a group G and actions of G on {1. By Cauchy’s theorem. K is also central and therefore normal.36 4. Then G contains a subgroup of order pm whenever pm ≤ |G|. By the orbit–stabilizer theorem. This shows that G has subgroups of any possible p-power order. (ii) Assume now that G is simple. • Show that G has exactly 20 Sylow 19 subgroups. Suppose that G is a p-group of order > 1 and that the theorem is true for p-groups of smaller order. (iv) Assume now that Q is a Sylow p-subgroup. . Therefore g PQ is a p-group. Then it follows that Q = g P. n}. the number r of conjugates of P is |G : NG (P)|. Let P be the set of all conjugates of P. By the correspondence theorem. THE SYLOW THEOREMS • if = m then we have proved the claim in part (i).15. Let Q be any p-subgroup. Each Q-orbit has size a power of P.
(i) Let G be a non-abelian group of order pq where p and q are primes. E XERCISE 4. . (i) Prove that every group of order 15 is cyclic. (iii) Show that the subgroup of S7 generated by σ = (1 2 3 4 5 6 7) and τ = (2 3 5)(4 7 6) has order 21 and is non-abelian. (ii) Show that the symmetric group S5 has no elements of order 15. (v) Write down a list all the numbers of the form pq where p < q are primes and q ≤ 13. [Note this cannot be deduced directly from Lagrange’s Theorem because 15 divides 60 = |A5 |. We have already seen that G has exactly q subgroups of order p. (i) Prove that every simple group of order 60 is isomorphic to a subgroup of the alternating group A6 . Use the Sylow theorems to prove that (ii) Deduce that every group of order 77 is abelian.18. E XERCISE 4.] (iv) Show that the subgroup of S11 generated by (1 2 3 4 5 6 7 8 9 10 11) and (2 5 6 10 4)(3 9 11 8 7) is non-abelian of order 55.16.17. There are 15 such numbers. (iv) Find an example of a group of order 3420 which is abelian but which has no elements of order 171. • Show that if P is a Sylow 19-subgroup then NG (P) has order 171. ρ : G → S20 .SIMPLE GROUPS 37 • By considering the action of G by conjugation on the set of Sylow 19-subgroups. Which of the numbers n in this list have the property that all groups of order n are abelian? q≡1 (mod p). (iii) Find an example of a group of order 3420 which has an element of order 171. (iii) Show that the alternating group A5 has no subgroups of order 15.] E XERCISE 4. (ii) Prove that every simple group of order 168 is isomorphic to a subgroup of the alternating group A8 . [Hint: show first that τ σ = σ 2 . show that there is a permutation representation • Prove that ρ is a monomorphism whose image lies in the alternating group A20 . • Show that G does not contain an element of order 171. | 3,194 | 10,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-30 | latest | en | 0.912372 |
https://www.physicsforums.com/threads/newtons-second-law-and-kinematics.595319/ | 1,534,290,665,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221209650.4/warc/CC-MAIN-20180814225028-20180815005028-00450.warc.gz | 967,047,897 | 14,511 | # Homework Help: Newtons second law and kinematics
1. Apr 10, 2012
1)A rock is dropped off a cliff and falls the first half of the distance in to the ground in t1 seconds. If it falls the second half of the distance in t2 seconds, what is the value of t2/t1?
Relevant equations: I think the kinematics equations of d=vt+(1/2)at^2
I have no idea how to solve this please help!! I tried to find t1 with the equation above and god t1=sq.rt(d/5) but i don't know if this is right :)
2)a bos of mass m slides on a horizontal surface with initial speed v. I feels no forces other than gravity and the force from the surface. If the coefficient of kinetic friction between the box and the surface is mu how far does the box slide before coming to rest?
Relevant equations: f=ma Ff=Fn*mu,
I basically drew a free body diagram so far
Thanks so much for any help!
2. Apr 10, 2012
### azizlwl
1.
You are using correct kinematics equations of d=vt+(1/2)at^2.
As you see here we have 2 unknowns, d and t with v=0
You need 2 equations to solve for unique solution.
3. Apr 11, 2012
### Rokas_P
Well you're asked for a ratio, t1/t2. Whenever you're asked for a ratio, chances are that something will cancel from both the numerator and the denominator. And I suspect this may just be the unknown distance/height the rock travels, d.
The equation for d is correct as azizlwl pointed out, but since the rock starts from rest, initial speed is zero. So you have d=1/2*at^2.
Okay, you go like this:
1. how long does in take for the rock to fall the first half?
$\dfrac{d}{2}=\dfrac{gt^2}{2}\quad\Rightarrow\quad t=\sqrt{\dfrac{d}{g}}$
and actually this is your t1.
2. how long does it take for the rock to fall down the other half?
You repeat the procedure again but you note that the free fall in the second half of the distance starts with non-zero initial velocity (that's obvious since the rock will have accumulated velocity by the time it is halfway to the ground).
That means the equation for distance travelled is this one now:
$\dfrac{d}{2}=\dfrac{gt^2}{2}+v_0t$
What's $v_0$?
4. Apr 14, 2012
### Rokas_P
OK, since there has been no activity here for quite some time, I'll finish this one off just for the fun of it.
Continuing from my previous post, v0 is the initial velocity for the second half of the free fall. It is non-zero and is the same as the final velocity the stone will have accumulated by the time it is halfway to the ground. The first half of the free fall started from rest and continued for $t_1=\sqrt{\frac{d}{g}}$ seconds (see my post above). Then the speed accumulated over this many seconds is,
$v_0=gt_1$
$v_0=g\sqrt{\dfrac{d}{g}}=\sqrt{gd}$
Then
$\dfrac{d}{2}=\dfrac{gt^2}{2}+\sqrt{gd}t$
This is a quadratic equation (with respect to t). Its solution gives us the time t2.
$gt^2+2\sqrt{gd}t-d=0$
$t^2+2\sqrt{\dfrac{d}{g}}t-\dfrac{d}{g}=0$
$t_{1,2}=-\sqrt{\dfrac{d}{g}}\pm\sqrt{2}\sqrt{\dfrac{d}{g}}$
We disregard the negative solution. Then we are left with
$t_2=\sqrt{\dfrac{d}{g}}\left(\sqrt{2}-1\right)$
Ultimately,
$\dfrac{t_2}{t_1}=\sqrt{2}-1$
5. Apr 14, 2012
### LawrenceC
The box problem is one of constant acceleration. Determine the acceleration. Since the problem is one of constant acceleration, it is very similar to an object being projected vertically. How high does it go?
Last edited: Apr 14, 2012
6. Apr 16, 2012
### azizlwl
$\frac {1}{2}d=\frac {1}{2}at{_1}^2........(1)$
$d=\frac {1}{2}a(t_1+t_2)^2........(2)$
from (1)
$d=at_1^2$
sub. in (2)
$at_1^2=\frac {1}{2}a(t_1+t_2)^2$
$2t_1^2=t_1^2+2t_1t_2+t_2^2$
$t_1^2=2t_1t_2+t_2^2$
$1=2\frac {t_2}{t_1} +\frac{t_2^2}{t_1^2}$
$let \ \frac{t_2}{t_1}=x$
$x^2+2x-1=0$
Taking positive root.
x= -1+ √2 | 1,203 | 3,721 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-34 | latest | en | 0.918632 |
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# Page 1-Mathematics Achievement Study Guide for the ISEE
## General Information
This math section of the ISEE is centered on your ability to “work out” problems, using procedures and calculations. It will be necessary to know common mathematical procedures to do the calculations. Knowing methods of checking your calculations will also help. You will not be able to use a calculator or scratch paper, but you may make notes in the test booklet.
Before test day, take some time to review the concepts and procedures below. You may want to seek additional online or paper/pencil practice with anything that gives you trouble, just to boost your memory. Keep in mind, however, that the school really needs to know the level at which you can do math, so trying to teach yourself new things just for the test will prove counter-productive to your overall school success.
The best preparation for this test would involve seeing how much you remember about the following concepts and the procedures involved in calculating within them.
## Numbers and Operations
### Representing Very Large and Very Small Numbers
Word problems that involve very large or very small numbers will usually represent these numbers using scientific notation. This is the standard manner of writing numbers of this kind and follows this form:
$N × 10^a$
Where $$N$$ is a number more than $$1$$ but less than $$10$$ ($$1 \lt N \lt 10$$) and $$a$$ is a positive or negative exponent representing how many times and in which direction the decimal point was moved to get to the number $$N$$.
An item may ask for the evaluation of numbers written in this notation, for example:
Solve for the value of:
$\frac {2.43 × 10^8}{2 × 10^{-3}}$
### Types of Numbers and Number Systems
The number system is divided into real numbers and complex numbers which are expressions containing imaginary numbers. Real numbers are further classified as rational numbers and irrational numbers. Rational numbers are numbers that can be written as ratios or fractions of two integers. Thus, $$25$$, $$-3$$, $$\frac{1}{4}$$, $$\sqrt4$$ and $$0.75$$ are rational numbers because:
$25 = \frac{25}{1}$ $-3 = -\frac{3}{1}$ $\sqrt{4} = 2 = \frac{2}{1}$ $0.75 = \frac{3}{4}$
and $$\frac{1}{4}$$ is a fraction, thus, rational.
Irrational numbers are numbers that cannot be written as fractions or ratios, such as:
• $$\pi$$ (pi) $$= 3.1415926$$…
• $$\sqrt{3} = 1.7320508$$…
• Euler’s number = $$2.718281828$$…
### Vectors and Matrices
Vectors and matrices are used in solving systems of equations, and the skill in calculating the determinant and performing basic operations is important.
Matrices are described by their number of rows $$M$$ and number of columns $$N$$, such that a $$3 \times 2$$ matrix has $$3$$ rows and $$2$$ columns.
To calculate the determinant of a matrix with its first row described as $$[a, b]$$ and second row as $$[c, d]$$, multiply $$a$$ and $$d$$ and subtract from this the product of $$b$$ and $$c$$. To find the determinant of:
$[3, 5]$ $[2, 4]$ $(3 \cdot 4) \;– \;(2 \cdot 5) = 12\; – 10 = 2$
The method will be different for a $$3 \times 3$$ matrix, so it is important to review this, as well as basic operations, such as: addition, subtraction, multiplication, and division that involve vectors or matrices.
### Number Theory
A solid foundation in number theory is essential before learning higher math theories. The study of numbers focuses on: natural numbers, integers, the number system, the number line, the properties of even and odd numbers, and other types of numbers. This also includes the relationship between numbers, the concepts of inequality and equality of numbers, ordering, absolute value, prime and composite, factorization, sequences, counting, and more.
### Multiples and Factors
Factors are numbers that divide another number, such that the result is a whole number. So $$3$$ is a factor of $$9$$, and $$4$$ is a factor of both $$12$$ and $$8$$. A related concept is the greatest common factor, or GCF, which is important when solving for the simplest form of a fraction or finding equivalent fractions.
Multiples are numbers that are repetitions of another number, for example, $$9$$ is a multiple of $$3$$ because $$3 \times 3 = 9$$ or it is $$3$$ repeated $$3$$ times. It is also important to understand the concept of least common multiple, or LCM, of several numbers, especially when adding or subtracting fractions.
### Permutations and Combinations
Permutations and combinations are counting techniques. Permutation refers to the number of ways in which numbers, people or things can be grouped when their order is taken into consideration, such as the number of possible 3-digit codes for a lock. The order of the digits in a code is important, so the number of ways can be determined by counting the permutations.
The number of permutations, $$P$$, can be solved using these formulas:
When repetition is not allowed:
$P = \frac{n!}{(n – r)!}$
Where $$n$$ represents the number of things to choose from, and $$r$$ represents the number of choices to be made
Or when repetition is allowed:
$P = n^r$
Combination, on the other hand, refers to the number of ways in which numbers, people, or things can be grouped when their order is not taken into consideration, such as the number of combinations of shirts you can make if you were to buy only $$10$$ from a hundred shirts on sale. Use this formula for computing the number of combinations:
$C = \frac{n!}{r!(n-r)!}$
### Using “Mental Math” and Reason
It is important to check your answers after computing. The quickest way to judge the correctness of your answer choice is by mentally computing or approximating. Round off numbers, and take note that you may reduce errors in estimating by keeping a balance between rounding up and rounding down. It’s usually safe to estimate when the answer choices given are spaced widely apart.
Use logic and reasoning to validate your answer. When asked about the area of a bathroom, for instance, and you come up with $$200$$ sq.ft., you have to inspect your computation to see why a bathroom should be so large. The ability to visualize word problems and question your own answer is a skill that will be very useful for this test, specifically, but your future success as a mathematician or scientist, as well.
## All Study Guides for the ISEE are now available as downloadable PDFs
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null | null | null | null | null | null | Re: Licence Expired - Very unhappy!
I got a pop-up message that ViEmu trial has expired so I went immediately to the linked "store" to buy the product.
Meanwhile it has left Visual Studio in an unusable state. It is neither in vi mode not in original mode and I am unable to edit a file. If I press 'o' to open a line, it inserts o into the text. If I press Return to open a line, it does nothing other than beep. I'm most unhappy about this, if you're going to disable the software without warning, it should disable cleanly and not leave me unable to work.
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Always include your reference number, REF #90602895, in
the subject line of any messages you send to share-it!
Dear Mr. Stanford,
share-it! processes orders and collects payments on behalf of NGEDIT
Unfortunately, we were unable to charge your credit card for your order
90602895. The order cannot be completed as submitted.
You can easily change the payment information for the existing order
90602895 by clicking the link below.
I don't want to change the payment information. It is correct. If you have decided you don't want to sell me the product or, more likely, have chosen an incompetent credit card agency, please advise how to put Visual Studio back into a working state.
It's a great shame if you don't want me to have your software, by the way. I find it very useful and it makes me much more productive. My advice would be to dump Share-It immediately. In the meanwhile, I shall try to pay you once more and then shall give up.
Re: Licence Expired - Very unhappy!
I apologize for the problems. Thank you for letting me know about this.
I understand that share-it is being unduly protective with your credit card. Please take into account that it's probably not even share-it, credit card companies apply heuristics to try to prevent fraud (which is rampant online), and sometimes they may refuse a credit card, for example, because there has been a batch of cards stolen in your area and they're being used online, etc... Their fraud prevention systems are often based on heuristics, so not even *they* know exactly why it has been rejected (and, of course, they are secretive about the exact measures, so no chance on knowing exactly what is happening).
I offer direct paypal payment for this kind of cases. I will send you the details right now, in case you are still interested.
With the exception of a couple of cases, one of which was a payment from China, and this other one, and with the sometimes annoying factor that they require a non-free email address (they won't accept @hotmail.com addresses, etc...), share-it has worked pretty well so far. They are among the most expensive payment processors, but I keep them over switching to paypal or another service because they are very professional, respond inquiries within the day, and they have an earned reputation. Please realize that setting up a fully featured merchant account wouldn't be justified by the amount of sales of ViEmu yet (it's not a huge market).
I also apologize for the faulty disabling of the software. It should just restore the usual Visual Studio behavior, but I realize it's not working properly in all cases. I will review this so that it doesn't annoy other people the same as you. ViEmu should have been warning about the imminent end of the trial period for about 5 days, but even that is not an excuse
I will get in touch with you by email immediately to send you a trial period extension and to try to fix the problems.
In any case, thanks for trying to buy the product (it shouldn't be so problematic), thanks too for letting me know about this, and I'm hoping we can fix it somehow.
My apologies and best regards,
Re: Licence Expired - Very unhappy!
The issue is that they check the apparent country of the browser (getting it wrong where a cache is involved) against the billing country and then tell the buyer that his credit card is invalid without giving a reason.
There is nothing in their terms and conditions that states that the IP address country and billing country should match and they tell lies when you talk to them. I have found them to be just about the worst of a bad bunch of credit card clearance houses.
I have received your mail and now purchased the product though Paypal with no issues. Jon's response was excellent and gave the sort of service that it is so rare to find nowadays on the web.
This is a great product and carries great service. Thank you.
Re: Licence Expired - Very unhappy!
Cliff, thanks for the kind note and for being so understanding. I hope to improve the payment processing in the future, whenever I can find another reasonable service. My main complaints with share-it are it's highish fees, the delay in paying me (15-45 days, average 30), this kind of problem, and the fact that I can't customize validation in the order form. But, in any case, my idea is to get a full merchant account as soon as the business grows enough to warrant it.
Thanks again and I'm happy it all got solved swiftly.
I do believe in any case that software without proper service is worth very little, and strive to provide the best I can.
Best regards,
Re: Licence Expired - Very unhappy!
Dear Sir:
I have encountered this same problem with Share-It. I have entered my payment details correctly (double and triple-checked) and it still will not allow processing of my payment - probably due to my CC company thinking it's fraudulent. Can I please use your PayPal service to process payment and get a license for Viemu?
Re: Licence Expired - Very unhappy!
Sure, I will get in touch with you by email in a couple of minutes and send you the link.
Thanks for getting in touch, | null | null | null | null | null | null | null | null | null |
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# report 4.2.docx - Megan McLoughlin Math 2228 Lab Report 4.2...
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Megan McLoughlin Math 2228 10/21/2019 Lab Report 4.2 Using calc streak, compute the streak lengths of sim basket. 1. Describe the distribution of streak lengths. What is the typical streak length for this simulated independent shooter with a 45% shooting percentage? How long is the player’s longest streak of baskets in 133 shots? > outcomes <- c("H", "M") > sim_basket <- resample(outcomes, size = 133, prob = c(0.45, 0.55)) >sim_streak <- calc_streak(sim_basket) The distribution of streak lengths is skewed to the right. They typical streak length for the simulated independent shooter with a 45% shooting percentage is between 0 and 1. The player’s longest streak of baskets in 133 shots is 7. 2. If you were to run the simulation of the independent shooter a second time, how would you expect its streak distribution to compare to the distribution from the question above?
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Unformatted text preview: Exactly the same? Somewhat similar? Totally different? Explain your reasoning. If we ran the stimulation for a second time I would expect the streak distribution to be slightly different then the distribution from question one. This is because we didn’t set a seed value and we are just sampling at random. 3. How does Kobe Bryant’s distribution of streak lengths compare to the distribution of streak lengths for the simulated shooter? Using this comparison, do you have evidence that the hot hand model fits Kobe’s shooting patterns? Explain. > barplot(table(kobe_streak)) > barplot(table(sim_streak)) The distribution of the streak lengths of Kobe Bryannt compared to the streak lengths of the simulated shooter are quite similar. This means that the hot hand model fits with Kobe’s shooting pattern....
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A vehicle enters the curve at g1 and after a distance of L it departs the curve at g2. The total grade change is (g2-g1). The grade change rate, k, is the total grade change divided by the distance, Equation B-3.
Equation B-3
For example, a -2.00% grade into a +1.00% grade connected with a 400.00 ft curve has a k of:
This tells us that the grade changes +0.75% per each 100 ft of curve. If we go 100 ft past the BVC, the curve grade is -1.00%+0.75% = +0.25%
Increasing the curve length to 900.00 ft changes k to +0.33%/sta.
That second smaller k means the longer curve is flatter since is spreads the total grade change over a longer distance, Figure B-4.
Figure B-4 Different Curve Lengths
Note that the example was a sag curve with a positive k: we went from a negative grade to a positive grade. A crest curve, on the other hand, would have a negative grade change rate, Figure B-5.
Figure B-5 Crest and Sag Curves
What about curves whose grades at both ends are different but have the same mathematical sign? Figure B-6 shows four different curve situations where the incoming and outgoing grades have the same mathematical sign.
Figure B-6 Grades with Same Mathematical Sign
The two curves on the left are both crest curves having a negative k since the outgoing grades are less than the incoming grades.
The two curves on the right are sag curves having a positive k since their outgoing grades are mathematically greater (less negative or more positive) than the incoming grades. | 369 | 1,523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-24 | latest | en | 0.939366 |
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## Presentation on theme: "1 So, what happens here? ABCDE 13475 22585 39577 43588 575 68599 79512 4 8=(A5+B4)*A7 9 =(B2+C1)*B4."— Presentation transcript:
1 So, what happens here? ABCDE 13475 22585 39577 43588 575 68599 79512 4 8=(A5+B4)*A7 9 =(B2+C1)*B4
2 How to work it out Cell references will change when you copy a formula. What they end up as depends on where you paste the formula to! To work it out: –First consider horizontal movement (across) –Do column arithmetic E.g. if you move across (to the right) 2 cells from column A, A + 2 is C (think A B C) E.g. if you move across (to the left) 2 cells from column D, D - 2 is B (think D C B) –Then consider vertical movement (up or down) –Do row arithmetic E.g. if you move down 2 cells from row 3, 3 + 2 is 5 E.g. if you move up 2 cells from row 8, 8 - 2 is 6 + (Add) - (Subtract) + (Add) - (Subtract)
3 Practise Copy formula in C2 to: a)D2______________ b)B2 ______________ c)C1______________ d)C4______________ e)D4______________ f)B1______________ =C3+F6 =A3+D6 =B2+E5 =B5 +E8 =C5 +F8 =A2 +D5
4 Absolute addressing What we’ve just seen is called relative addressing At times we don’t want cell references to change \$ signs function like pegs to hold row numbers or column letters in place and stop them being translated relative to their new positions when copied In the formula =\$A1* 4 the column letter is fixed and the row number can change
5 When is absolute addressing useful?
6 Think you’ve got it? ABCDE 1=E5+D7 2 3=C4/A2 4 5=\$A7+D5 6 7 8=D\$1-\$E2 = G5+F7 = C7/A5 = \$A3+F1 =B\$1-\$E1 1 2 3 4
7 Fill handles and AutoFill Fill Handle Drag fill handle in the direction you want to AutoFill Excel can make intelligent guesses at what you want to enter into cells and fill these cells for you E.g. a range of days or months, repeated entries, etc
8 Custom AutoFill Lists (1) If you want Excel to be able to AutoFill customised data, you need to create a custom AutoFill list Office button Excel Options Popular, and then under Top options for working with Excel Edit Custom Lists
9 Custom AutoFill Lists (2) NEW LIST type the entries in the List entries box Add
10 Using Custom AutoFill Lists On the worksheet, click a cell, and then type the item in the custom fill series that you want to use to start the list. Drag the fill handle across the cells that you want to fill
11 Edit Fill (1) The Fill option on the Editing group of the Home tab can be used to automatically fill more complex series of numbers, dates and so on.
12 Edit Fill (2) E.g., it is possible to fill a selected area with a series of weekday dates, leaving out weekends. This can be achieved as follows: –Enter a date into the first cell and then select the chosen area –Choose Fill Series from the Editing group of the Home tab
13 Edit Fill (3) –You are filling a column so select Series in Columns, Type Date, Date Unit Weekday, Step Value 1 (i.e. fill every weekday rather than every 2nd or 3rd weekday). Click OK.
14 Edit Fill (4) Results look as follows Consulting a calendar shows the tool to be accurate
15 More on Edit Fill Fill Series offers a lot of flexibility- addition, multiplication, dates, using a selection or a Stop value
16 Section 5 –Manipulating Data: Functions [f(x)] Function formulas are specialised formulas that perform calculations of varying complexity Save time and effort Can be used for a range of financial, accounting, and other functions Accessible from the Function Library group of the Formulas tab on the Ribbon
17 Using functions Functions can be described and searched for, or selected from a drop-down list of categories
18 AutoSum as a function A closer look at AutoSum reveals the function formula created when the AutoSum button was clicked Note that A1:A6 represents a cell range Equals signFunction NameArguments
19 Naming Ranges Ranges can be named/labelled as follows: –Select the cell, range of cells, or nonadjacent selections that you want to name. –Click the Name box at the left end of the formula bar. Type the name for the cells. –Press ENTER. Formulas can use range names rather than ranges specified by using cell references. Here B4:B11 is the same as the name January. THIS HAS NOTHING TO DO WITH THE COLUMN HEADING.
20 Text Functions (1) =REPT(string, number_of_repetitions) =LEFT(string, number_of_characters) =RIGHT(string, number_of_characters) =MID(string, start position, number_of_characters)
21 Text Functions (2) =LEN(string) =TRIM(string) =UPPER (string) =LOWER (string) =PROPER(string)
22 Text Functions (3) =REPLACE(whole string, position of text to replace, number of chars to replace, new text) =SUBSTITUTE(whole string, text to replace, new text)
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posted by .
I am a bit confused about conjugates in algebra. I am supposed to multiply 8/(the square root of two + 4) by the square root of 2 -4. My math book then goes on to simplify this to (8 times the square root of two minus 32) divided by (2-4 times the square root of two plus 4 times the square root of 2 minus 16). I do not see how the bottom part came to be. Could someone offer an explanation, please?
• Algebra -
if a=4
b=20
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10. ### Math simplifying mixed radicals
Please help me simplify these following mixed radicals:: and show me how you did it cause I need to learn : I don't know how to type a square root sign so I just wrote square root and the numbers infront of the square root are supposed …
More Similar Questions | 655 | 2,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-34 | latest | en | 0.910916 |
https://origin.geeksforgeeks.org/alpha-and-beta-test/ | 1,653,021,683,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531352.50/warc/CC-MAIN-20220520030533-20220520060533-00097.warc.gz | 506,210,885 | 28,529 | # Alpha and Beta test
• Last Updated : 26 Nov, 2020
In the previous article, we discussed hypothesis testing which is the backbone of inferential statistics. We previously discussed the basic hypothesis testing including Null and Alternate Hypothesis, z-test, etc. Now, in this discussed more Type I and Type II error, level of significance (alpha), and Power(beta).
#### P-value:
The p-value is defined as the probability of obtaining a result or more extreme than what was actually observed in the normal distribution. Generally, we take the level of significance=0.05, it means if the observed p-value is less than the level -of-significance then, we reject the null hypothesis.
To calculate the p-value, we need the table particular test statistics (t-test, z-test, f-test) and whether it is a one-tailed, two-tailed test.
p-value
#### Alpha and Beta Test:
• Type I Error (Alpha): Now, if we reject the null hypothesis based on the level of significance p-value calculations, there is a possibility that the samples, in reality, belong to the same (null) distribution, and we incorrectly rejected it, this is called Type I error and it is denoted by alpha
• Type II Error (Beta): Now, on the basis of the level of significance and p-value, if we accept a sample that does not really belong to the same distribution then, it is referred to as Type II error
#### Power and Confidence Interval:
• Confidence Interval: The confidence interval is the region in which is we can confidently reject the null hypothesis. It is calculated by subtracting alpha and 1
• Power: Power is the probability of correctly rejecting the null hypothesis and accepting the Alternative Hypothesis (HA). Power can be calculated by subtracting beta from 1.
The higher the power makes lower the probability of making a Type II error. Lower power means a higher risk of performing a Type II error and vice-versa. Generally, 0.80 power is considered good enough. Power is also dependent on the following factors:
• Effect size: Effect size is simply the way of measuring the strength of the relationship between two variables. There are many ways of calculating effect sizes such as Pearson correlations for calculating correlations between two variables, Cohen’s d test for measuring the difference between groups, or simply by calculating the difference between means of different groups.
• Sample Size: The number of observations that are included in the statistical sample.
• Significance: Level of significance used in the test (alpha).
Steps to Perform Power Analysis
• State the Null Hypothesis (H0) and Alternative Hypothesis (HA).
• State the alpha risk level (level of significance).
• Choose the appropriate statistical test.
• Decide the Effect size.
• Create sampling plans and determine the sample size. After that gather the sample.
• Calculate the test statistic by determining the p-value.
• If p-value < alpha, then we reject the null hypothesis.
• Repeat the above steps a few times.
#### Examples
Special diet distribution vs Normal diet distribution
• Suppose there is two distribution representing the weights of two groups of people, the left representing people on diet and right representing people who take normal food.
• We take some samples from both the distribution and calculate their means.
• Here, our null hypothesis will be both samples are from the same distribution (no effect of diet plan) and the alternate hypothesis will be that both samples are from a different distribution.
• Now, we calculate the p-value from these samples.
• If our p-value is smaller than the level of significance then we correctly reject the null hypothesis that both these samples are from the same distribution.
• else, we don’t reject the null hypothesis.
• Now, we repeat the above steps numerous times (i.e 1000, 10000), etc. and we calculate the probability of correctly rejecting the null hypothesis i.e. Power.
## Python3
# Necessary Imports import numpy as np from statsmodels.stats.power import TTestIndPower import matplotlib.pyplot as plt # here effect size is taken as (u1-u2) /sd effect_size = (60-50)/10 alpha = 0.05 samples =20 p_analysis = TTestIndPower() power = p_analysis.solve_power(effect_size=effect_size, alpha=alpha, nobs1 = samples, ratio =1) print("Power is ",power)
0.8689530131730794
My Personal Notes arrow_drop_up
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https://api-project-1022638073839.appspot.com/questions/how-do-you-find-an-equation-of-the-line-containing-the-given-pair-of-points-7-4- | 1,718,490,891,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861618.0/warc/CC-MAIN-20240615221637-20240616011637-00034.warc.gz | 90,379,964 | 6,925 | # How do you find an equation of the line containing the given pair of points (-7, -4) and ( -2, -6)?
May 20, 2017
See a solution process below:
#### Explanation:
First, we need to determine the slope of the line running through the two points. The slope can be found by using the formula: $m = \frac{\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}}{\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}}$
Where $m$ is the slope and ($\textcolor{b l u e}{{x}_{1} , {y}_{1}}$) and ($\textcolor{red}{{x}_{2} , {y}_{2}}$) are the two points on the line.
Substituting the values from the points in the problem gives:
$m = \frac{\textcolor{red}{- 6} - \textcolor{b l u e}{- 4}}{\textcolor{red}{- 2} - \textcolor{b l u e}{- 7}} = \frac{\textcolor{red}{- 6} + \textcolor{b l u e}{4}}{\textcolor{red}{- 2} + \textcolor{b l u e}{7}} = \frac{- 2}{5} = - \frac{2}{5}$
We can now use the point-slope formula to write and equation for the line. The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$
Where $\textcolor{b l u e}{m}$ is the slope and $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through.
Substituting the slope we calculated and the values from the first point in the problem gives:
$\left(y - \textcolor{red}{- 4}\right) = \textcolor{b l u e}{- \frac{2}{5}} \left(x - \textcolor{red}{- 7}\right)$
$\left(y + \textcolor{red}{4}\right) = \textcolor{b l u e}{- \frac{2}{5}} \left(x + \textcolor{red}{7}\right)$
We can also substitute the slope we calculated and the values from the second point in the problem giving:
$\left(y - \textcolor{red}{- 6}\right) = \textcolor{b l u e}{- \frac{2}{5}} \left(x - \textcolor{red}{- 2}\right)$
$\left(y + \textcolor{red}{6}\right) = \textcolor{b l u e}{- \frac{2}{5}} \left(x + \textcolor{red}{2}\right)$
May 20, 2017
See the explanation.
#### Explanation:
First find the slope , $m$ of the line, using the two points.
$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$
Either point can be 1 or 2. I'm going by the order listed in the question. Point 1$=$$\left(- 7 , - 4\right)$, Point 2$=$$\left(- 2 , - 6\right)$
Insert the points into the equation.
$m = \frac{- 6 - \left(- 4\right)}{- 2 - \left(- 7\right)} = - \frac{2}{5}$
Now use the point slope form for a straight line.
$y - {y}_{1} = m \left(x - {x}_{1}\right)$
Insert either point as $\left({x}_{1} , {y}_{1}\right)$. I'm going to use Point 1 from the first part of this answer: $\left(- 4 , - 7\right)$.
$y - \left(- 4\right) = - \frac{2}{5} \left(x - \left(- 7\right)\right)$
Simplify.
$y + 4 = - \frac{2}{5} \left(x + 7\right)$
$y + 4 = - \frac{2}{5} x - 14$
Solve for $y$ to get the slope intercept form for a straight line: $y = m x + b$, where $m$ is the slope and $b$ is the y-intercept.
Subtract $4$ from both sides and simplify.
$y + 4 = - \frac{2}{5} x - 14$
$y = - \frac{2}{5} x - 14 - 4$
$y = - \frac{2}{5} x - 18$ graph{y=-2/5x-18 [-16.42, 15.6, -25.96, -9.94]} | 1,144 | 3,050 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 33, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.9375 | 5 | CC-MAIN-2024-26 | latest | en | 0.708414 |
https://www.studymode.com/essays/Doc-x-57276399.html | 1,603,451,349,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881369.4/warc/CC-MAIN-20201023102435-20201023132435-00073.warc.gz | 886,193,296 | 26,041 | Doc X
Topics: Regression analysis, Statistics, Econometrics Pages: 2 (728 words) Published: September 16, 2014
Study Guide for Exam IV
Econ 302/BUS 302
The exam is similar in length and form to the previous 3 exams. Bring a tall, green skinny scantron to the exam, a calculator, and a pencil. Scratch paper will be provided, if needed. You will be provided with a copy of the t-distribution on page 920 in the text, and the F-distribution on page 925. The formulas for the full F test statistic and the partial F test statistic will be given on the exam. All other formulas you will be expected to know (excluding the adjusted R2).
Exam will cover chapter 16 and anything on multiple regression. It is assumed that past material, such as population model, regression equation, estimated regression equation, error term, population parameters, etc. are already known.
Previous Information – Know these concepts from the last exam • Population model, regression equation, estimated regression equation, error term, population parameters
• When regressors are added to the model, what happens to the SSE, SSR, or SST? What happens to the R2? What is the adjusted R2 and why is it necessary? The formula for the R2adj will be on the exam. How do you interpret the adjusted R2?
• What is a quadratic term? When should you add a quadratic term to the model? Know how to interpret the estimated coefficients to a quadratic term.
• What is an interaction term? When should you add an interactive term to the model? Know how to interpret the estimated coefficients to an interactive term. • What are dummy variables? When should you add them to the model? Make sure you know how to add qualitative data to a regression, such as we did with Gender or Hospital Location. How do interpret the estimated slope coefficients to Dummy Variables? Be able to find the different population models for different values of the qualitative data. • You should know how to do individual significance test and a Full F-test. • What is multicollinearity? What does it cause?
• Be able to take results from a regression in... | 453 | 2,103 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-45 | latest | en | 0.866356 |
http://slideplayer.com/slide/5986412/ | 1,713,309,541,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817112.71/warc/CC-MAIN-20240416222403-20240417012403-00589.warc.gz | 26,442,831 | 21,283 | # Adding Integers with Different Signs
## Presentation on theme: "Adding Integers with Different Signs"— Presentation transcript:
Adding Integers with Different Signs
Warm Up -5 + (-4) = 5. -1 + (-10) = -9 + (-1) = 6. -90 + (-20) =
-5 + (-4) = (-10) = -9 + (-1) = (-20) = -52 + (-48) = = -4 + (-5) + (-6) = (-175) + (-345) =
Adding on a Number Line
Start on the first addend.
The number of units you move on the number line is equal to the absolute value of the second addend. If the second addend is positive, you move to the right on the number line, which is the positive direction. If the second addend is negative, you move to the left on the number line, which is the negative direction.
Explain your prediction and check it using a number line.
Predict the sum of Explain your prediction and check it using a number line.
Modeling Sums of Integers with Different Signs
+ You can use colored counters (or positive and negative symbols) to model adding integers with different signs. When you add a positive integer (yellow counter or positive symbol) and a negative integer (red counter or negative symbol), the result is zero. One positive and one negative form a zero pair.
Zero Pairs The opposite of any real number (a) is (-a).
The additive inverse Property tells us that the sum of any number and its opposite is zero. Zero is neither positive or negative, and zero is its own opposite. Zero pairs are formed by combining opposite integers. When zero is added or subtracted from any number, that number is unchanged. This is known as the identity property of addition or subtraction. a + 0 = a and a – 0 = a
5 + (-1) = 4 + (-6) = = = Kyle models a sum of two integers. He uses more negative counters than positive counters. What do you predict about the sign of the sum? Will the sum be positive or negative? Explain.
Using Absolute Values
= 10 + (-18) = 13 + (-13) = 25 + (-26) = | 476 | 1,902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2024-18 | latest | en | 0.866319 |
http://all-funny.info/page/541 | 1,524,687,840,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947957.81/warc/CC-MAIN-20180425193720-20180425213720-00530.warc.gz | 14,452,778 | 36,201 | ## City of Los Angeles
### Los AngelesHigh School Math Proficiency Exam
Name:____________________
Gang:________________________
1. Duane has an AK47 with a 30 round clip. If he misses 6 out of 10 shots and shoots 13 times at each drive by shooting, how many drive by shootings can he attempt before he has to reload?
2. If Jose has two ounces of cocaine and he sells an 8 ball to Jackson for \$320 and 2 grams to Billy for \$85 per gram, what is the street value of the balance of the cocaine if he doesn’t cut it?
3. Rufus is pimping for three girls. If the price is \$65 for each trick, how many tricks will each girl have to turn so Rufus can pay for his \$800 per day crack habit?
4. Jarome wants to cut his 1/2 pound of Heroin to make 20% more profit. How many ounces of cut will he need?
5. Willie gets \$200 for stealing a BMW, \$50 for a Chevy and \$100 for a 4×4. If he has stolen 2 BMW’s and 3 4×4’s, how many Chevy’s will he have to steal to make \$800?
6. Raoul is in prison for 6 years for murder. He got \$10,000 for the hit. If his common law wife is spending \$100 per month, how much money will he have left when he gets out of prison and how many years will he get for killing her since she spent his money?
7. If the average spray paint can covers 22 square feet and the average letter is 3 square feet, how many letters can a tagger spray with 3 cans of paint?
8. Hector knocked up six girls in his gang. There are 27 girls in the gang. What percentage of the girls in the gang has Hector knocked up?
### Survey for nerds – nerdity test
INTRODUCTION:
Hello, and welcome to the nerdity test. This test is designed to help you determine your nerdity quotient. In the past, someone may have watched you, or listened to something you said and then exclaimed, “You geek! What do you think you are doing?” Or maybe it’s just us. In any event, we here at the nerdity testing lab were prompted to ask “just what is a nerd?” In response, we came up with this test. By taking it, you will determine your current nerdity quotient (from 0% to 100%), with 100% roughly corresponding to a pile of sludge unable to communicate with anything human except through a device that is a miracle of modern medicine and engineering, and whose only connection to the outside world is through the computer internet system.
As this test is being distributed primarily in places of high concentrations of known nerds, and nerds in turn tend to have nerd friends, that someone who has never heard of or seen the nerdity test is assumed to be 0% nerd. However, once such knowledge comes to them, they are immediately placed in the 100% nerdity category. This is done because it is also assumed that only a true geek would utter something to the effect of: “Nerdity test?!? What a stupid concept! I’m too cool to take something as dumb as that.” The values in between are determined by taking the test and scoring it as follows.
For each question below for which you can answer “yes” or “true”, take one point. At the end of the test, divide the total number of points you scored by the total number of questions in the test. Treat this number as a percentage that represents your nerdity quotient.
Some of the questions will have parentheticals at the end of them. What is contained within the parentheticals is a short list of examples relating to the given question. The list is not to be taken as all inclusive but merely as suggestions that might apply.
All technicalities count – after all, being technical is half of what being a nerd is all about.
RECOMMENDATIONS and HINTS:
It is felt that for maximum enjoyment, you should respond out loud with your answers. You should treat each “yes” that you say as a personal catharsis of what you are doing wrong (or right depending on your opinion of nerdity) and each “no” may then be disputed by your peers. In this way, errors due to lying or personal oversight are avoided and the test also has a therapeutic effect for the closet nerd. As an aside, information gleaned about others should be treated confidentially. Each of us has a dork-side that we don’t want others to know about.
Experiment shows that nerdity CAN be cured! With effort and personal sacrifice… The nerdity quotient is a cross between proclivity toward as well as actual current status in nerddom. Some questions are “have you ever…” while others are “do you now…”. The former register the fact that you have a propensity toward nerdity, while the later acknowledge the fact that you are currently geeking. Obviously, as your answers toward the “do you now” type questions change, so will your nerd quotient.
Please use only a number two pencil. Mark all answers in your blue book. Shake well before using. Lather. Rinse. Repeat as desired. Show all work. Refrigerate after opening. No partial credit will be given. A table of useful formulas is included at the end. You may begin…. NOW!
SECTION 1: Education and Schooling
1. Have you ever taken a “higher” math course? (Trig, Calculus)
2. …at the college level?
4. Are you still capable of doing what you learned in the course of #1?
5. Have you ever taken a science course? (Biology, Physics, Chemistry)
6. …at the college level?
8. Are you still capable of doing what you learned in the course of #5?
9. Have you ever majored in the “hard sciences”? (engineering, physics chemistry, etc. but excluding psychology, economics, etc.)
10. Have you ever taken Latin?
11. Have you ever asked a question in lecture?
13. Have you ever corrected a professor in lecture?
14. Have you ever answered a rhetorical question?
15. Have you ever given a lecture?
16. Do you sit in the front row more than 20% of the time?
17. Have you ever had a “perfect attendance record”?
18. Have you ever verified an equation in a science text on your own? (i.e. experimental proof)
19. Have you ever derived an equation you found in a science text?
20. …when you didn’t have to?
21. …using other principles? (starting from a different equation than the text did)
22. Do you take notes in more than one color?
23. Do you use other props when taking notes? (ruler, compass, protractor)
24. Have you ever tutored someone else?
25. Have you ever done homework on a Friday night?
26. Have you ever pulled an all-nighter?
27. Have you taken any classes pass/fail just to preserve your GPA?
28. Have you ever known more about the subject material than the lecturer?
29. …but continued in the class because you “needed the grade?”
31. Have you ever had an argument with a professor?
32. Did you win?
33. Has a lecturer ever referred someone to you as being more knowledgeable?
34. Did you apply to any college merely for the sake of “seeing if I would get in”?
36. Did you score higher than 1200 combined on the SAT?
SECTION 2: Knowledge
37. Can you count in binary? (up to decimal 10)
38. Can you count in hexadecimal? (up to decimal 20)
39. Can you count in Roman numerals?
40. Do you know Maxwell’s equations? (integral or differential form)
41. Do you know Schroedinger’s Equation?
42. Have you ever solved Schroedinger’s Equation?
43. …for fun?
44. Do you know the difference between a scalar and a vector?
45. Do you know the difference between a vector and a tensor?
46. Do you know the right-hand-rule for cross-products?
47. Do you know the Latin name (genus and species) for anything? (fruit fly, human being)
48. Can you understand the owner’s manual for electronic equipment?
49. Can you understand the electronic schematic for electronic equipment?
50. Do you know what a “reverse polish notation” calculator is?
51. Can you name the first nine elements of the periodic table in order?
52. Can you translate more than half the chemical symbols into the name of the element they represent?
53. Do you know the wavelengths in the visible spectrum?
54. Are you bilingual?
55. …and not an immigrant or child of an immigrant?
56. Can you briefly outline the biological process that occur due to alcohol when it is consumed by a human?
57. …while drunk?
58. Do you know how your car’s engine works?
59. Have you ever interpolated?
60. Have you ever extrapolated?
61. Do you know the difference between interpolation and extrapolation?
62. Have you ever integrated numerically?
63. …and known the result ahead of time?
64. …and complained about how slow the computer was?
65. Have you ever seen or utilized the spherical harmonic functions?
66. …and found them aesthetically pleasing?
67. Do you know most of the words to “The Lumberjack Song” by Monty Python?
68. Do you own an encyclopedia?
69. Have you ever read an encyclopedia entry that you weren’t researching?
70. Have you ever wanted to know something for no apparent reason?
71. Have you ever been laughed at for wanting to know something?
72. Can you program the time on a VCR?
73. Has anyone ever asked you to program their VCR time for them?
74. Have you ever used the word “asymptotic”?
75. Have you ever referred to something as an L.E.D.?
76. Have you ever referred to a ruler as a “straight-edge”?
77. Have you ever said “quartz crystal”?
78. Have you ever called something a “print out” or “hard copy”?
79. Have you ever referred to a curve/object as hyperbolic, parabolic, etc.?
80. Do you feel your vocabulary is larger than most people’s?
Answer YES if you know what the following acronyms stand for. Note: it may be useful to actually state out loud what you think the acronym stands for as your interpretation may be wrong or not the nerdy one being sought after.
83. …MODEM?
84. …RAM?
85. …DNA?
86. …ATP?
88. …CRT?
89. …CRC?
90. …STP?
92. …NASA?
93. …MUD?
94. …LED?
95. …AI?
96. …LASER?
97. …RPG?
98. …TLA?
99. …SCUBA?
100. …WYSIWYG?
101. …DAT?
102. …PINE?
103. …JOVE?
104. Did you not know one of the above, but took a wild guess at in anyway?
105. Have you ever created an acronym in order to simplify your writing?
The next few questions deal with physical constants. Mark yes for any that you can give the value (2 or more significant digits) for. Knowledge of the units attached is NOT necessary, just the numeric portion.
106. gravitational constant? (G)
107. earth’s gravity near the surface? (g)
108. mass of an electron?
109. charge of an electron?
110. speed of light in vacuum?
111. speed of sound at STP?
112. Planck’s constant? (h or h-bar)
113. permittivity of free space? (epsilon naught)
114. permeability of free space? (mu naught)
116. molar gas constant?
117. pi? (exception: must know more than 3 digits)
118. Mark this true if you are presently the person knowing the most digits of pi in the room.
119. e? (exception: must know more than 3 digits)
Can you give the conversion factor between… (2 or more sig. digits)
120. …centimeters and inches?
121. …kilometers and miles?
122. …joules and electron-volts?
123. …atomic mass units and kilograms?
124. …Celsius and Kelvin?
125. …Celsius and Fahrenheit?
126. …meters and Astronomical Units (AU)?
127. …AU and light years?
128. …light years and parsecs?
129. If, while answering any question in this section, you said someone else’s answer was wrong and were right, mark this question true. (e.g. “you nob! Pi isn’t 3.1425. It’s 3.1415!”)
130. If while answering any question in this section, you checked a reference book to find out the correct answer, mark this question true. (e.g. “AARRGGH! What’s that last R in radar stand for?”)
SECTION 3: Computers
132. …for more than 4 hours continuously?
133. …for more than 8 hours continuously?
134. …past 4 a.m.?
135. …as a source of income?
136. …on Friday, Saturday and Sunday of the same weekend?
137. …with someone you were physically attracted toward?
138. …for money?
139. …in the last 24 hours?
140. …in the last half hour?
141. …as a source of entertainment? (computer game)
142. …in the last three months?
143. …in the last three weeks?
144. Have you ever programmed a computer?
145. …to write a computer game?
146. …to write a computer virus?
147. …to write a shell script?
148. Do you still own any computer with less than 512k of RAM? (e.g. Commodore 64, Apple II +/e/c, TRS 80, ad infinitum)
149. …that is still in working condition?
150. …and still buy software for it?
151. Do you own more than one computer with at least a megabyte of RAM?
152. Do you own any computer which would be classified as a work station?
153. Have you ever taken your computer on vacation with you?
154. Have you ever lost sleep over a computer game?
Have you ever used a …
155. mouse?
156. hard disk drive?
157. light-pen?
158. computer with a touch sensitive monitor?
159. track-ball?
160. …for something other than a video game?
161. Devorak keyboard? (as opposed to QWERTY)
162. modem?
163. Have you ever seen a magnetic tape reel?
164. Have you ever mounted a magnetic tape reel?
165. Have you ever seen a computer punch card?
166. Have you ever programmed using punch cards?
167. Are you still capable of programming with punch cards?
168. Do you have any “pirated” software? (i.e. second-hand copywritten)
169. Do you have any “public-domain” software?
170. Do you have any “shareware”? (i.e. software author requests a fee be sent to them for its use)
171. Do you currently own a modem capable of 14.4kbs or faster?
172. Do you still own any modem whose top speed is 300 baud or less?
173. Have you ever telnet’ed from one computer system to another?
175. …to call a government computer? (NASA, FBI, NORAD, etc.)
176. …to call a research institution? (CERN, JPL, etc.)
177. …where the other machine was outside of your native country?
178. Do you have an electronic mail address?
179. …more than one e-mail address?
180. Have you ever sent e-mail?
181. …to yourself?
182. …to someone who was in the same room as you at the time?
183. …with a .sig file appended to the end of it?
184. …in the last week?
185. Have you ever set up and run a mailing list for e-mail?
186. Do you receive more e-mail than you send?
187. Have you ever FTP’d?
188. …anonymously?
191. Have you ever multi-tasked? (ran 2+ applications concurrently)
192. Have you ever set up a kill file?
193. …that does more than simply ‘kill’?
194. Do you have a .plan or similar file for when people finger you?
195. Have you set up a login.com or similar file for auto-execution on logging unto a computer system? (autoexec.bat, login.com…)
196. Do you use alias/batch commands to standardize your OS? (e.g. alias dir ls)
197. Have you ever read the postings on USENET?
198. …in the last week?
199. Have you posted to USENET?
200. …and gotten a response?
201. …from someone you knew outside of the net?
202. …and gotten a “flame”?
Have you ever posted to…
203. …a science fiction news group? (rec.arts.sf)
204. …a sex news group? (alt.sex)
205. …talk.bizarre?
206. …rec.humor?
207. …a sci. or science-related news group?
208. Have you ever written a FAQ for a USENET news group?
209. Have you ever run a vote for a USENET news group?
210. Have you ever moderated a USENET news group?
211. Have you played any MUD’s, MUSH’s or other multi-user games?
212. …in the last week?
213. …today?
214. Do you consistently play more than one MUD, MUSH, etc.?
215. Are you a “wizard/implementor/immortal” on any MUD’s, MUSH’s, etc.?
216. Do you have GIF files as wallpaper?
218. Have you ever built a computer?
219. …from chips?
220. Do you have a favorite computer language?
221. …that you’ve had to defend in verbal debate?
Which of the following computer languages do you know…
222. …BASIC?
223. …PASCAL?
224. …FORTRAN?
225. …assembly language?
226. …C?
227. Have you ever forgotten a person’s name but not their e-mail address?
229. Do you tend to remember the IP numbers instead of the alpha address for computer sites? (128.253.232.63 vs. crux3.cit.cornell.edu)
230. Do you find that you type more often than you write longhand?
231. Have you ever forgotten how to write longhand?
232. Have you ever used computer symbology elsewhere? (goto, *, etc.)
233. Have you ever spoken internet-ese? (btw, imho, :), brb, afk)
234. Have you ever blown off doing something you were supposed to do in order to work on the computer?
235. Have you ever felt jealous of someone merely because they owned a better computer system than you?
SECTION 4: Possessions
236. Do you frequently find yourself with more plugs than outlets?
237. Do you currently own a can of WD-40?
238. Do you currently own a can of compressed air?
239. Do you have a personal copy of any version of the nerdity test?
240. …in space allocated to you on a computer system?
241. Have you ever owned a light saber (Star Wars)?
242. …that wasn’t made of plastic?
243. Do you own an 8-track tape player or any 8-track tapes?
244. Do you own an almanac? (World, Farmer’s)
245. Do you own an atlas?
246. Do you own a globe?
247. …and have it on display? (on a desk, bookshelf…)
248. …that has bumps corresponding to mountain ranges?
249. …that lights up?
250. Do you own any “maps of the ancient world”?
251. …and have them on display?
252. Do you have any “mathematical” artwork? (Escher, fractals)
253. Have you ever faxed something?
254. Have you ever received a fax?
255. Do you own a cellular phone? (car phone)
256. Do you own a non-standard calculator? (scientific, programmable)
257. Do you own a “reverse polish notation” calculator?
258. Do you own a slide rule?
259. …and know how to use it?
260. Other than a thermometer, do you own any meteorological equipment?
261. Do you own any orienteering equipment? (compass, sextant, etc.)
262. Do you own a pencil case?
263. Do you own any mechanical pencil?
264. …and have refills for it?
265. Do you own an electric pencil sharpener?
266. Do you own a laboratory notebook?
267. Do you own any graph paper? (quad-ruled)
268. Do you own any log or semi-log paper?
269. Do you own a table of integrals?
270. Have you ever stolen scientific (radiation, biohazard) warnings for personal use?
SECTION 5: Leisure Time
271. Have you ever taken something apart?
272. …and put it back together correctly?
273. …without worrying about voiding the warranty?
274. Do friends and/or family ask you to fix things?
276. …because you are the only person they know who OWNS that tool?
277. Have you ever put something together without reference to the assembly instructions?
278. Have you ever bought something primarily for the pleasure of taking it apart to “see how it works”?
279. Have you ever rewired something?
280. Have you ever played a non-sexual role-playing game? (D&D)
281. …since leaving high school?
282. Have you ever been to a RPG convention? (GenCON, etc.)
283. …in the last six months?
284. Have you ever taken a “self help” test?
285. Do you derive perverse pleasure from self-help tests?
286. Do you ever lord your scores on such tests over people around you?
287. Have you ever dissected something?
288. …while not involved in a biology class?
289. Do you play chess?
290. Were you ever on a chess team?
291. …on a math team?
292. …on a debate team?
293. …on a “trivia” team? (college bowl, JEOPARDY)
294. …the captain for any of the teams listed above?
295. …the coach for any of the teams listed above?
296. Did you ever join one of the above teams for the purpose of picking up members of the opposite sex?
297. Were you ever in a science fair?
298. …that you placed in the top three?
299. Are you a member of Mensa?
300. Have you ever made a technical joke?
301. …in the last week?
302. …that no one around you understood?
303. …and you found yourself trying to explain it?
304. …that everyone around you understood?
305. …but their reason for laughing was not yours?
SECTION 6: Leisure Time – Nerd Toys
306. Have you ever bought something from Radio Shack?
307. Do you know what an oscilloscope does?
308. Have you ever used an oscilloscope?
309. Do you own an oscilloscope?
310. Have you ever used a microscope?
311. Do you own a microscope?
312. Have you ever used a telescope?
313. …not for peering through someone’s bedroom window?
314. Do you own a voltmeter?
315. Do you own any remote controlled vehicles?
316. Do you own a CB radio?
320. Have you ever used a chemistry set?
321. …since the age of 13?
322. Have you ever used a rare earth element?
323. Do you own a slinky?
325. Do you own a Rubik’s cube?
326. Are you able to solve Rubik’s Cube?
327. …without using the book?
328. …in less than two minutes?
329. Have you ever tried to calculate the number of possible permutations a Rubik’s Cube can have?
SECTION 7: Leisure Time – TV and Movies
330. Do you watch more than 4 hours of TV on any given day of the week?
331. Can you name more than 5 shows on PBS? (inc.:A&E, Discovery Channel)
332. Have you ever watched a PBS documentary?
333. …voluntarily?
334. …in the last three weeks?
335. Have you ever watched C-Span for more than 5 minutes?
Have you ever watched a complete episode of…
336. …Dr. Who?
337. …Battlestar Galactica?
338. …Space: 1999?
339. …Starblazers? (cartoon about the WWII carrier flying through space)
Can you whistle, hum, sing or snap the theme songs to…
340. …Gilligan’s Island?
341. …Flintstones?
343. …The Jetson’s?
344. …Dobbie Gillis?
346. …I Dream of Genie?
347. Have you ever seen any of the “Revenge of The Nerd” movies more than once?
348. Have you seen all of the Star Wars movies?
349. …in one 24 hour period?
350. Have you ever watched something and stated “that’s physically impossible” (due to Newton’s laws, etc.)?
SECTION 8: Leisure Time – Books and Magazines
Have you ever read anything by…
352. …Isaac Asimov?
353. …Arthur C. Clarke?
354. …Robert H. Heinlein?
355. …Piers Anthony?
356. …J.R.R. Tolkein?
358. …Richard Feynman? (e.g. his lectures, etc.)
359. …Stephen Hawkings?
360. …Carl Sagan?
361. Have you ever read -Cultural-Literacy- or any other book on “what you, as an intelligent person, should know”?
363. Do you read books on a daily basis?
364. Have you finished a book in the last week?
365. Have you finished more than one book in the last week?
366. Have you ever bought a book of crossword puzzles/logic problems?
367. Do you read archaic computer manuals for pleasure?
Do you have magazine subscriptions to…
368. …Popular Mechanics?
369. …Popular Science?
370. …Omni?
371. …Scientific America?
372. …any computer oriented magazine? (MacWorld, PCWorld, etc.)
373. …Computer Gaming World or other “video game” magazine?
374. …Discover?
375. …any medical journals? (New England Journal of Medicine)
376. …any science periodicals?
377. …National Geographic?
378. …any comic book or “graphic novel”? (X-Men, Superman, Heavy Metal)
SECTION 9: Star Trek
379. Can you name or discuss the plots of more than 10 Star Trek episodes?
380. Have you seen all of the Star Trek films?
381. …in one 48 hour period?
382. Do you refer to the various “Treks” as “TOS” (The Original Series), “TNG” (The Next Generation) and “DS9” (Deep Space 9) or similar?
383. Have you ever argued with someone over which “Trek” is better?
384. Have you ever argued over who was a better commander of the Enterprise?
385. Have you ever felt the urge to learn the Klingon language?
386. Have you ever been to a trek convention?
387. …in the last six months?
388. Have you ever owned a pair of Spock ears (Star Trek)?
389. …and worn them in public?
SECTION 10: Clothing and Apparel
391. Do you own a digital watch?
392. …that plays music?
393. …that’s currently set to chime on the hour?
394. …that has a calculator built in?
395. Do you own a pocket protector?
396. …and are you wearing it?
397. Do you have acne?
398. Do you have greasy hair?
399. …without realizing it?
400. Do you own any clothing with scientific knowledge printed on it? (e.g. t-shirts with Maxwell’s equations)
401. …which you still wear from time to time?
403. …while not in the laboratory?
404. …and described what it was to someone, who then backed away in fear?
405. Are your pants too short?
407. Is your outfit uncoordinated? (have someone else evaluate this)
408. Have you ever worn a button-down shirt and left the tails hanging out?
409. Have you ever bought similar looking shirts/pants in order to save time when dressing because “everything goes together”?
410. Do you wear glasses?
411. …held together by adhesives? (tape, glue, boogers)
412. Is your vision worse than 20/40? (in either eye)
413. Is your vision worse than 20/80? (in either eye)
414. Do your glasses weigh more than one pound?
SECTION 11: Personality and Lifestyle
415. Have you ever slept an inverted day? (sleep at dawn, wake at dusk)
416. …for more than one day in a row?
417. Have you ever slept round the clock? (24 continuous hours in bed)
Which of the following have you used to prevent sleep?
418. …Caffeine?
419. …exercise?
420. …Vivarin?
421. …NoDoz?
422. …something you made in chem. lab?
423. …something you found in chem. lab?
424. Have you worked for an engineering or manufacturing firm?
425. …in the last 3 months?
426. …and gotten credit at a school for doing so?
427. Have you worked in a research lab?
428. …and been more interested in the work than the pay?
429. Have you ever visited a power plant? (Hoover Dam, nuclear plant, etc.)
430. …and not been bored?
431. Are you socially inept?
432. Was the last naked person you saw a hi-res computer scan?
433. Do you talk to yourself?
434. …when other people are around?
435. Do you talk to imaginary people?
436. …do they talk back?
437. …do they seem to be more/less intelligent than you?
438. Do you have a tough time remembering people’s names?
439. …but no trouble with their numeric data? (phone#, SS#)
440. Have you ever played mathematical games with other numbers you see to pass the time? (square/cube root, prime factors of phone#)
441. Do you see everyday situations as representing mathematical concepts?
442. Do you look at quantitative factors when participating in social events? (ex: choosing drinks by % alcohol rather than taste)
443. Mark this true if you did NOT go to your senior prom.
444. Did you go stag to your Senior Prom?
445. Have you ever found a grammatical error in a published book?
446. Have you ever quoted a piece of literature from memory? (poem, quote)
447. Have you ever eaten pizza cold?
448. …do you like it that way?
449. …because you’re too lazy to reheat it?
450. Have you ever gotten pizza delivered to the lab/office/science building?
451. Is any leftover delivery food currently residing in your refrigerator?
452. …that’s been there so long, you can’t remember ordering it?
453. …that’s been there so long, it’s become mobile/sentient?
454. Is any food in your refrigerator moldy?
455. Have you ever commented on the lack of intellectual ability found in a “JEOPARDY” contestant?
456. Have you ever contemplated the meaning of life/existence of God?
457. …while not drunk?
458. …while alone?
459. Have you ever thought about extra dimensions/parallel universes?
460. …and discussed their possibilities with others?
461. Have you come to any conclusions about UFO’s/life on other planets?
462. …and used Time-Life’s “Mysteries of the Unknown” series as a factual reference to support your claim?
463. Have you ever commented: “If I drive fast enough at the red light, it’ll appear green.”
464. Have you ever found yourself discussing one of the popular scientific theories of the day with someone you just met? (cold fusion)
465. …did they bring it up because they thought you incapable of talking about non-technical topics?
466. Have you ever taken part in an experiment to prove/disprove one of the popular scientific theories of the day? (cold fusion, big bang)
468. …for sexual purposes?
469. …and had some degree of success?
470. …but been laughed at by a leading medical institution?
471. Have you ever given an inanimate object a name? (inc.: stuffed animal)
472. Was the object something electronic or mechanical?
473. Did the object also have a “personality”?
474. Have you ever compared and contrasted two scientists? (Einstein vs. Newton, etc.)
475. Have you ever argued with someone else over which of two scientists was better?
476. Have you ever argued with someone over which of two computer types/OS’s is better? (Macintosh vs. IBM, UNIX vs. VMS)
477. Have you ever laughed out loud at a joke written in a serious scientific paper? (Feynman’s lectures, textbook)
478. Has anyone ever called you a geek/nerd?
479. …in the last two weeks?
480. …for doing/saying something you knew to be geeky?
481. Have you ever intentionally done something that you consider geeky?
482. …in the last month?
483. …today?
SECTION 12: The Nerd Test
484. Are you taking this test alone?
485. Are you currently reading this test on a computer screen?
487. Do you feel the need (or are you currently using) a calculator to score the test?
488. Are you computing your score in scientific notation?
489. Have you contemplated writing a computer program that would ask and/or tabulate questions found on this test?
490. Are you currently scoring this test in reverse? (i.e. Assuming 100% nerd and deducting for each ‘no’?)
491. Have you come across copies of this test from two separate sources?
492. If you are still reading this test, do you really need a test score to prove you are a nerd?
493. Is your nerdity test score higher than your purity test score?
494. Did you feel offended by any of the questions on this test?
495. Did you resort to lying in order to raise your score?
496. Did you resort to lying in order to lower your score?
497. Are you currently competing with someone else for the highest score on this test (or were contemplating it)?
498. …did you come up second best and challenge them to a rematch?
499. Have you asked for a technical clarification of anything on this test?
500. Have you ever thought of a question that belongs on this test?
To analyze your Nerdity Quotient, divide your total number of “yes/true” responses by the total number of questions and compare to this list.
Ranking:
0 – 20 Nerd-wannabe
21 – 30 Nerd-in-Training
31 – 35 Closet nerd
36 – 40 You dress like people in Walmart ads
41 – 45 You refuse to live anywhere without pizza delivery service
46 – 50 Your social life needs some serious help
51 – 55 YOU need some serious help
56 – 60 You are on first name basis with Radio Shack employees
61 – 65 Your best friend is a microchip
66 – 70 Bill Gates and E. Gary Gygax are your heroes
71 – 75 You own more surge protectors than cooking utensils
76 – 80 “Revenge of the Nerds” poster-child
81 – 85 Hoping to invent Warp Field Theory or transporter technology
86 – 90 Desperately seeking cybernetic interface implanted in your brain
91 – 99 Move over, Einstein
100 Hail, O Nerd Master, virgin sliderulers I sacrifice unto you
### Arkansas Residency Application
```Name: ________________
(_) Billy-Bob
(_) Billy-Joe
(_) Billy-Ray
(_) Billy-Sue
(_) Billy-Mae
(_) Billy-Jack
(Check appropriate box)
Age: ____
Sex: ____ M _____ F _____ N/A
Shoe Size ____ Left ____ Right
Occupation:
(_) Farmer
(_) Mechanic
(_) Hair Dresser
(_) Un-employed
Spouse's Name: __________________________
Relationship with spouse:
(_) Sister
(_) Brother
(_) Aunt
(_) Uncle
(_) Cousin
(_) Mother
(_) Father
(_) Son
(_) Daughter
(_) Pet
Number of children living in household: ___
Number that are yours: ___
Mother's Name: _______
Father's Name: _______(If not sure, leave blank)
Education: 1 2 3 4 (Circle highest grade
completed)
Do you (_)own or (_)rent your mobile home?
___ Total number of vehicles you own
___ Number of vehicles that still crank
___ Number of vehicles in front yard
___ Number of vehicles in back yard
___ Number of vehicles on cement blocks
Firearms you own and where you keep them:
____ truck
____ bedroom
____ bathroom
____ kitchen
____ shed
Model and year of your pickup: ______ 194_
Do you have a gun rack?
(_) Yes (_) No; please explain:
Newspapers/magazines you subscribe to:
(_) The National Enquirer
(_) The Globe
(_) TV Guide
(_) Soap Opera Digest
(_) Rifle and Shotgun
___ Number of times you've seen a UFO
___ Number of times you've seen Elvis
___ Number of times you've seen Elvis in a UFO
How often do you bathe:
(_)Weekly
(_)Monthly
(_)Not Applicable
Color of teeth:
(_)Yellow
(_)Brownish-Yellow
(_)Brown
(_)Black
(_)N/A
Brand of chewing tobacco you prefer:
(_)Red-Man | 8,131 | 32,569 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-17 | longest | en | 0.960794 |
https://python-book.softuni.mn/chapter-05-loops-exam-problems.html | 1,679,592,268,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945182.12/warc/CC-MAIN-20230323163125-20230323193125-00795.warc.gz | 530,053,488 | 15,141 | # Chapter 5.2 Loops - Exam Problems
In the previous chapter, we learned how to run a command block more than once. That's why we implemented for loop and covered some of its main applications. Our task in the current chapter is to hone our knowledge by solving a couple of more complex problems with loops, which appear in exams. For some of them, we’ll show detailed solved examples, while for others there’ll be tips only. Before we begin, we’ll recall the for loop construction:
for loops consist of:
• Initialization block, where the variable-counter (i) is declared, and with the help of the range(…) function built into Python, we define what its starting and ending value will be.
• Updating the counter – we implement it as a third parameter in the range(…) function, and it shows with how many steps the variable-counter should be updated.
• Loop body - it has a random block full of source code.
## Exam Problems
Let’s solve a couple of problems with loops in SoftUni’s exams.
## Problem: Histogram
We’re given n-count integers in the range [1 … 1000]. A percent of them, p1, is under 200, p2 percent are between 200 and 399, p3 percent are between 400 and 599, p4 percent are between 600 and 799, and the remaining p5 percent begin at 800. Write a program that calculates and prints the percentages p1, p2, p3, p4, and p5.
Example: we have n = 20 integers: 53, 7, 56, 180, 450, 920, 12, 7, 150, 250, 680, 2, 600, 200, 800, 799, 199, 46, 128, 65. We get the following distribution and visualization:
Group Numbers Number count Percentage
< 200 53, 7, 56, 180, 12, 7, 150, 2, 199, 46, 128, 65 12 p1 = 12 / 20 * 100 = 60.00%
200 … 399 250, 200 2 p2 = 2 / 20 * 100 = 10.00%
400 … 599 450 1 p3 = 1 / 20 * 100 = 5.00%
600 … 799 680, 600, 799 3 p4 = 3 / 20 * 100 = 15.00%
≥ 800 920, 800 2 p5 = 2 / 20 * 100 = 10.00%
### Input Data
On the first line of the input is an integer n ( 1 <= n <= 1000 ), which stands for the number of lines with numbers, which will be given to us. On the next n lines, there is one integer in the range [1 … 1000] – the numbers that the histogram will be based on.
### Output Data
In the console, print a histogram of 5 lines, each of them containing a number between 0% and 100%, formatted with two-digit precision after the decimal point (for example, 25.00%, 66.67%, 57.14%).
### Sample Input and Output
Input Output Input Output
3
1
2
999
66.67%
0.00%
0.00%
0.00%
33.33%
4
53
7
56
999
75.00%
0.00%
0.00%
0.00%
25.00%
Input Output Input Output
7
800
801
250
199
399
599
799
14.29%
28.57%
14.29%
14.29%
28.57%
9
367
99
200
799
999
333
555
111
9
33.33%
33.33%
11.11%
11.11%
11.11%
Input Output
14
53
7
56
180
450
920
12
7
150
250
680
2
600
200
57.14%
14.29%
7.14%
14.29%
7.14%
### Hints and Guidelines
The program that solves this problem can be divided theoretically into three parts:
• Reading the input data – in the current problem, this means reading the integer n, followed by a count of n integers, each on a new line.
• Processing The Input Data – in this case, this means dividing the numbers into groups and calculating the division percentage by those groups.
• Outputting the final result – printing the histogram in the console, in the given format.
#### Processing The Input Data
Before we transition to the real reading of the input, we have to declare our variables, in which the data will be stored:
We declare variables p1_percentage, p2_percentage, etc., in which we’ll store the percentages, as well as cnt_p1, cnt_p2, etc., in which we’ll keep the count of numbers for the respective group.
After we’ve declared the needed variables, we can move on to reading the number n from the console:
#### Processing The Output Data
To read and assign each number to its respective group, we’ll use a for-loop from 0 to n (the count of the numbers). Each iteration of the cycle will read and assign only one number (current_number) to its respective group. So that we can decide if a selected number belongs to a group, we check its range. If it passes, we increase the count of this group’s numbers (cnt_p1, cnt_p2, etc.) by 1:
After we’ve found out how many numbers there are in each group, we can move on to calculating the percentages, which is also the main part of the problem. We’ll use the following formula:
(Group percentage) = (Group number count) * 100 / (Count of all numbers)
It doesn’t matter whether we’ll divide by 100 (an integer type), or 100.0(a float type) since the division will take place and the result will be saved to the variable. Example: 5 / 2 = 2.5, and 5 / 2.0 = 2.5. In Python 3, there’s no difference whether we’ll be dividing by an integer or a real number - if the result is a real number itself, then it will be saved in the variable as a floating-point number. But in Python 2.7 we have to convert the numbers to a float type, to get the correct result – a real number. Having that in mind, the first variable’s formula will look like this:
To better understand what’s happening, let’s look at the following example:
Input Output
3
1
2
999
66.67%
0.00%
0.00%
0.00%
33.33%
In this case, n = 3. The cycle consists of:
• i = 0 - we read the number 1, which is lower than 200 and belongs to the first group, thus increasing the counter of the number (cnt_p1) by 1.
• i = 1 – we read the number 2, which, again, belongs to the first group and we increase the group’s counter(cnt_p1) by 1.
• i = 2 – we read the number 999, which belongs to the last group(p5), because it’s bigger than 800, and we increase its group counter (cnt_p5) by 1.
After reading the numbers, we have two of them in the first group, and we have only one in the last group. There are no numbers in the other groups. After we apply the aforementioned formula, we calculate the percentage of each group. It doesn’t matter whether we multiply by 100 or 100.0 – we’ll get the same result: the first group has 66.67%, and the last group – 33.33%. We have to mention that this is valid only for Python 3.
#### Printing The Final Result
The last step is to print the calculated results. In the problem’s description, it’s said that the percentages have to be with 2-digit precision after the decimal point. To achieve this, we have to write .2f after the placeholder.
## Problem: Smart Lily
Lily is N years old. Each birthday she receives a gift. For her odd birthdays (1, 3, 5, …, n) she receives toys, and for each even birthday (2, 4, 6, …, n) she receives money. For her second birthday she receives 10.00 USD, and the sum increases by 10 USD with each following even birthday (2 -> 10, 4 -> 20, 6 -> 30, etc.). Lily has secretly been saving the money for years. Her brother, in the years when she receives money, takes 1.00 USD. Lily sold the toys received with the years, each for P USD, and added the sum to the saved money. With them, she wants to buy a washing machine for X USD. Write a program that calculates how much money she has saved and whether it's enough to buy a washing machine.
### Input Data
3 numbers are read from the console, each on a new line:
• Lily's ageinteger in the range [1 … 77].
• The price of the washing machine – a number in the range [1.00 … 10 000.00].
• The price of a single toyinteger in the range [0 … 40].
### Output Data
Print a single line in the console:
• If Lily's money is enough:
• "Yes! {N}" – where N is the remaining money after the purchase
• If it is not:
• "No! {M}" – where M is the amount of money lacking
• The numbers N and M should be formatted with 2-digit precision after the decimal point.
### Sample Input and Output
10
170.00
6
Yes! 5.00
On the first birthday she receives a toy; Second -> 10 USD; 3rd -> toy; 4th -> 10 + 10 = 20 USD; 5th -> toy; 6th -> 20 + 10 = 30 USD; 7th -> toy; 8th -> 30 + 10 = 40 USD; 9th -> toy; 10th -> 40 + 10 = 50 USD
She has saved -> 10 + 20 + 30 + 40 + 50 = 150 USD. She has sold 5 toys by 6 USD = 30 USD.
Her brother took 1 USD for 5 years = 5 USD. Remaining money -> 150 + 30 – 5 = 175 USD. 175 >= 170 (washing machine's price) She has succeeded buying it and there remain 175-170 = 5 USD.
21
1570.98
3
No! 997.98
She has saved 550 USD. She's sold 11 toys by 3 USD each = 33 USD. Her brother has taken 1 USD for 10 years = 10 USD. There remain 550 + 33 – 10 = 573 USD.
573 < 1570.98She's failed buying a washing machine. She needs 1570.98–573 = 997.98 USD more.
### Hints and Guidelines
Solving this problem, like the previous one, again can be divided into three parts – reading the input data, processing it, and outputting a result.
As we already know, like in most scripting languages, in Python as well, we don't bother defining the types of the variables that we declare. The interpreter decides on its own what it'll be. For Lily's (age) and a single toy's price (present_price) in the problem's description, it's said that they'll be integers. That's why we'll use the built-in function int() to convert the read value from string to integer. When the input() function is used, the input's value in the console is always (string), that's why if a conversion to another type is needed, we can use the built-in functions of Python for this problem. For the washing machine's price, (price_of_washing_machine), we know that it's a fractional number and we choose the float type. In the code below we declare and initialize (assign a value) to the variables:
To solve the problem, we'll need a couple of helper variables – for the toys' count (number_of_toys) for the saved money (saved_money) and the money received on each birthday (money_for_birthday). We initially assign 10 to money_for_birthday, because in the description it's said that the first sum received by Lily is 10 USD:
With a for loop, we go through each of Lily's birthdays. When a loop variable is an even number, it means that Lily has received money and we add them to her savings. At the same time, we subtract 1 USD - the money taken by her brother. After that we increase the value of the variable money_for_birthday, meaning we increase the sum by 10 for the next time she receives money for her birthday. Contrary, when the loop variable is an odd number, we increase the toys' count. Checking whether it's even or odd happens with a division with the remainder (%) by 2 – when the remainder is 0, the number is even, and when the remainder's 1 - it's odd:
We add the money from the sold toys to Lily's savings:
At the end we print the results, taking into account the required formatting, meaning the sum has to be rounded to 2 digits after the decimal point:
In some programming languages there's a construction called conditional operator (?:) (also known as ternary operator), as it's shorter to write. It has the following syntax in Python: operand1 if operand2 else operand3. The second operand is our condition and it has to be of bool type (meaning it has to return true/false). If operand2 returns true, it'll execute operand1, and if it returns falseoperand3. In our case, we check whether Lily's saved money is enough to buy a washing machine. If it's higher or equal to its price, the check saved_money >= price_of_washing_machine will return true and it'll print "Yes! …", while if it's lower – the result will be false, and "No! …" will be printed. Of course, instead of the ternary operator, we can use simple if expressions.
## Problem: Back to The Past
John is 18 years old and receives an inheritance of X USD and a time-traveling machine. He decides to travel back to 1800, but he doesn't know whether the money is enough to live without working. Write a program that calculates whether John will have enough money, to live without working until a given year, including the year itself. We accept that each even year (1800, 1802, etc.) he'll spend 12 000 dollars. For each odd year (1801, 1803, etc.) he'll spend 12 000 + 50 * [John's age in the given year].
### Input Data
The input is read from the console and contains exactly 2 lines:
• The inherited money – a real number in the range [1.00 … 1 000 000.00].
• The year until he has to live (inclusive) – a real number in the range [1801 … 1900].
### Output Data
Print to the console 1 line. The sum has to be formatted with 2-digit precision after the decimal point:
• If the money is enough:
• "Yes! He will live a carefree life and will have {N} dollars left." – where N is the remaining money.
• If the money is NOT enough:
• "He will need {M} dollars to survive." – where M is the amount lacking.
### Sample Input and Output
Input Output Explanation
50000
1802
Yes! He will live a carefree life and
will have 13050.00 dollars left.
1800 → even
→ Spends 12000 dollars
→ Left 50000 – 12000 = 38000
1801 → odd
→ Spends 12000 + 19*50 = 12950 dollars
→ Left 38000 – 12950 = 25050
1802 → even
→ Spends 12000 dollars
→ Left 25050 – 12000 = 13050
100000.15
1808
He will need 12399.85 dollars
to survive.
1800 → even
→ Left 100000.15 – 12000 = 88000.15
1801 → odd
→ Left 88000.15 – 12950 = 75050.15
1808 → even → -399.85 - 12000 = -12399.85
12399.85 lacking
### Hints and Guidelines
The method of solving this problem isn't unlike the previous ones, so we begin by declaring and initializing the needed variables. In the problem's description, it's said that John's age is 18, so we declare the variable years with the initial value of 18. We read other variables from the console:
With the help of a for loop, we loop through all years. We begin at 1800 – the year when John travels back in time, and we end at the year until he has to live. In the loop, we check whether the current year is even or odd. We check it with division with a remainder (%) by 2. If the year is even, from the inheritance (inheritance) we subtract 12000, while if it's odd, from the inheritance (inheritance) we subtract 12000 + 50 * (John's age):
In the end, we print the results, and we do a check whether the inheritance (inheritance) has been enough for him to live without working or not. If the inheritance (inheritance) is a positive number, we print: "Yes! He will live a carefree life and will have {N} dollars left.", while if it's a negative number: "He will need {M} dollars to survive.". We don't forget to format the sum with 2-digit precision after the decimal point.
Hint: Think about using the function abs(…) when printing the output and the inheritance is not enough.
## Problem: Hospital
For a given amount of time, patients arrive for a checkup in the hospital every day. She initially has 7 doctors. Each of them can check one patient a day only, but sometimes there's a shortage of doctors, so the other patients are sent to other hospitals. Every third day the hospital calculates whether the count of patients that haven't been examined is higher than those that have been and if so, an additional doctor is assigned. The assignment happens before the start of the day.
Write a program that calculates the count of treated and untreated patients for the given period.
### Input Data
The input is read from the console and contains:
• On the first line – the period for which we'll be calculating. Integer in the range [1 … 1000].
• On the following lines (equal to the number of days) – the count of patients that arrive for a checkup on the current day. Integer in the range [0 … 10 000].
### Output Data
Print to the console 2 lines:
• On the first line: "Treated patients: {count of treated patients}."
• On the second line: "Untreated patients: {count of untreated patients}."
### Sample Input and Output
Input Output Explanation
4
7
27
9
1
Treated patients: 23.
Untreated patients: 21.
Day 1: 7 treated and 0 untreated patients for the day
Day 2: 7 treated and 20 untreated patients for the day
Day 3: Until now the treated patients are 14,
and the untreated – 20 –> A new doctor is assigned
–> 8 treated and 1 untreated patients for the day
Day 4: 1 treated and 0 untreated patients for the day
Total: 23 treated and 21 untreated patients.
Input Output
6
25
25
25
25
25
2
Treated patients: 40.
Untreated patients: 87.
3
7
7
7
Treated patients: 21.
Untreated patients: 0.
### Hints and Guidelines
Again, we begin by declaring and initializing the needed variables. The period, for which we have to do our calculations, we read from the console and assign to the variable period. We'll need a couple of additional variables: the count of treated patients (treated_patients), the count of untreated patients (untreated_patients), and the count of doctors (count_of_doctors), which is initially 7:
With the help of the for loop, we go through all days in the given period (period). For each day we read the number of patients from the console (current_patients). The addition of doctors is said in the problem's description to happen every third day, BUT only if the untreated patients' count is higher than the treated's count. That's why we check whether the day is the third one – with the arithmetic operator for division with a remainder (%): day % 3 == 0.
Example:
• If the day is a third one, the remainder of division by 3 will be 0 (3 % 3 = 0) and the check day % 3 == 0 will return true.
• If the day is a second one, the remainder of division by 3 will be 2 (2 % 3 = 2) and the check will return false.
• If the day is a fourth one, the remainder of the division will be 1 (4 % 3 = 1) and the check will again return false.
If the conditional check day % 3 == 0 returns true, there'll also be a check whether the count of untreated patients is higher than the treated's count: untreated_patients > treated_patients. If the result is again true, then the count of doctors (count_of_doctors) will increase.
After that, we check whether the patients' count for the current day (current_patients) is higher than the doctors' count (count_of_doctors). If the patients' count is higher:
• We increase the value of the variable treated_patients with the doctors' count (count_of_doctors).
• We increase the value of the variable untreated_patients with the count of patients left, which we calculate by subtracting the doctors' count from the patients' count (current_patients - count_of_doctors).
If the patients' count is lower, we increase only the variable treated_patients with the current day's count of patients (current_patients):
In the end, we only have to print the count of treated and untreated patients.
## Problem: Division
We're given n integers in the range [1 … 1000]. A percentage of them, percent p1, are divided by 2 without remainder, percent p2 are divided by 3 without remainder, percent p3 are divided by 4 without remainder. Write a program that calculates and prints the percentages p1, p2, and p3. Example: we have n = 10 integers: 680, 2, 600, 200, 800, 799, 199, 46, 128, 65. We get the following distribution and visualization:
Division without remainder by: Numbers Count Percentage
2 680, 2, 600, 200, 800, 46, 128 7 p1 = (7 / 10) * 100 = 70.00%
3 600 1 p2 = (1 / 10) * 100 = 10.00%
4 680, 600, 200, 800, 128 5 p3 = (5 / 10) * 100 = 50.00%
### Input Data
On the first line of the input stands the integer n (1 ≤ n ≤ 1000) – count of numbers. On the next n lines, each stands a single integer in the range [1 … 1000] – the numbers for which we'll check their divisors.
### Output Data
Print to the console 3 lines, each of them containing a percentage between 0% and 100%, with 2-digit precision after the decimal point, for example: 25.00%, 66.67%, 57.14%.
• On the first line – the percentage of numbers divisible by 2.
• On the second line – the percentage of numbers divisible by 3.
• On the third line – the percentage of numbers divisible by 4.
### Sample Input and Output
Input Output Input Output Input Output
10
680
2
600
200
800
799
199
46
128
65
70.00%
10.00%
50.00%
3
3
6
9
33.33%
100.00%
0.00%
1
12
100.00%
100.00%
100.00%
### Hints and Guidelines
For this and the next problem, you'll have to write the program's code on your own, with the help of the following advice.
The program that solves the current problem is similar to the one from the problem Histogram, which we viewed previously. That's why we can begin by declaring our needed variables. For example, a couple of variable names can be n – count of numbers (which we'll read from the console) and divisible_by_2, divisible_by_3, divisible_by_4 – additional variables, storing the count of numbers in their respective groups.
To read and distribute each number in its respective group, we'll have to start ourfor loop from 0 and end at n (the numbers' count). Each of the loop's iterations has to read and distribute a single number. The difference is that a single number can be distributed to more than one group simultaneously, so we have to do three different if checks for each number – whether it's divided by 2, 3, and 4, and to increase the value of the variable that stores the count of numbers in the respective group.
Warning: an if-elif construction won't be of use to us, since when it detects a match, the loop gets broken before checking the following conditions.
In the end, print the found results while keeping the given in the problem's description format.
## Problem: Logistics
You're responsible for the logistics of different cargos. Depending on the weight of each load, a different vehicle is needed and costs a different price per ton:
• Up to and including 3 tonsvan (200 USD per ton).
• Over 3 and up to 11 tonstruck (175 USD per ton).
• Over 11 tons – train (120 USD per ton).
Your task is to calculate the average price per ton of transported load, as well as what percentage of the overall cargo is transported by each vehicle.
### Input Data
A sequence of numbers is read from the console, each on a different line:
• First line: count of loads for transportation – integer in the range [1 … 1000].
• On each following line, the weight in tons of the current load is written – integer in the range [1 … 1000].
### Output Data
Print 4 lines to the console, as given:
• Line #1the average price per ton of transported cargo (rounded to the second digit after the decimal point).
• Line #2the percentage of cargo, transported with a van (between 0.00% and 100.00%, rounded to the second digit after the decimal point).
• Line #3the percentage of cargo, transported with a truck (between 0.00% and 100.00%).
• Line #4the percentage of cargo, transported with a train (between 0.00% and 100.00%).
### Sample Input and Output
Input Output Explanation
4
1
5
16
3
143.80
16.00%
20.00%
64.00%
Two of the cargos are transported with a van: 1 + 3, total 4 tons.
One cargo is transported with a truck: 5 tons.
One cargo is transported with a train: 16 tons.
The sum of all cargos is: 1 + 5 + 16 + 3 = 25 tons.
The percentage of van-transported cargo: 4/25*100 = 16.00%
The percentage of truck-transported cargo: 5/25*100 = 20.00%
The percentage of train-transported cargo: 16/25*100 = 64.00%
Average price per ton of transported cargo: (4 * 200 + 5 * 175 + 16 * 120) / 25 = 143.80
Input Output Input Output
5
2
10
20
1
7
149.38
7.50%
42.50%
50.00%
4
53
7
56
999
120.35
0.00%
0.63%
99.37%
### Hints and Guidelines
First, we'll read the weight of each load and we'll sum how many tons are being transported by a van, truck, and train respectively, and we'll additionally calculate the total tons of transported cargos. We'll calculate the prices of each transport type according to the total tons and the total price. In the end, we'll calculate and print the total average price per ton and what part of the overall load is transported by each transport type, in percentages.
We declare our variables, for example: count_of_loads – the count of cargos to be transported (we read them from the console), sum_of_tons – the sum of the overall load's weight, microbus_tons, truck_tons, train_tons – variables, holding the sum of the weights transported respectively by a van, truck, and train.
We'll need a for loop from 0 to count_of_loads - 1, to go through all loads. For each load, we read its weight (in tons) from the console and assign it to our variable, for example, tons. To the sum of all loads (sum_of_tons), we add our current load's weight (tons). After we've read our current load's weight, we have to decide which transport vehicle will be used for it (van, truck or train). For this we'll need some if-elif checks:
• If the value of the variable tons is lower than 3, we increase the value of the variable microbus_tons by the value of tons:
microbus_tons += tons;
• Else, if the value of tons is up to 11, we increase truck_tons by tons.
• If tons is higher than 11, we increase train_tons by tons.
Before we print our output, we have to calculate the percentage of tons transported by each vehicle and the average price per ton. For the average price, we'll declare another additional variable total_price, in which we'll sum the total price of all transported loads (with a van, truck, and train). The average price will be calculated by dividing total_price by sum_of_tons. You're left with calculating by yourself the percentage of tons transported by each vehicle and printing the output, adhering to the format as shown in the problem's description. | 6,697 | 25,130 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2023-14 | latest | en | 0.872835 |
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