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https://itstillruns.com/convert-gallon-km-per-litre-5027208.html | 1,721,926,010,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763860413.86/warc/CC-MAIN-20240725145050-20240725175050-00550.warc.gz | 280,500,461 | 26,663 | # How to Convert Miles Per Gallon to Km Per Litre
by Johnno CaryUpdated August 02, 2023
Convert miles into kilometers. Since a kilometer is shorter than a mile, and there are 1.61 kilometers in every mile, you will have to multiply to get the answer. Doing the math (25 x 1.61 = 40.25) shows that 25 miles is equivalent to 40.25 kilometers.
Convert gallons into litres. Since a litre is smaller than a gallon, you must again multiply to get the answer. There are 3.79 litres in every gallon. Here the math is simple, because you are using only 1 gallon. 1 x 3.79 = 3.79, so a gallon of gas is equal to 3.79 litres.
Perform the calculation to determine how many kilometers are attributable to each litre of fuel consumed. Since you have determined that the vehicle traveled 40.25 kilometers on 3.79 litres of fuel, simple division will tell you how many kilometers are traveled on 1 litre. 40.25 ÷ 3.79 = 10.62. 25 mpg is therefore equivalent to 10.62 km/l.
Create a conversion factor, based on the above calculations. One mile is equal to 1.61 kilometers. One gallon is equal to 3.79 litres. Simple division will give you a conversion factor. 1.61 ÷ 3.79 = 0.425. This means that 0.425 km/l equals 1 mpg. You can use the newly calculated conversion factor by multiplying miles per gallon by 0.425. Checking your previous work, 25 mpg x 0.425 = 10.62 km/l, which is identical to the answer in the three-step calculation above. | 380 | 1,430 | {"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-2024-30 | latest | en | 0.891931 |
http://www.solutioninn.com/are-students-who-participate-in-sports-more-extraverted-than-those | 1,506,022,905,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687837.85/warc/CC-MAIN-20170921191047-20170921211047-00522.warc.gz | 573,793,523 | 8,088 | # Question: Are students who participate in sports more extraverted than those
Are students who participate in sports more extraverted than those who do not? A random sample of students at a small university were asked to indicate whether they participated in sports (yes or no) and to rate their level of extraversion. Extraverts are outgoing, are talkative, and don't mind being the center of attention. Students were asked whether they agreed with the statement that they were extraverts, using a scale of 1-5 with 1 meaning "strongly disagree" and 5 meaning "strongly agree." There were 51 students who participated in sports and 64 who did not.
The mean extraversion score for those who participated in sports was 3.618, and the mean for those who did not participate was 3.172, a difference of 0.446 point. To determine whether the mean difference was significant, we performed a randomization test to test whether the mean extraversion level was greater for athletic students.
a. The histogram shows the results of 1000 randomizations of the data.
In each randomization, we found the mean difference between two groups that were randomly selected from the combined group of data: the combined data for the sporty and nonsporty. Note that, just as you would expect under the null hypothesis, the distribution is centered at about 0. The red line shows the observed sample mean difference in extraversion for the sporty minus the nonsporty. From the graph, does it look like the observed mean difference is unusual for this data set? Explain.
b. The software output gives us the probability of having an observed difference of 0.446 or more. (See the column labeled "Proportion = 7 Observed"). In other words, it gives us the right tail area, which is the p-value for a one-sided alternative that the mean extraversion score is higher for athletes than for nonathletes. State the p-value.
c. Using a significance level of 0.05, can we reject the null hypothesis that the means are equal?
d. If you did not have the computer output, explain how you would use the histogram to get an approximate p-value.
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5000 | 486 | 2,197 | {"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-2017-39 | latest | en | 0.969899 |
http://slideplayer.com/slide/3615727/ | 1,529,433,922,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863109.60/warc/CC-MAIN-20180619173519-20180619193519-00243.warc.gz | 307,095,365 | 26,769 | # Chapter 20: Testing Hypotheses About Proportions
## Presentation on theme: "Chapter 20: Testing Hypotheses About Proportions"— Presentation transcript:
Chapter 20: Testing Hypotheses About Proportions
Unit 5 AP Statistics
Hypotheses Hypotheses are working models that we adopt temporarily.
Our starting hypothesis is called the null hypothesis. The null hypothesis, that we denote by H0, specifies a population model parameter of interest and proposes a value for that parameter. We usually write down the null hypothesis in the form H0: parameter = hypothesized value. The alternative hypothesis, which we denote by HA, contains the values of the parameter that we consider plausible if we reject the null hypothesis.
Hypotheses (cont.) The null hypothesis, specifies a population model parameter of interest and proposes a value for that parameter. We might have, for example, H0: p = 0.20. We want to compare our data to what we would expect given that H0 is true. We can do this by finding out how many standard deviations away from the proposed value we are. We then ask how likely it is to get results like we did if the null hypothesis were true.
A Trial as a Hypothesis Test
Think about the logic of jury trials: To prove someone is guilty, we start by assuming they are innocent. We retain that hypothesis until the facts make it unlikely beyond a reasonable doubt. Then, and only then, we reject the hypothesis of innocence and declare the person guilty.
A Trial as a Hypothesis Test (cont.)
The same logic used in jury trials is used in statistical tests of hypotheses: We begin by assuming that a hypothesis is true. Next we consider whether the data are consistent with the hypothesis. If they are, all we can do is retain the hypothesis we started with. If they are not, then like a jury, we ask whether they are unlikely beyond a reasonable doubt.
P-Values The statistical twist is that we can quantify our level of doubt. We can use the model proposed by our hypothesis to calculate the probability that the event we’ve witnessed could happen. That’s just the probability we’re looking for—it quantifies exactly how surprised we are to see our results. This probability is called a P-value.
P-Values (cont.) When the data are consistent with the model from the null hypothesis, the P-value is high and we are unable to reject the null hypothesis. In that case, we have to “retain” the null hypothesis we started with. We can’t claim to have proved it; instead we “fail to reject the null hypothesis” when the data are consistent with the null hypothesis model and in line with what we would expect from natural sampling variability. If the P-value is low enough, we’ll “reject the null hypothesis,” since what we observed would be very unlikely were the null model true.
What to Do with an “Innocent” Defendant
If the evidence is not strong enough to reject the presumption of innocent, the jury returns with a verdict of “not guilty.” The jury does not say that the defendant is innocent. All it says is that there is not enough evidence to convict, to reject innocence. The defendant may, in fact, be innocent, but the jury has no way to be sure.
What to Do with an “Innocent” Defendant (cont.)
Said statistically, we will fail to reject the null hypothesis. We never declare the null hypothesis to be true, because we simply do not know whether it’s true or not. Sometimes in this case we say that the null hypothesis has been retained.
What to Do with an “Innocent” Defendant (cont.)
In a trial, the burden of proof is on the prosecution. In a hypothesis test, the burden of proof is on the unusual claim. The null hypothesis is the ordinary state of affairs, so it’s the alternative to the null hypothesis that we consider unusual (and for which we must marshal evidence).
Examples A research team wants to know if aspirin helps to thin blood. The null hypothesis says that it doesn’t. They test 12 patients, observe the proportion with thinner blood, and get a P-value of They proclaim that aspirin doesn’t work. What would you say? An allergy drug has been tested and found to give relief to 75% of the patients in a large clinical trial. Now the scientists want to see if the new, improved version works even better. What would the null hypothesis be? The new drug is tested and the P-value is What would you conclude about the new drug?
The Reasoning of Hypothesis Testing
There are four basic parts to a hypothesis test: Hypotheses Model Mechanics Conclusion Let’s look at these parts in detail…
The Reasoning of Hypothesis Testing (cont.)
Hypotheses The null hypothesis: To perform a hypothesis test, we must first translate our question of interest into a statement about model parameters. In general, we have H0: parameter = hypothesized value. The alternative hypothesis: The alternative hypothesis, HA, contains the values of the parameter we consider plausible when we reject the null.
Hypothesis Example A 1996 report from the U.S. Consumer Product Safety Commission claimed that at least 90% of all American homes have at least one smoke detector. A city’s fire department has been running a public safety campaign about smoke detectors consisting of posters, billboards, and ads on radio and TV and in the newspaper. The city wonders if this concerted effort has raised the local level above the 90% national rate. Building inspectors visit 400 randomly selected homes and find that 376 have smoke detectors. Is this strong evidence that the local rate is higher than the national rate? Set up the hypotheses.
The Reasoning of Hypothesis Testing (cont.)
Model To plan a statistical hypothesis test, specify the model you will use to test the null hypothesis and the parameter of interest. All models require assumptions, so state the assumptions and check any corresponding conditions. Your model step should end with a statement such Because the conditions are satisfied, I can model the sampling distribution of the proportion with a Normal model. Watch out, though. It might be the case that your model step ends with “Because the conditions are not satisfied, I can’t proceed with the test.” If that’s the case, stop and reconsider.
The Reasoning of Hypothesis Testing (cont.)
Model (cont.) Each test we discuss in the book has a name that you should include in your report. The test about proportions is called a one-proportion z-test.
One-Proportion z-Test
The conditions for the one-proportion z-test are the same as for the one proportion z-interval. We test the hypothesis H0: p = p0 using the statistic where When the conditions are met and the null hypothesis is true, this statistic follows the standard Normal model, so we can use that model to obtain a P-value.
The Reasoning of Hypothesis Testing (cont.)
Mechanics Under “mechanics” we place the actual calculation of our test statistic from the data. Different tests will have different formulas and different test statistics. Usually, the mechanics are handled by a statistics program or calculator, but it’s good to know the formulas.
The Reasoning of Hypothesis Testing (cont.)
Mechanics (continued) The ultimate goal of the calculation is to obtain a P-value. The P-value is the probability that the observed statistic value (or an even more extreme value) could occur if the null model were correct. If the P-value is small enough, we’ll reject the null hypothesis. Note: The P-value is a conditional probability—it’s the probability that the observed results could have happened if the null hypothesis is true.
P-value Example A large city’s DMV claimed that 80% of candidates pass driving tests, but a survey of 90 randomly selected local teens who had taken the test found only 61 who passed. Does this finding suggest that the passing rate for teenagers is lower than the DMV reported? What is the P-value for the one-proportion z-test? Don’t forget to check the conditions for inference!
The Reasoning of Hypothesis Testing (cont.)
Conclusion The conclusion in a hypothesis test is always a statement about the null hypothesis. The conclusion must state either that we reject or that we fail to reject the null hypothesis. And, as always, the conclusion should be stated in context.
The Reasoning of Hypothesis Testing (cont.)
Conclusion Your conclusion about the null hypothesis should never be the end of a testing procedure. Often there are actions to take or policies to change.
Conclusion Example Recap: A large city’s DMV claimed that 80% of candidates pass driving tests. Data from a reporter’s survey of randomly selected local teens who had taken the test produced a P-value of What can the reporter conclude? And how might the reporter explain what the P-value means for the newspaper story?
Alternatives Hypotheses
There are three possible alternative hypotheses: HA: parameter < hypothesized value HA: parameter ≠ hypothesized value HA: parameter > hypothesized value
Alternatives Hypotheses (cont.)
HA: parameter ≠ value is known as a two-sided alternative because we are equally interested in deviations on either side of the null hypothesis value. For two-sided alternatives, the P-value is the probability of deviating in either direction from the null hypothesis value.
Alternatives Hypotheses (cont.)
The other two alternative hypotheses are called one-sided alternatives. A one-sided alternative focuses on deviations from the null hypothesis value in only one direction. Thus, the P-value for one-sided alternatives is the probability of deviating only in the direction of the alternative away from the null hypothesis value.
Steps for Hypothesis Testing for One-Proportion z-Tests
Check Conditions and show that you have checked these! Random Sample: Can we assume this? 10% Condition: Do you believe that your sample size is less than 10% of the population size? Success/Failure: 𝒏𝒑 𝟎 ≥𝟏𝟎 and 𝒏𝒒 𝟎 ≥𝟏𝟎 State the test you are about to conduct Ex) One-proportion z-test Set up your hypotheses H0: HA:
Steps for Hypothesis Testing for One-Proportion z-Tests (cont.)
Calculate your test statistic 𝒛= 𝒑 − 𝒑 𝟎 𝒑 𝟎 𝒒 𝟎 𝒏 Draw a picture of your desired area under the Normal model, and calculate your P-value. Make your conclusion. When your P-value is small enough (or below α, if given), reject the null hypothesis. When your P-value is not small enough, fail to reject the null hypothesis.
Testing a Hypothesis Example
Home field advantage –teams tend to win more often when the play at home. Or do they? If there were no home field advantage, the home teams would win about half of all games played. In the 2007 Major League Baseball season, there were 2431 regular-season games. It turns out that the home team won 1319 of the 2431 games, or 54.26% of the time. Could this deviation from 50% be explained from natural sampling variability, or is it evidence to suggest that there really is a home field advantage, at least in professional baseball?
Graphing Calculator Shortcuts
One Proportion Z-Test: Stat TESTS 5: 1-Prop ZTest Po = hypothesized proportion x = number of successes n = sample size Determine the tail Calculate
P-Values and Decisions: What to Tell About a Hypothesis Test
How small should the P-value be in order for you to reject the null hypothesis? It turns out that our decision criterion is context-dependent. When we’re screening for a disease and want to be sure we treat all those who are sick, we may be willing to reject the null hypothesis of no disease with a fairly large P-value (0.10). A longstanding hypothesis, believed by many to be true, needs stronger evidence (and a correspondingly small P-value) to reject it. Another factor in choosing a P-value is the importance of the issue being tested.
P-Values and Decisions (cont.)
Your conclusion about any null hypothesis should be accompanied by the P-value of the test. If possible, it should also include a confidence interval for the parameter of interest. Don’t just declare the null hypothesis rejected or not rejected. Report the P-value to show the strength of the evidence against the hypothesis. This will let each reader decide whether or not to reject the null hypothesis.
Examples A bank is testing a new method for getting delinquent customers to pay their past-due credit card bills. The standard way was to send a letter (costing about \$0.40) asking the customer to pay. That worked 30% of the time. They want top test a new method that involves sending a DVD to customers encouraging them to contact the bank and set up a payment plan. Developing and sending the video costs about \$10 per customer. What is the parameter of interest? What are the null and alternative hypotheses? The bank sets up an experiment to test the effectiveness of the DVD. They mail it out to several randomly selected delinquent customers and keep track of how many actually do contact the bank to arrange payments. The bank’s statistician calculates a P-value of What does this P-value suggest about the DVD? The statistician tells the bank’s management that the results are clear and they should switch to the DVD method. Do you agree? What else might you want to know?
What Can Go Wrong? Hypothesis tests are so widely used—and so widely misused—that the issues involved are addressed in their own chapter (Chapter 21). There are a few issues that we can talk about already, though:
What Can Go Wrong? (cont.)
Don’t base your null hypothesis on what you see in the data. Think about the situation you are investigating and develop your null hypothesis appropriately. Don’t base your alternative hypothesis on the data, either. Again, you need to Think about the situation.
What Can Go Wrong? (cont.)
Don’t accept the null hypothesis. If you fail to reject the null hypothesis, don’t think a bigger sample would be more likely to lead to rejection. Each sample is different, and a larger sample won’t necessarily duplicate your current observations.
Recap We can use what we see in a random sample to test a particular hypothesis about the world. Hypothesis testing complements our use of confidence intervals. Testing a hypothesis involves proposing a model, and seeing whether the data we observe are consistent with that model or so unusual that we must reject it. We do this by finding a P-value—the probability that data like ours could have occurred if the model is correct. | 3,056 | 14,332 | {"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} | 4.375 | 4 | CC-MAIN-2018-26 | latest | en | 0.906831 |
https://physics.stackexchange.com/questions/701647/how-can-one-detect-a-device-phone-falling-using-its-3-axis-accelerometer | 1,696,292,776,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511023.76/warc/CC-MAIN-20231002232712-20231003022712-00526.warc.gz | 482,547,124 | 43,535 | # How can one detect a device (phone) falling using its 3-axis accelerometer?
I am working on an Android app to use its 3-axis inbuilt accelerometer to find if the phone falls.
I googled and found that if the following value is less than 2:
$$a = \sqrt{x^2 + y^2 + z^2}$$
(where $$x, y, z$$ are the readings from accelerometer)
Then the device is considered to be free-falling.
I'm not sure how does this approach detect free-falling and where does the condition $$a < 2$$ come from.
Could anyone explain it?
• Hello! To have a discussion with an answerer, please comment under their answer. Check the edit history (by clicking the “edited X ago” link) to find your remarks.
– rob
Apr 3, 2022 at 12:36
TL;DR: Accelerometer records the $$x,y,z$$ projections of acceleration $$\vec{a_r} = \vec{a} - \vec{g}$$, where $$\vec{a}$$ is the total acceleration of the object.
When the object is free-falling, the recorded acceleration $$\vec{a_r}$$ is zero until air resistance force is large enough. Speed increase could lead to air resistance acceleration being as much as 1 $$\frac{m}{s^2}$$ (0.1 G) on the drops around 1 - 1.5 meters. Recorded acceleration modulo is calculated as $$\sqrt{x^2 + y^2 + z^2}$$ ($$x,y,z$$ - the readings from accelerometer). This is essential info to detect free-fall of your phone.
When your device rests on your desk, or in the palm of your hand, or on anything which causes it not to fall, there're two forces applied to the object: the force of gravity and the normal (ground reaction) force. Having opposite direction and being equal by modulo, these forces essentially compensate each other (so the object doesn't move) while "squeezing" the object, sort of.
I mean, consider this simple image, which you could have seen in a school book:
#### How can we describe these pics?
1. The object is static. If this object is you, you can feel that you're pushed against the ground. That's the normal force. You can definitely feel the gravity, but you do feel it just because the normal force is present.
2. The object is falling. Again, if that's you, you would feel like you're floating (if we don't account the air resistance for simplicity). You can't feel the gravity, since there's nothing that stops your movement - all the gravity force goes into accelerating you and nothing prevents it.
That's because we're under a constant (for a fixed height above the Earth) gravitational force. What does it imply?
1. If an object doesn't move, that means something prevents it. In this case, something stops an object with an acceleration of $$g = 9.8$$ $$\frac{m}{s^2}$$, with the direction of acceleration vector being opposite to the vector of gravity.
2. If an object falls, that means that either nothing stops it (that's free-falling), or something stops the object with an acceleration less than $$g = 9.8$$ $$\frac{m}{s^2}$$ (to be more specific, the projection of stopping acceleration on the $$z$$ axis is less than $$g = 9.8$$ $$\frac{m}{s^2}$$)
Thanks to this definition, we could now move on to the next step:
### What data an accelerometer gives to us?
Basically, accelerometer is giving us an information for the 3 projections (x, y, z) of the recorded acceleration.
Important note: although I've stated that already, it's necessary to understand the actual acceleration of the object is not the acceleration, recorded by the accelerometer. Accelerometer records the force $$\vec{F} = (\vec{a} - \vec{g})m$$ that stops it from free-falling, that could include the ground reaction force or the air resistance force, or any other force applied to the object.
But for the sake of simpleness let's consider a 2D world, with just two axis: x and y. Here you can see what data captures an accelerometer when an object stays on the ground and when it's tilted by 45 degrees (although still static):
There's much fun drawing this and all, but an important conclusion would be: the accelerometer will calculate an acceleration of approx $$g = 9.8$$ $$\frac{m}{s^2}$$ (vector parallel to gravity force) if it's static, i.e. when it's not falling
Let's consider another two examples:
1. An object is moving towards the ground with an acceleration less than $$g = 9.8$$ (falling). This means that the platform under the object is also accelerating towards the ground, but not free-falling either.
2. An object is moving towards the ground with an acceleration of approx $$g = 9.8$$ (free-falling). This means there's no platform or any support under the object.
In the first case, the total acceleration obtained from accelerometer would be approx $$6.8$$ $$\frac{m}{s^2}$$, in the second case - approx $$0$$ $$\frac{m}{s^2}$$.
In the both cases the object can be considered as falling, since the acceleration is less than $$g = 9.8$$ $$\frac{m}{s^2}$$
### So, what's the formula?
Now we can finally understand a formula.
An total modulo acceleration, obtained from the acceleration, would be:
$$|a| = \sqrt{x^2 + y^2 + z^2}$$
If the formula is unclear, you could refer Pythagorean theorem, one of our favorite school theorems:
In the same way you can calculate the acceleration vector length, where x, y, z are the projections. If the term "projection" is unclear, here's a quick explanation: take a stick in the sunny day, its shadow will be its projection.
As we remember, we can consider that the device is falling if $$|a|$$ is less than 9.8, but that doesn't account for a calculation error and air resistance.
### Experiment
In your case, you're mentioning that a phone can be considered to be falling if $$|a|$$ is less than $$2$$. That is valid for a rough approximation (20%), but not necessary at least for iPhones on small drops (without much air resistance). I have dropped my iPhone on the bed and measured the following accelerations during the 1.5 meter fall:
X Y Z t (time)
-0.016 -0.03 -0.151 0.00s
-0.021 0.015 -0.13 0.04s
-0.023 0.013 -0.202 0.08s
-0.016 0.014 -0.245 0.12s
-0.017 0.017 -0.308 0.16s
-0.028 0.02 -0.399 0.20s
-0.017 0.026 -0.462 0.24s
-0.019 0.023 -0.545 0.28s
-0.023 0.023 -0.622 0.32s
-0.031 0.021 -0.717 0.36s
Table 1. Raw X/Y/Z accelerometer input.
Upon calculating the acceleration with formula, we get these values for the beginning of the fall:
Acceleration Mistake or/and air resistance hit
0.154 ~ 1.5%
0.132 ~ 1.3%
0.203 ~ 2%
0.245 ~ 2.4%
0.308 ~ 3.1%
0.4 ~ 4%
0.463 ~ 4.7%
0.545 ~ 5.5%
0.622 ~ 6.3%
0.739 ~ 7.2%
Table 2. Calculated acceleration and its mistake.
Mistake (Y) and the time (X) plotted
As we can see, the mistake varies with a increasing pattern.
### Interesting observations
I believe that an increasing mistake is present because of air resistance. My phone is huge, and I dropped it flat parallel to the ground, that accounts for the maximal possible air resistance - that's exactly what the $$Z$$ axis values tell us (I mean, check the last column of the table 1).
I realize I could also drop it on the side to prove the point, but I'm just afraid that it jumps of the bed (or could it? that's another question...).
### Conclusion
A mistake of 20% sounds like an overkill to detect a start of falling.
Although, if you would like to account the whole process of falling, it could be just enough - it just depends on how much the acceleration slows down during the fall. That said, I doubt that there's any point accounting for mile-height drops, because well... the phone's not likely to last.
Another interesting task would be to record the end of falling. That's not that difficult - an impact will make an acceleration to suddenly jump to 5G's (50 $$\frac{m}{s^2}$$) and even more, which would make it easy to detect. I have got about 7.5 G's peak while dropping my phone onto the bed from the height of about 1 meter.
• 888 words, that would likely make a longest answer of mine Apr 1, 2022 at 12:54
• I haven't managed to understand why the error/time graph looks pretty linear at least during the 0.12 - 0.36s of the fall (isn't air resistance supposed to be like v^2?). As for the first 0.00s - 0.12s, I believe there maybe there's a little friction width my fingers after I've almost released the iPhone but it hasn't left the "starting" position completely. Apr 1, 2022 at 17:06
• OK, thanks! I upvoted. It is a thorough answer.
– hft
Apr 1, 2022 at 19:26
• @Rhino hello! Unfortunately I do not have an android phone, I have downloaded "Accelerometer" app on my iPhone. But you could search Play Store (or any other app store that your phone suppors) for the application with a name similar to "Accelerometer" Apr 1, 2022 at 19:35
• @nicael. Thank you very much for your well elaborate answer and it means a lot to me , I have summarise my understanding in my question , can you please correct me if I understood wrong Apr 2, 2022 at 23:55 | 2,354 | 8,791 | {"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": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 34, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2023-40 | longest | en | 0.939458 |
https://homework.cpm.org/category/MN/textbook/cc3mn/chapter/10/lesson/10.3.3/problem/10-107 | 1,723,658,359,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641118845.90/warc/CC-MAIN-20240814155004-20240814185004-00547.warc.gz | 230,967,871 | 15,612 | ### Home > CC3MN > Chapter 10 > Lesson 10.3.3 > Problem10-107
10-107.
Solve the equations below by first changing each equation to a simpler equivalent equation. Check your solutions.
1. $3000 x - 2000 = 10{,}000$
• Divide by a common factor, then solve.
1. $\frac { - 2 x } { 3 } - \frac { x } { 7 } = 17$
• Multiply by the Least Common Multiple of $3$ and $7$.
1. $\frac { 5 } { 2 } x - \frac { 1 } { 3 } = 13$
• Multiply the equation by a common denominator.
1. $\frac { 3 } { 10 } + \frac { 2 x } { 5 } = \frac { 1 } { 2 }$
• See part (c) for help.
$x = \frac{1}{2}$ | 212 | 581 | {"found_math": true, "script_math_tex": 7, "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-2024-33 | latest | en | 0.628557 |
https://www.scribd.com/document/22171447/Sequences-Worksheet-01-Number-from-GCSE-Maths-Tutor | 1,542,449,599,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743351.61/warc/CC-MAIN-20181117082141-20181117104141-00091.warc.gz | 983,693,638 | 28,978 | You are on page 1of 3
# Number - sequences
GCSE Maths Tutor
Sequences
question sheet N-SEQ-01
www.gcsemathstutor.com
[email protected]
1. Write down the 6th and 8th terms in each case.
(a) 2, 4, 6, 8 …
(b) 3, 6, 9, 12 …
(c) 4, 8, 12, 16 …
(d) 5, 7, 9, 11 …
(e) 4, 7, 10, 13 …
(f) 7, 12, 17, 22 …
(g) 2, 4, 8, 14 …
(h) 3, 5, 9, 15 …
(i) 4, 7, 13, 22 …
(j) 8, 11, 17, 26 …
(k) 5, 10, 20, 35 …
(l) 6, 10, 18, 30 …
(m) 13, 14, 16, 19 …
(n) 13, 20, 34, 55 …
(o) 11, 20, 38, 65 …
2. If n =1 is the first term, find the formula for the n th term in each case.
(a) 5, 10, 15, 20 …
(b) 4, 7, 10, 13 …
(c) 2, 7, 12, 17 …
(d) 1, 9, 17, 25 …
(e) 6, 11, 16, 21 …
(f) 5, 17, 29, 41 …
(g) 9, 12, 15, 18 …
(h) 21, 25, 31, 36 …
(i) 10, 17, 24, 31 …
(j) 19, 25, 31, 37 …
(k) 13, 20, 27, 34 …
(l) 29, 36, 43, 50 …
(m) 11, 15, 19, 23 …
(n) 9, 22, 35, 48 …
(o) 3, 30, 57, 84 …
3. If n =1 is the first term, the n th term can be witten as a + b(n - 1) where a and b are
constants. Find a and b
(a) 2, 4, 6, 8 …
(b) 4, 8, 12, 16 …
(c) 5, 7, 9, 11
(d) 6, 19, 32, 45 …
(e) 17, 22, 27, 32 …
(f) 11, 15, 19, 23 …
(g) 19, 25, 31, 37 …
(h) 57, 64, 71, 78 …
(i) 101, 103, 105, 107 …
(j) 2, 58, 114, 170 …
(k) 29, 42, 55, 68 …
(l) 4, 31, 58, 85 …
(m) 11, 22, 33, 44 …
(n) 100, 200, 300, 400
(o) 17, 40, 63, 86 …
GCSE Maths Tutor
www.gcsemathstutor.com
[email protected]
Number - sequences
GCSE Maths Tutor
Sequences
www.gcsemathstutor.com
[email protected]
1.
(a) 12, 16
(b) 18, 24
(c) 24, 32
(d) 15, 19
(e) 19, 25
(f) 32, 42
(g) 32, 58
(h) 33, 59
(i) 49, 88
(j) 53, 92
(k) 80, 145
(l) 66, 118
(m) 28, 41
(n) 118, 208
(o) 146, 279
(b) 4 + 3(n-1)
(c) 2 + 5(n-1)
(d) 1 + 8(n-1)
(e) 6 + 5(n-1)
(f) 5 + 12(n-1)
(g) 3 + 7(n-1)
(h) 1 + 6(n-1)
(i) 1 + 11(n-1)
(j) 19 + 6(n-1)
(k) 13 + 7(n-1)
(l) 29 + 7(n-1)
(m) 11 + 4(n-1)
(n) 9 + 13(n-1)
(o) 3 + 27(n-1)
3.
(a) 2, 2
(b) 4, 4
(c) 5, 2
(d) 6, 13
(e) 17, 5
(f) 11, 4
(g) 19, 6
(h) 57, 7
(i) 101, 2
(j) 2, 56
(k) 29, 13
(l) 4, 27
(m) 11, 11
(n) 100, 100
(o) 17, 23
www.gcsemathstutor.com
[email protected]
2.
(a) 5n
GCSE Maths Tutor
Number - sequences
GCSE Maths Tutor
GCSE Maths Tutor
Sequences
www.gcsemathstutor.com
www.gcsemathstutor.com
question sheet N-SEQ-01
[email protected]
[email protected] | 1,297 | 2,358 | {"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} | 3.96875 | 4 | CC-MAIN-2018-47 | latest | en | 0.473818 |
https://www.mooc-list.com/course/what-are-chances-probability-and-uncertainty-statistics-coursera | 1,686,336,135,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656788.77/warc/CC-MAIN-20230609164851-20230609194851-00189.warc.gz | 1,006,210,552 | 16,423 | # What are the Chances? Probability and Uncertainty in Statistics (Coursera)
##### Start Date
Jun 5th 2023
Course Auditing
Categories
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41.00 EUR/month
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Familiarity with descriptive statistics and basic regression models
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This course focuses on how analysts can measure and describe the confidence they have in their findings. The course begins with an overview of the key probability rules and concepts that govern the calculation of uncertainty measures. We’ll then apply these ideas to variables (which are the building blocks of statistics) and their associated probability distributions. The second half of the course will delve into the computation and interpretation of uncertainty. We’ll discuss how to conduct a hypothesis test using both test statistics and confidence intervals. Finally, we’ll consider the role of hypothesis testing in a regression context, including what we can and cannot learn from the statistical significance of a coefficient. By the end of the course, you should be able to discuss statistical findings in probabilistic terms and interpret the uncertainty of a particular estimate.
Course 4 of 5 in the Data Literacy Specialization
### Syllabus
WEEK 1
Probability Theory
The Monty Hall problem is a classic brain teaser that highlights the often counterintuitive nature of probability. The problem is typically stated as follows: Suppose you're a contestant on a game show and asked to select one of three doors for your prize. Behind one door is a car and behind the other two doors are goats. You pick one door. The host, who knows what's behind each door, opens another, which has a goat. He then gives you the option to stick with your selected door or switch to the other closed door. What should you do? The answer is that, under these circumstances, you should always switch. There is a 2/3 chance of winning the car if you switch and a 1/3 chance of winning if you stick with your original selection. Most people, however, assume that there is only a 50/50 chance of winning if you switch. Hopefully this brain teaser, and content we cover in this module, will help you better approach probabilistic problems.
WEEK 2
Random Variables and Distributions
In this module, we'll dive into a topic you've likely encountered all of your adult life but perhaps have never explored from a statistical perspective: the normal curve. More generally, we'll discuss probability distributions, including their key features and relevance to quantifying uncertainty. Although studying probability theory can sometimes feel detached from applied statistics, it's valuable to develop a foundational understanding of probability to be able to critically evaluate statistical models. An appreciation for probability, and its counter-intuitive nature, will help you interpret the uncertainty of a statistical result as accurately as possible. This is particularly important when the stakes are high and policy makers want to know whether or not to act based on a statistical finding.
WEEK 3
Confidence Intervals and Hypothesis Testing
In this module we will apply the concepts of probability, random variables and distributions to measuring and interpreting uncertainty. In particular, we'll focus on statistical significance. A relationship is statistically significant if it can be distinguished from zero. Suppose you want to examine the effect of exposure to negative campaign ads on one's likelihood of voting. The independent variable is one's exposure to negative campaign ads and the dependent variable is one's likelihood of voting. If we find that exposure to negative campaign ads has no relationship with the likelihood of voting, we would say that this is a statistically insignificant relationship. If, instead, we find that exposure to negative campaign ads leads to a decline in one's likelihood of voting, we have uncovered a statistically significant (i.e., non-zero) relationship.
WEEK 4
Quantifying Uncertainty in Regression Analysis and Polling
In this final module of the course, we'll cover how to measure the uncertainty of regression estimates and poll results. It is often the case that a regression model will reveal a non-zero relationship, but it's important to determine whether that relationship sufficiently different from zero such that we can conclude that the relationship is statistically significant. For example, suppose a regression model reveals that a drug improves patient outcomes by 3.2%. Is 3.2% statistically different from 0? A statistical significance test will answer this question. This module, however, will also discuss some of the drawbacks of relying a statistical significance for data-driven decision making. While statistical significance is an important consideration, it is not the only criterion one should use when determining whether to act on a set of a statistical findings.
MOOC List is learner-supported. When you buy through links on our site, we may earn an affiliate commission. | 984 | 5,128 | {"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} | 3.625 | 4 | CC-MAIN-2023-23 | latest | en | 0.940921 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-6-section-6-4-graphs-of-polar-equations-exercise-set-page-754/42 | 1,726,766,177,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652055.62/warc/CC-MAIN-20240919162032-20240919192032-00046.warc.gz | 735,610,026 | 13,663 | ## Precalculus (6th Edition) Blitzer
The graph does not necessarily have symmetry with respect to the polar axis, the line $\theta =\frac{\pi }{2}$, or the pole.
We look for symmetry by making the following substitutions: (a) $\theta \to - \theta$ :$$r=\frac{3\sin 2(-\theta )}{\sin ^3 (-\theta )+\cos ^3 (-\theta )} \quad \Rightarrow \quad r=-\frac{3 \sin 2 \theta}{-\sin ^3 \theta +\cos ^3 \theta }$$Thus, the graph does not necessarily have symmetry with respect to the polar axis. (b) $r \to -r, \quad \theta \to -\theta$ :$$-r =\frac{3 \sin 2 (-\theta )}{\sin ^3 (-\theta )+ \cos ^3(-\theta )} \quad \Rightarrow \quad r=\frac{3\sin 2\theta}{-\sin ^3 \theta +\cos ^3 \theta }$$Thus, the graph does not necessarily have symmetry with respect to the line $\theta=\frac{\pi}{2}$. (c) $r \to -r$ :$$-r =\frac{3\sin 2\theta }{\sin ^3 \theta +\cos ^3 \theta } \quad \Rightarrow \quad r=-\frac{3\sin 2 \theta}{\sin ^3 \theta +\cos ^3 \theta }$$Thus, the graph does not necessarily have symmetry with respect to the pole. | 351 | 1,018 | {"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} | 4.34375 | 4 | CC-MAIN-2024-38 | latest | en | 0.662197 |
https://www.gradesaver.com/textbooks/math/other-math/basic-college-mathematics-9th-edition/chapter-5-ratio-and-proportion-review-exercises-page-374/52 | 1,527,364,831,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867859.88/warc/CC-MAIN-20180526190648-20180526210648-00077.warc.gz | 741,936,494 | 12,481 | Basic College Mathematics (9th Edition)
We can take the cross products of the given proportion to determine whether or not the proportion is true. If the cross products are equal, then we know that the proportion is true. $\frac{\frac{1}{5}}{2}=\frac{1\frac{1}{6}}{11\frac{2}{3}}=\frac{\frac{7}{6}}{\frac{35}{3}}$ $\frac{1}{5}\times\frac{35}{3}=2\times\frac{7}{6}$ Left side: $\frac{35}{15}=\frac{35\div5}{15\div5}=\frac{7}{3}$ Right side: $\frac{14}{6}=\frac{14\div2}{6\div2}=\frac{7}{3}$ Therefore, the proportion is True. | 186 | 525 | {"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} | 4.25 | 4 | CC-MAIN-2018-22 | longest | en | 0.653183 |
http://acscihotseat.org/index.php?qa=54&qa_1=question-2-of-tut-2-total-accumulated-value-at-any-time-t-0 | 1,537,577,897,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158001.44/warc/CC-MAIN-20180922005340-20180922025740-00408.warc.gz | 7,706,358 | 9,371 | # Question 2 of tut 2: Total accumulated value at any time t>0
97 views
The question states:
"The force of interest at any time $$t$$, (measured in years) is given by
$$f(t) = \begin{cases}0.07-0.005t & 0 \le t < 5 \\ 0.06-0.003t & 5 \le t < 10 \\ 0.03 & t \ge 10 \end{cases}$$
What is the total accumulated in value at any time $$t > 0$$ of investments of R100 at times 0, 4 and 6?"
I understand getting to first 'time period' i.e. for $$0 \le t \lt 4$$ but from there on I get confused. The memo splits the answer into $$0 \le t \lt 4$$, $$0 \le t \lt 5$$, $$0 \le t \lt 6$$, $$0 \le t \lt 10$$, $$t \ge 10$$. Why is it not split from 0 to 4, 4 to 5, 5 to 6, 6 to 10 and greater than 10?
commented Apr 10, 2016 by (4,220 points)
reshown Apr 10, 2016
I think there was a mistake in that memo hey. The way you understand it is correct.
commented Apr 10, 2016 by (2,770 points)
Okay great, thanks Simon. And for question 4 (or question 2, since they're basically identical but with different values), how do they simplify the two terms into one? I seem to get two terms, one for each payment when $$5 \le t < 6$$ but the memo has only one term.
commented Apr 10, 2016 by (2,610 points)
From what I understand, you still kind of have separate values for each cashflow (summed together of course at correct time intervals, when the cashflows become relevant) at different times (which should be correct, if I understand your question correctly). The memo simply combines those cashflows at certain times, for example in question 4, at time 5, it will combine the accumulated value of the one rand invested at time 0, with the one rand invested at time 5, and then accumulate it in value further. I hope that answers your question.
commented Apr 10, 2016 by (4,220 points)
Oh ya, I had a second look and the memo is right. But it's a lot of working out! :/
commented Apr 10, 2016 by (2,770 points)
Thank you, that does make sense. However, I'm still struggling a bit with the manipulation.
In question 4, for $$5 \le t \lt 6$$ I get the following by accumulating the first payment to $$t=4$$, adding the payment at $$t=5$$ and then accumulating further:
$$(e^{0.04\times5+0.0025\times25}+1)e^{\int_5^t0.04+0.005sds}$$
Simplifying this I get:
$$e^{0.04t+0.0025t^{2}}(1+e^{\frac {-21}{80}})$$
I notice that $$e^{0.570488}$$ (from the memo) is roughly the same as my last bracket in the line above, but I am unsure as to how to simplify $$(1+e^{\frac {-21}{80}})$$ to $$e^{0.570488}$$.
And w.r.t question 2, I am very confused as to how they combine the intervals i.e. how $$0 \le t \lt 4$$ and $$4 \le t \lt 5$$ are combined into the interval $$0 \le t \lt 5$$.
Thanks very much for the help so far.
answered Apr 10, 2016 by (4,220 points)
selected May 6, 2016 by Pandy
Sorry for all the confusion. I've looked even more carefully, and the bounds are wrong. They should read $$0 \le t < 4$$, $$4 \le t < 5$$, $$5 \le t < 6$$, $$6 \le t < 10$$, and $$t \ge 10$$.
e.g. If $$4 \leq t < 5$$ then
$$A(t) = 100\exp[\int_{0}^{t}(0.07 - 0.005s)ds] + 100\exp[\int_{4}^{t}(0.07 - 0.005s)ds]$$
$$= 100\exp[0.07t - 0.0025t^{2}](1 + e^{-0.24}) = 178.663\exp[0.07t - 0.0025t^{2}]$$
Obviously it doesn't make sense if we substitute some $$t < 4$$ in the above. And similarly for the rest.
Regarding Q4, you're spot on. What's $$\log(1 + e^{-21/80})$$?
commented Apr 10, 2016 by (2,770 points)
Awesome! Thanks very much Simon. | 1,172 | 3,448 | {"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": 2, "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-2018-39 | longest | en | 0.944745 |
https://math.stackexchange.com/questions/4457035/calculate-the-integral-by-riemann-int-01-sqrtx-dx | 1,716,568,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058721.41/warc/CC-MAIN-20240524152541-20240524182541-00700.warc.gz | 339,425,297 | 42,179 | # Calculate the integral by Riemann $\int_{0}^{1}\sqrt{x}\,dx$
Calculate the integral by Riemann $$\displaystyle \int _{0}^{1}\sqrt{x} \, dx$$
$$\displaystyle a_{k} =0+k\cdot \frac{1}{n} =\frac{k}{n}$$
$$\displaystyle \Delta x=\frac{1-0}{n} =\frac{1}{n}$$
\begin{align*} \lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left(\sqrt{\frac{k}{n}}\right) \cdotp \frac{1}{n} & = \lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left(\frac{\sqrt{k}}{\sqrt{n}}\right) \cdotp \frac{1}{n} \\[1ex] & = \lim_{n\rightarrow \infty }\sum_{k=1}^{n}\left(\frac{1}{n\sqrt{n}} \cdotp \sqrt{k}\right) \\[1ex] & = \lim_{n\rightarrow \infty }\frac{1}{n \sqrt{n}} \cdotp \sum_{k=1}^{n}\left(\sqrt{k}\right) \end{align*}
I am stuck on the sum of $$\sqrt k$$ maybe have another way to solve this question?
• Can you use, that root is integrable? Then you'll be able to choose specific points for partition. May 23, 2022 at 17:02
• Does this answer your question? For Riemann sums involving square roots, why do we let $c_{i} = \frac{i^{2}}{ n^{2}}$? May 23, 2022 at 17:02
• @user170231 well thanks but I tried right now and I didn't succeed I write $\dfrac{k^2}{n^2}$ and calculate the sum and the limit and I got an answer of 0.5 with is wrong May 23, 2022 at 17:13
• You can use the method here to do all general power integrals (continuity needed for irrational powers) May 23, 2022 at 17:27
As suggested in the linked question and other answers posted here, you need not use equally-spaced subintervals in the partition of $$[0,1]$$. Instead, consider the sequence of intervals
$$\left\{\left[\left(\frac{i-1}n\right)^2, \left(\frac in\right)^2\right] \,\bigg| \, 1 \le i \le n \right\}_{n\in\Bbb N}$$
each with length $$\frac{i^2}{n^2} - \frac{(i-1)^2}{n^2} = \frac{2i - 1}{n^2}$$.
Then the (right-endpoint) Riemann sum is
$$\sum_{i=1}^n \sqrt{\left(\frac in\right)^2} \frac{2i-1}{n^2} = \sum_{i=1}^n \frac{2i^2-i}{n^3} = \frac{(n+1)(4n-1)}{6n^2}$$
As $$n\to\infty$$, the sum converges to the definite integral, and we have
$$\int_0^1 \sqrt x \, dx = \lim_{n\to\infty} \frac{(n+1)(4n-1)}{6n^2} = \frac23$$
which agrees with the known antiderivative result,
$$\int_0^1 x^{\frac12} \, dx = \frac23 x^{\frac32}\bigg|_0^1 = \frac23$$
In the spirit of Reimann sums, you do not need to make the sequence $$a_k$$ arithmetic, you can use any sequence that covers the interval.
In this case, since you need to take the square root, it will make things simpler if you make $$a_k$$ a quadratic sequence...
$$a_k = 0 + \frac {k^2 }{n^2}$$
Then $$\sqrt{a_k}$$ simplifies to $$\frac kn$$ , but your $$\Delta x$$ will now depend of $$k$$... $$\Delta x_k = a_{k+1}-a_k = \frac {2k+1}{n^2}$$
So $$\int_0^1 \sqrt x dx = \lim_{n \to \infty} \sum_{k=1}^n a_k \Delta x_k$$
$$= \lim_{n \to \infty} \frac 1 {n^3} \sum_{k=1}^n k(2k+1)$$
$$= 2\lim_{n \to \infty} \frac 1 {n^3} \sum_{k=1}^n k^2 + \lim_{n \to \infty} \frac 1 {n^3} \sum_{k=1}^n k$$
Now you can use the formulas...
$$\sum_{k=1}^n k^2 = \frac n6 (2n+1)(n+1) \text{ and } \sum_{k=1}^n k=\frac n2(n+1)$$
to obtain the required result
• Note that it is not quite enough that the partition cover the interval. You also need the length of the largest subinterval to converge to $0$ as $n$ grows to $\infty$. That does happen here with the largest subinterval being the last one, which is $(2n-1)/n^2$ wide. May 23, 2022 at 17:59
• That was what I was trying to evoke using the term "cover" - but I accept that my use of the term was not correct. Perhaps an alternate formulation of what constitutes a valid Reimann partition could be that every open cover of the sequence is also an open cover of the interval.
– WW1
May 23, 2022 at 18:22
I thought it might be instructive to present a way forward that uses creative telescoping and simple estimates. To that end we proceed.
Note that we can write
\begin{align} n^{3/2}-1&=\sum_{k=1}^n \left((k+1)^{3/2}-k^{3/2}\right)\\\\ &=\frac32 \sum_{k=1}^n \left(\sqrt{k} +O(k^{-1/2})\right)\tag1 \end{align}
Rearranging $$(1)$$ reveals
$$\frac1{n^{3/2}}\sum_{k=1}^n \sqrt k=\frac23 \left(1-n^{-3/2}\right)+\frac1{n^{3/2}}\underbrace{\sum_{k=1}^n O(k^{-1/2})}_{\le O(n)}$$
Letting $$n\to \infty$$ yields the coveted result
$$\lim_{n\to\infty}\frac1{n^{3/2}}\sum_{k=1}^n \sqrt k=\frac23$$
Here is a direct way to look at this if you would like to stick with a Riemann sum that uses equally spaced subintervals. The strategy is, in small doses, to replace the irrational $$\sqrt{k}$$ values with nearby integers.
\begin{align} \lim_{n\to\infty}\frac{1}{n\sqrt{n}}\sum_{k=1}^n\sqrt{k} &=\lim_{n\to\infty}\frac{1}{n^2\sqrt{n^2}}\sum_{k=1}^{n^2}\sqrt{k} \end{align}
Already this might give you pause. But for one thing, as $$n\to\infty$$, so does $$n^2$$. The sum on the right is like the sum on the left but between one value of $$n$$ and the next, you add
$$\frac{1}{n^2\sqrt{n^2}}\left(\sqrt{(n-1)^2+1}+\sqrt{(n-1)^2+2}+\cdots+\sqrt{(n-1)^2+(2n-1)}\right)$$
That is, you add $$2n-1$$ terms that are each less than or equal to $$n$$, and the result is divided by $$n^3$$. So this tail that you add with each step, all by itself, converges to $$0$$. So the two sums have the same limit. Starting again:
\begin{align} \lim_{n\to\infty}\frac{1}{n\sqrt{n}}\sum_{k=1}^n\sqrt{k} &=\lim_{n\to\infty}\frac{1}{n^2\sqrt{n^2}}\sum_{k=1}^{n^2}\sqrt{k}\\ &=\lim_{n\to\infty}\frac{1}{n^3}\left(\sqrt{1}+\sqrt{2}+\cdots+\sqrt{n^2}\right)\\ &\leq\lim_{n\to\infty}\frac{1}{n^3}\left(\sqrt{1}+\sqrt{4}+\sqrt{4}+\sqrt{4}+\sqrt{9}+\sqrt{9}+\cdots+\sqrt{n^2}\right)\\ \end{align} Here we have one $$\sqrt{1}$$s, then three $$\sqrt{4}$$s, then five $$\sqrt{9}$$s, and so on.
\begin{align} &=\lim_{n\to\infty}\frac{1}{n^3}\left(1+3(2)+5(3)+\cdots+(2n-1)n\right)\\ &=\lim_{n\to\infty}\frac{1}{n^3}\sum_{k=1}^n(2k-1)k\\ &=\lim_{n\to\infty}\frac{1}{n^3}\left(2\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}2\right)\\ &=\lim_{n\to\infty}\left(\frac13(1+1/n)(2+1/n)-\frac{(1+1/n)}{2n}\right)\\ &=\frac23 \end{align}
So we have an upper bound and the limit you are after is at most $$\frac23$$. (Of course we know independently that the limit actually is $$\frac23$$.)
Do you see how to modify this to get a lower bound? Hint: $$1,1,1,2,2,2,2,2,3,3,3,3,3,3,3,\ldots$$.
Maybe this will help you (i hope this is true).
First let's make a change of variable: $$t=\sqrt{x}$$ so we get $$\int_{0}^{1}\sqrt{x}dx=\int_{0}^{1}2t^2dt$$
Now lets compute: $$\int_{0}^{1}2t^2dt$$ using Riemman integrable.
Thus by doing exactly as you did we get: $$lim_{n \to \infty } \frac{2}{n^3} \sum_{k=1}^{n}k^2$$
And we know that the sum of squares of n natural numbers can be calculated using the formula $$[n(n+1)(2n+1)] / 6$$. So: $$lim_{n \to \infty } \frac{2}{n^3} [n(n+1)(2n+1)] / 6 = 2/3$$
This is correct because: $$\int_{0}^{1}\sqrt{x}dx = 2/3$$
So we did exactly what it was asked:
Calculate the integral by Riemann $$\int_{0}^{1}\sqrt{x}dx$$
But by calculating integral by Riemman on a simplest function.
• This is not a clear answer May 23, 2022 at 17:27
• @FShrike why? how can i improve it? May 23, 2022 at 17:31
• They want square roots May 23, 2022 at 17:34
• @FShrike It is not say in the question that we have to calculate the Riemman sum directly on the root function given. In the question it is only asked to compute the given integrable using a Riemman integrable, and this is exactly what we did. If not the op must be more precise. May 23, 2022 at 17:38
If the answer upper isn't satisfying here an oher.
Choose a partition as follow: $$x_0=0 , x_1 = 1/n^2 ,x_2=(2^2)/n^2 ,...,x_n=n^2/n^2=1$$ so $$x_k-x_{k-1}=\Delta x_k=\frac{2k-1}{n^2}$$.
$$\sum_{k=1}^{n}\frac{2k-1}{n^2}\frac{k}{n}=\sum_{k=1}^{n}\frac{2k^2-k}{n^3}$$
And we know a formula for each of those sums. | 3,010 | 7,714 | {"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": 58, "wp-katex-eq": 0, "align": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-22 | latest | en | 0.687373 |
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South African Journal of Science
Print version ISSN 0038-2353
When to stay, when to go: trade-offs for southern African arid-zone birds in times of drought
W.R.J. DeanI, *; P. BarnardI, II; M.D. AndersonIII
IDST/NRF Centre of Excellence at the Percy FitzPatrick Institute of African Ornithology, University of Cape Town, Rondebosch, 7701 South Africa
IIBirds and Environmental Change Partnership, Climate Change and Bio-Adaptation Division, South African National Biodiversity Institute, Private Bag X7, Claremont, South Africa
IIIBirdLife South Africa, P.O. Box 515, Randburg 2125, Gauteng, South Africa
Key words: avian-nomadism, arid environments, Karoo-Namib Desert, Kalahari, breeding
Prolonged spells of drought pose dilemmas for most organisms, even those adapted to hyper-aridity. For birds, reproduction, feeding, dispersal and moult are all activities which may need careful timing in relation to fluctuations in ecological productivity. In arid and semi-arid ecosystems, rainfall often triggers events in the life cycle which may be suppressed for months or even years during dry periods.1,2,3,4 In times of very low productivity or harsh conditions, birds, like many other animals, can move to escape local conditions and improve their chances of feeding or reproduction.5 However, some resident species may not move, but simply adjust their activities to 'ride out' difficult periods.
Semi-arid ecosystems are characterised by wet or dry states that are patchy in time and space. This patchiness is particularly true of southern Africa, where much of the region is semi-arid6 and the environment experiences extremes in weather, from periods of intense and prolonged drought to exceptionally high rainfall events.7 Rainfall in arid and semi-arid ecosystems in the southern hemisphere, including southern Africa, is greatly affected by El Niño Southern Oscillations (ENSO), leading to large variability in rainfall and prolonged droughts.6 Similarly, extensive wet periods (La Niña events) have concomitant effects on ecosystem functioning.8 The effects of El Niño and La Niña have been well studied in a few organisms, particularly birds, where changes in species richness and density9 and reproduction10 are correlated with the wet–dry cycles of ENSO years.
Birds, other animals and plants respond to increased rainfall in similar ways, and both birds and plants tend to increase reproductive effort with more rain.11–14 The response by biota to extended dry periods in semi-arid environments, however, differs markedly between and within invertebrate and vertebrate phyla, and markedly between most animals and plants. Plants cope with droughts by dying after depositing dormant propagules (ephemerals) or becoming dormant and restricting their internal water use by discarding leaves or leaves and stems.13,15 Although death and propagule dormancy are options taken by plants and some invertebrate taxa, viz. brine shrimp (Artemia), these options are not available to vertebrates, leaving escape and water conservation as the alternatives.5,16
Drought-induced dormancy in vegetation therefore has effects on animals. Animals in general, and birds in particular, cope with droughts and changes in vegetation by using behavioural and physiological tactics, including opportunistic movement away (in birds), shifts in habitat5 and deferred hatching or dormancy in eggs (in locusts).2,3,4 It is not known in detail whether birds make dietary shifts during extended droughts, but many species are opportunistic in their foraging and feeding in the Karoo17 and it is very likely that such shifts do occur.
Our questions in this review are: (1) what is the influence of variability in rainfall on avian populations, including breeding and movements? and (2) how resilient are bird populations to extended dry periods? While we do not explicitly review climate change per se, our conclusions should be of value in predicting species-specific responses and levels of vulnerability to changing rainfall patterns in Africa.
The influence of rainfall on population dynamics
Exceptionally high rainfall may stimulate rare breeding events. For example, banded stilts (Cladorhynchus leucocephalus) in arid Australia arrive and breed in very large colonies following exceptionally high rainfall at ephemeral pans that may have been dry for decades.1,18 Variability in rainfall has strong effects on clutch sizes and population dynamics in birds, as observed in unrelated species in different arid parts of the world, such as Galapagos ground finches on the Galápagos Islands,10 galahs (Cacatua roseicapilla) in Australia19 and larks,20 thrushes and helmet-shrikes in arid southern Africa.12,21 Droughts have different effects on bird populations, often stimulating movements into better-watered areas, but also, conversely and seemingly inexplicably, stimulating movements into dry areas.22 Crowned hornbills (Tockus alboterminatus) wander from their forest habitat in the non-breeding season,23 some years reaching far into the semi-arid Karoo.24,25 These movements are thought to be related to unusual aridity or cold in their coastal habitats,23 although these are probably always more mesic than the areas to which the birds disperse.
The movements of nomadic birds in arid and semi-arid ecosystems throughout the world have been reviewed by Dean.5 We know that some birds have evolved to cope with stochastic weather events in their habitat. What is less clear, however, is what resident bird species do when faced with increasing aridity in their environment. For those species that move, the benefits must outweigh the costs.26 Similarly, for those species that are resident, the benefits of being resident must outweigh the cost of moving. For example, in partially-migratory rock kestrels (Falco rupicolus), individual males used one or the other of these two tactics.27 Males departed for non-breeding areas later than did females, and returned to their territories earlier than did females. Not all males migrated, and those that remained successfully retained their territories and had a higher probability of breeding in the next breeding season. However, there was a cost: males that stayed on their territories during the winter faced increased competition for food. Males that migrated probably had higher survival, but risked losing their territories. Male rock kestrels in the Karoo thus faced a trade-off between increased chances of breeding and the risk of mortality.27
Trade-offs are necessary where rainfall is highly unpredictable. Unlike the short breeding season in the temperate northern hemisphere, the concept of extended 'equally good months' for breeding (EGMs)28 often applies to bird populations in drier parts of the southern hemisphere, where weather patterns are less predictable and suitable or unsuitable conditions for breeding may arise at any time of the year. EGMs are calculated as the months in which there are breeding records equal to or above the average number of breeding records per month for the species. Africa and Australia both have a mean EGM 1.9 months a year longer than anywhere else,29 presumably caused by out-of-season rainfall, or rainfall in areas that have been dry for a long period. Even within South Africa, areas with higher coefficients of variation in total annual rainfall,30 such as the Karoo, have more EGMs than areas with more sharply seasonal, but more predictably timed, rainfall such as the southwestern Cape (Table 1).
The most frequently measured responses by birds in arid regions to above average rainfall or isolated rainfall events are breeding out of season, breeding in large numbers, and increases in clutch size or number of young produced. For example, 'heavy rains' following a five-year drought in the western Karoo stimulated bird breeding activity, with a number of species recorded with eggs and young, and all collected specimens in breeding condition31. Similarly, 170 mm of rain between January and March 1985, following a severe drought in the central Karoo, was believed to have stimulated breeding in 27 species of birds32. However, no studies of breeding activity in either case were made during the 'severe drought' broken by the rains, so the level of increase in breeding activity was difficult to evaluate.
Better data on the influence of rainfall on population dynamics are provided by long-term studies. Verreaux's eagles (Aquila verreauxii) breeding in the Matobo Hills, Zimbabwe showed responses to variations in annual rainfall amounts. The number of resident pairs of Verreaux's eagles increased during the high rainfall years and correlated with increases in the number of hyraxes (Procavia capensis and Heterohyrax brucei)33. Both the number of pairs in the local population that bred and their production of young fit a negative exponential curve (Fig. 1), suggesting that the eagles benefit from both lower and higher rainfall years, but not from average rainfall years. Lower rainfall years have as many pairs breeding as higher rainfall years, with relatively higher breeding success. This suggests that breeding success is not only correlated with rainfall, but that other factors also influence the population dynamics of this species at this locality. Average rainfall years may increase competition for hyraxes, or higher rainfall years may lead to increased chick mortality.
In deciduous woodland habitats that come into leaf before the rains, there is some evidence that the level of breeding activity in birds is correlated with rainfall during the previous year. A study in the Miombo woodland in Zimbabwe21 showed that rainfall during both the previous breeding season and current breeding season influences the number of nests and clutch sizes in Kurrichane thrushes (Turdus libonyanus) and white-crested helmet-shrikes (Prionops plumatus) (Tables 2 and 3). Both are resident species, generally maintaining territories all year round. The difference in breeding effort between years was not very marked in the Kurrichane thrush following the dry year of 1972, but breeding effort following the wettest year (1973) increased markedly, with an earlier start to breeding, increases in nest density per hectare (Table 2) and increases in the number of breeding attempts.21,34 With lower rainfall in the previous season there were decreases in the number of breeding groups and total number of birds, and increases in home range and territory sizes in white-crested helmet-shrikes (Table 3).21
In a more arid environment, Maclean35 showed that resident and nomadic larks (Alaudidae) and resident sociable weavers (Philetarius socius) in the Kgalagadi Transfrontier Park responded quickly to rain, and that the duration of the breeding 'season' varied according to the amount of rain. Table 4 gives data for one species, the grey-backed sparrowlark (Erempoterix verticalis) and Table 5 presents data for the sociable weaver. The average annual rainfall for this area is about 180 mm per year so the amount of rain that fell in April–November 1965 and January–April 1966 was more than 70% of the average annual rainfall. The relationship between rainfall, breeding season and breeding success, however, is not linear. Maclean36 suggested that the duration of the breeding season is not directly governed by the amount of rain, but rather by the time of year in which rain falls.
This suggestion is supported by Lloyd20 who noted that the timing and length of the breeding season in Bushmanland, Northern Cape Province depended on the integrated effect of rainfall and temperature on the growing season of the vegetation. Higher breeding success, measured as the percentage of young that left the nest, apparently depends on time of year and not the amount of rain per se (Table 5). The breeding success of sociable weavers was markedly lower during the hot summer months of January to April than the cooler, longer period from April to November36 (Table 5).
Clutch size may vary with rainfall and aridity. In sociable weavers35 there was some variation in mean clutch size per month, but specific rainfall data for those months were unavailable. However, a study of a bird community in Bushmanland, Northern Cape, showed that breeding activity of five species increased markedly after rainfall (Table 6), and more than half the species showed an increase in average clutch size compared to drier conditions.20 Two species, black-eared sparrowlarks (Eremopterix australis) and grey-backed sparrowlarks, showed a rapid response to rain and began laying larger clutches within seven days of 78 mm of rain.20
The influence of drought on bird populations
Movements in birds may be in response to rain or in response to drought. In general, drought, or extended dry periods, has the effect of reducing bird diversity and reducing the number of individuals,37 either through emigration5 or lowered survival.38 We do not know how resilient bird populations are to drought in southern Africa, because there are no intensive long-term studies of birds that have covered a full cycle of dry and wet years. There are, however, studies showing that resident birds may be quite resistant to drought. A long-term study at Tierberg, near Prince Albert in the Karoo, showed that the number of individuals of resident long-billed larks (Certhilauda subcoronata), spike-heeled larks (Chersomanes albofasciata), Karoo chats (Cercomela schlegelii) and rufous-eared warblers (Malcorus pectoralis) showed relatively small annual fluctuations regardless of rainfall39 (Table 7). This data support the idea that in some cases the benefits of remaining in an area outweigh the costs of moving.
However, in the Kalahari, Botswana, resident species may show less resistance to drought, suggesting that resistance may depend to some extent on more subtle factors. For example, dry-season bird diversity, and number of individuals, were high following a higher rainfall wet season, and markedly lower following a poor rainfall season.37 Of 39 resident and nomadic species, populations of three resident species were stable and nine resident and nomadic species increased in abundance during the wet year following a dry year. But 20 species, including some nomads, decreased in abundance during that year. In the dry year following a wet year, all species had decreased markedly in number, and 12 species were no longer locally present. All nomadic species showed large increases in number in the wet year. Only one of the nomads, Temminck's courser (Cursorius temminckii), was present during the second dry year. There was no general pattern apparent in the residents or nomads, except that bird numbers were generally lower in the wet season following a dry season than in the dry season following a wet season. Both groups of birds showed fluctuations in number that cannot be entirely explained by rainfall amount. Some residents (but not all) showed some initial resilience to the dry conditions. Such resilience, however, did not last over the entire period of the study.
Studies of birds under drought conditions show that responses to changes in environmental conditions may be species-specific. In Bushmanland, Sclater's lark (Spizocorys sclateri) showed some resistance to drought, nesting at a higher density during low rainfall than during a wetter spell (Table 6). During the dry period, the larks fed on, amongst others, the large seeds of a grass (Enneapogon desvauxii), a dependable resource in dry times because the seeds are held almost below ground at the base of the plant. Because of the difficulty of extracting the seeds, this resource is used only by Sclater's larks and Stark's larks (S. starki). Stark's larks are nomadic and absent during drier periods; Sclater's larks are thus able to use the resource without competition during droughts.
In species for which the costs of moving may be high, for example, dune larks (Calendulauda erythrochlamys) on sparsely vegetated 'islands' in the Namib Desert dune sea, some individuals nevertheless moved away during a drought, with the population on some islands reduced to about one half or one third of their former size.40 There was also a reduction in foraging group size during the drought. Foraging group sizes in wetter periods were 2–6 individuals,41 whereas during the drought no groups larger than two birds were seen.40 No shifts in diet or changes in foraging methods and patterns between the wetter period and the drought were apparent, but there were seasonal differences in foraging patterns, and presumably diet, regardless of whether there was a drought or not.40 Whether the larks remained on their patch or moved away was strongly dependent on plant composition and the state of the vegetation and invertebrate populations. Safriel40 suggested that in patches where plant and invertebrate resources remained rich, foraging patterns (and presumably diet) would remain similar all year round. Dune larks can probably only be resident in the absence of any competition. The Namib Desert supports large numbers of nomadic granivores following rainfall.42 Some nomadic species overlap temporarily in habitat with dune larks, but depart as the ecosystem dries out.
Dune larks are thus able to remain on their patches provided there is no potential competition.
Decision making: when to stay and when to go?
We have shown, in this brief review, that birds use two main strategies for coping with the environment getting wetter or drier. In general, 'wet' years stimulate higher densities of nests (i.e. smaller territories), larger clutch sizes, unseasonal breeding, and, depending on the time of year, higher breeding success. Rainfall over a certain amount20,35,36,43 triggers breeding in resident species and an influx of nomadic species that breed and then move on. The amount of rainfall needed to trigger breeding varies between areas and seasons; 60 mm of rain in two days, followed by 22 mm eight days later in summer (January) stimulated breeding in the Kalahari,35 but 41 mm in Bushmanland over several days in summer (February) without any follow-up rains did not trigger breeding.20 It did, however, lead to an influx of nomadic granivores which did not breed at that time, whereas 54 mm of rain in winter (July) in the same area triggered breeding in all species.20
Nomads move in response to environmental cues, which are poorly understood for small terrestrial granivores,5,44 but better understood for larger aquatic species.45 Environmental cues for aquatic bird species may include distant thunderstorms, indicating heavy rain and thus the formation of temporary pans.45 This scenario may also hold for small terrestrial insectivores, but for small avian granivores the lag between rainfall and the response by the plants may be a week or more. It has been suggested5 that drifts of awns of grasses (used by nomadic larks and buntings for nest linings) may provide a strong visual cue that the area is suitable for nesting, but which other cues draw the birds initially to the area is not known.
Coping with an environment that is drying out may be easier for nomadic species. The environment dries out, particular food items become scarce, and provided that their young are large and able to fly, the nomads move on. What they must decide is where, and not when, to go. For resident species, however, whether to go or stay is a more difficult and apparently individual process. The examples we give suggest that in resident bird communities, there is a community-wide response to below-average rainfall, with fewer birds breeding in dry years and lower than usual production of young. But whether to breed or not, to stay or not, or to join a larger group remain individual decisions, possibly influenced by the decisions made by conspecifics.
If birds make the decision to remain in their area during a drought, they have several options to increase their chances of survival. They can increase territory size if neighbouring territories have become vacant or poorly-defended; they can shift their diet to eat a wider range of items; or they can abandon territories and join mixed-species foraging flocks. There is very little evidence of any of these tactics being particularly obvious in drought years. Furthermore, the benefits of all tactics have to be traded off against the losses. Territories that have increased in size might be too large to defend when good times arrive, territories that are abandoned need to be re-established when the drought is over, and shifts in diet may have consequences for the health or reproductive success of individual birds. But more importantly, resident species usually know their territory or home range very well.46 They know good places to forage, to nest and to avoid predators. This knowledge will increase in quality over the years.47 This fact is supported by Hanmer38 who shows that adult birds survive droughts better than immature ones, and suggests that it is due to the adults' better knowledge of parts of the habitat that gives them an advantage. Resident bird species are thus more likely to be resistant to drought and to use available resources as best they can without giving up their patch.
This review is based on several talks given at the Arid Zone Ecology Forum, Sutherland, in September 2007. We thank the Critical Ecosystem Partnership Fund for sponsoring the attendance of W.R.J.D. at the forum, and we thank the committee of the Arid Zone Ecology Forum for funding towards writing this review. We are grateful to Sue Milton for comments and suggestions on the manuscript.
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36. Maclean G.L. (1973). The sociable weaver, part 3: breeding biology and moult. Ostrich 44, 219–240. [ Links ]
37. Herremans M. (2004). Effects of drought on birds in the Kalahari, Botswana. Ostrich 75, 217–227. [ Links ]
38. Hanmer D.B. (1997). Bird longevity in the eastern highlands of Zimbabwe – drought survivors. Safring News 26, 47–54. [ Links ]
39. Dean W.R.J. and Milton S.J. (2001). The density and stability of birds in shrublands and drainage line woodland in the southern Karoo, South Africa. Ostrich 72, 185–192. [ Links ]
40. Safriel U.N. (1990). Winter foraging behaviour of the Dune Lark in the Namib Desert, and the effect of prolonged drought on behaviour. Ostrich 61, 77–80. [ Links ]
41. Cox G.W. (1983). Foraging behaviour of the dune lark. Ostrich 54, 113–120. [ Links ]
42. Willoughby E.J. (1971). Biology of larks (Aves: Alaudidae) in the central Namib Desert. Zool. Afr. 6, 133–176. [ Links ]
43. Simmons R.E. (1996). Population declines, viable breeding areas and management options for flamingos in southern Africa. Cons. Biol. 10, 504–514. [ Links ]
44. Dean W.R.J. and Williams J.B. (2004). Adaptations of birds for life in deserts with particular reference to larks (Alaudidae). Trans. Roy. Soc. S. Afr. 59, 79–91. [ Links ]
45. Simmons R.E., Barnard P. and Jamieson I.G. (1999). What precipitates influxes of wetland birds to ephemeral pans in arid landscapes? Observations from Namibia. Ostrich 70, 145–148. [ Links ]
46. Hinde R.A. (1956). The biological significance of the territories of birds. Ibis 98, 340–369. [ Links ]
47. Andersson M. (1980). Nomadism and site tenacity as alternative reproductive tactics in birds. J. Anim. Ecol. 49, 175–184. [ Links ]
Received 3 June 2008. Accepted 12 February 2009.
* Author for correspondence. E-mail: | null | null | null | null | null | null | null | null | null |
https://nrich.maths.org/public/topic.php?code=71&cl=4&cldcmpid=2817 | 1,597,059,240,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738674.42/warc/CC-MAIN-20200810102345-20200810132345-00011.warc.gz | 412,005,347 | 9,639 | # Resources tagged with: Mathematical reasoning & proof
Filter by: Content type:
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### There are 173 results
Broad Topics > Thinking Mathematically > Mathematical reasoning & proof
### Plus or Minus
##### Age 16 to 18 Challenge Level:
Make and prove a conjecture about the value of the product of the Fibonacci numbers $F_{n+1}F_{n-1}$.
### Polynomial Relations
##### Age 16 to 18 Challenge Level:
Given any two polynomials in a single variable it is always possible to eliminate the variable and obtain a formula showing the relationship between the two polynomials. Try this one.
### Pair Squares
##### Age 16 to 18 Challenge Level:
The sum of any two of the numbers 2, 34 and 47 is a perfect square. Choose three square numbers and find sets of three integers with this property. Generalise to four integers.
##### Age 16 to 18 Challenge Level:
Find all positive integers a and b for which the two equations: x^2-ax+b = 0 and x^2-bx+a = 0 both have positive integer solutions.
### Telescoping Functions
##### Age 16 to 18
Take a complicated fraction with the product of five quartics top and bottom and reduce this to a whole number. This is a numerical example involving some clever algebra.
### The Clue Is in the Question
##### Age 16 to 18 Challenge Level:
Starting with one of the mini-challenges, how many of the other mini-challenges will you invent for yourself?
### Diverging
##### Age 16 to 18 Challenge Level:
Show that for natural numbers x and y if x/y > 1 then x/y>(x+1)/(y+1}>1. Hence prove that the product for i=1 to n of [(2i)/(2i-1)] tends to infinity as n tends to infinity.
### Thousand Words
##### Age 16 to 18 Challenge Level:
Here the diagram says it all. Can you find the diagram?
### Golden Eggs
##### Age 16 to 18 Challenge Level:
Find a connection between the shape of a special ellipse and an infinite string of nested square roots.
### Fractional Calculus III
##### Age 16 to 18
Fractional calculus is a generalisation of ordinary calculus where you can differentiate n times when n is not a whole number.
### Interpolating Polynomials
##### Age 16 to 18 Challenge Level:
Given a set of points (x,y) with distinct x values, find a polynomial that goes through all of them, then prove some results about the existence and uniqueness of these polynomials.
### Particularly General
##### Age 16 to 18 Challenge Level:
By proving these particular identities, prove the existence of general cases.
### Unit Interval
##### Age 14 to 18 Challenge Level:
Take any two numbers between 0 and 1. Prove that the sum of the numbers is always less than one plus their product?
##### Age 16 to 18 Short Challenge Level:
Can you work out where the blue-and-red brick roads end?
### Rational Roots
##### Age 16 to 18 Challenge Level:
Given that a, b and c are natural numbers show that if sqrt a+sqrt b is rational then it is a natural number. Extend this to 3 variables.
### How Many Solutions?
##### Age 16 to 18 Challenge Level:
Find all the solutions to the this equation.
### Proof Sorter - Quadratic Equation
##### Age 14 to 18 Challenge Level:
This is an interactivity in which you have to sort the steps in the completion of the square into the correct order to prove the formula for the solutions of quadratic equations.
### Proof Sorter - Geometric Sequence
##### Age 16 to 18 Challenge Level:
Can you correctly order the steps in the proof of the formula for the sum of the first n terms in a geometric sequence?
### To Prove or Not to Prove
##### Age 14 to 18
A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.
### Leonardo's Problem
##### Age 14 to 18 Challenge Level:
A, B & C own a half, a third and a sixth of a coin collection. Each grab some coins, return some, then share equally what they had put back, finishing with their own share. How rich are they?
### Dalmatians
##### Age 14 to 18 Challenge Level:
Investigate the sequences obtained by starting with any positive 2 digit number (10a+b) and repeatedly using the rule 10a+b maps to 10b-a to get the next number in the sequence.
### Impossible Triangles?
##### Age 16 to 18 Challenge Level:
Which of these triangular jigsaws are impossible to finish?
### Generally Geometric
##### Age 16 to 18 Challenge Level:
Generalise the sum of a GP by using derivatives to make the coefficients into powers of the natural numbers.
##### Age 16 to 18 Challenge Level:
Find all real solutions of the equation (x^2-7x+11)^(x^2-11x+30) = 1.
### Janine's Conjecture
##### Age 14 to 16 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
### Mind Your Ps and Qs
##### Age 16 to 18 Short Challenge Level:
Sort these mathematical propositions into a series of 8 correct statements.
### Proof: A Brief Historical Survey
##### Age 14 to 18
If you think that mathematical proof is really clearcut and universal then you should read this article.
### Water Pistols
##### Age 16 to 18 Challenge Level:
With n people anywhere in a field each shoots a water pistol at the nearest person. In general who gets wet? What difference does it make if n is odd or even?
### Contrary Logic
##### Age 16 to 18 Challenge Level:
Can you invert the logic to prove these statements?
### Direct Logic
##### Age 16 to 18 Challenge Level:
Can you work through these direct proofs, using our interactive proof sorters?
### Iffy Logic
##### Age 14 to 18 Challenge Level:
Can you rearrange the cards to make a series of correct mathematical statements?
### Binomial
##### Age 16 to 18 Challenge Level:
By considering powers of (1+x), show that the sum of the squares of the binomial coefficients from 0 to n is 2nCn
### Mechanical Integration
##### Age 16 to 18 Challenge Level:
To find the integral of a polynomial, evaluate it at some special points and add multiples of these values.
### There's a Limit
##### Age 14 to 18 Challenge Level:
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?
### Target Six
##### Age 16 to 18 Challenge Level:
Show that x = 1 is a solution of the equation x^(3/2) - 8x^(-3/2) = 7 and find all other solutions.
### Sixational
##### Age 14 to 18 Challenge Level:
The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. . . .
### Three Ways
##### Age 16 to 18 Challenge Level:
If x + y = -1 find the largest value of xy by coordinate geometry, by calculus and by algebra.
### Little and Large
##### Age 16 to 18 Challenge Level:
A point moves around inside a rectangle. What are the least and the greatest values of the sum of the squares of the distances from the vertices?
### Notty Logic
##### Age 16 to 18 Challenge Level:
Have a go at being mathematically negative, by negating these statements.
### Sperner's Lemma
##### Age 16 to 18
An article about the strategy for playing The Triangle Game which appears on the NRICH site. It contains a simple lemma about labelling a grid of equilateral triangles within a triangular frame.
### More Dicey Decisions
##### Age 16 to 18 Challenge Level:
The twelve edge totals of a standard six-sided die are distributed symmetrically. Will the same symmetry emerge with a dodecahedral die?
### Euclid's Algorithm II
##### Age 16 to 18
We continue the discussion given in Euclid's Algorithm I, and here we shall discover when an equation of the form ax+by=c has no solutions, and when it has infinitely many solutions.
### Tetra Inequalities
##### Age 16 to 18 Challenge Level:
Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which could be the sides of a triangle.
##### Age 14 to 18 Challenge Level:
Which of these roads will satisfy a Munchkin builder?
### Continued Fractions II
##### Age 16 to 18
In this article we show that every whole number can be written as a continued fraction of the form k/(1+k/(1+k/...)).
### Proof Sorter - Sum of an Arithmetic Sequence
##### Age 16 to 18 Challenge Level:
Put the steps of this proof in order to find the formula for the sum of an arithmetic sequence
### Without Calculus
##### Age 16 to 18 Challenge Level:
Given that u>0 and v>0 find the smallest possible value of 1/u + 1/v given that u + v = 5 by different methods.
### Dodgy Proofs
##### Age 16 to 18 Challenge Level:
These proofs are wrong. Can you see why?
### Proof of Pick's Theorem
##### Age 16 to 18 Challenge Level:
Follow the hints and prove Pick's Theorem.
### An Introduction to Number Theory
##### Age 16 to 18
An introduction to some beautiful results of Number Theory (a branch of pure mathematics devoted primarily to the study of the integers and integer-valued functions) | 2,161 | 9,139 | {"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} | 3.78125 | 4 | CC-MAIN-2020-34 | latest | en | 0.81725 |
null | null | null | null | null | null | universalize in a sentence
Example sentences for universalize
That's because the filmmaker never finds a way to universalize the characters' experiences.
Well, there isn't one, and it was tough to find one that would universalize the concept of cupcake.
The only way to eliminate this bias would be to universalize all means-tested benefits currently targeted toward single mothers.
It may include a call to action, or universalize the points in the text, or provide a brief summary.
Famous quotes containing the word universalize
It is, most fundamentally because moral judgments are universalizable that we can speak of moral thought as rational (to... more
Copyright © 2014 Dictionary.com, LLC. All rights reserved. | null | null | null | null | null | null | null | null | null |
http://maeckes.nl/Decimaal%20getal%20GB.html | 1,603,346,437,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878921.41/warc/CC-MAIN-20201022053410-20201022083410-00008.warc.gz | 62,554,981 | 2,628 | < 1 2 3 4 5 >
### Decimal number
With the ten digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 we make the decimal numbers.
That is so because we have ten fingers.
##### Explanation
The number 237 is composed of
2 × 100 = 2 × 102 3 × 10 = 3 × 101 7 × 1 = 7 × 100
Because 100 is equal to 10 × 10 = 102 you can write 10 as 101, and 1 like 100. If there are digits after decimal point this continues. The power decreases, reaches less than zero, which means it becomes negative. Thus 0.1 = 10–1. The number 4267.893 is composed of
4 × 1000 = 2 × 103 2 × 100 = 2 × 102 6 × 10 = 6 × 101 7 × 1 = 7 × 100 8 × 0,1 = 8 × 10−1 9 × 0,01 = 9 × 10−2 3 × 0,001 = 3 × 10−3
Behind every digit in a number is a power of 10, that depends on its position. Because of that you must write a 0 if a certain place has no value. The number 3600.102 is composed of
3 × 1000 = 3 × 103 6 × 100 = 6 × 102 1 × 0,1 = 1 × 10−1 2 × 0,001 = 2 × 10−3
Now we understand why is
And we understand the number 1 better, as
Numbers with a negative exponent are just fractions. You see it clearly in | 486 | 1,122 | {"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} | 4.4375 | 4 | CC-MAIN-2020-45 | longest | en | 0.444746 |
https://oneconvert.com/unit-converters/specific-heat-capacity-converter/kilojoule-kilogram-k-to-btu-th-pound-176-f | 1,713,848,715,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818464.67/warc/CC-MAIN-20240423033153-20240423063153-00483.warc.gz | 377,734,289 | 24,993 | # Convert kilojoule/kilogram/K to Btu (th)/pound/°F
## How to Convert kilojoule/kilogram/K to Btu (th)/pound/°F
To convert kilojoule/kilogram/K to Btu (th)/pound/°F , the formula is used,
$\mathrm{Btu \left(th\right)/pound/°F}=\mathrm{kilojoule/kilogram/K}×0.238845897$
where the kilojoule/kilogram/K to Btu (th)/pound/°F value is substituted to get the answer from Specific Heat Capacity Converter.
1 kilojoule/kilogram/K
=
0.239 Btu (th)/pound/°F
1 Btu (th)/pound/°F
=
4.184 kilojoule/kilogram/K
Example: convert 15 kilojoule/kilogram/K to Btu (th)/pound/°F:
15 kilojoule/kilogram/K
=
15
x
0.239 Btu (th)/pound/°F
=
3.5851 Btu (th)/pound/°F
## kilojoule/kilogram/K to Btu (th)/pound/°F Conversion Table
kilojoule/kilogram/K Btu (th)/pound/°F
0.01 kilojoule/kilogram/K 0.002390057 Btu (th)/pound/°F
0.1 kilojoule/kilogram/K 0.023900574 Btu (th)/pound/°F
1 kilojoule/kilogram/K 0.239005736 Btu (th)/pound/°F
2 kilojoule/kilogram/K 0.478011472 Btu (th)/pound/°F
3 kilojoule/kilogram/K 0.717017208 Btu (th)/pound/°F
5 kilojoule/kilogram/K 1.195028681 Btu (th)/pound/°F
10 kilojoule/kilogram/K 2.390057361 Btu (th)/pound/°F
20 kilojoule/kilogram/K 4.780114723 Btu (th)/pound/°F
50 kilojoule/kilogram/K 11.95028681 Btu (th)/pound/°F
100 kilojoule/kilogram/K 23.90057361 Btu (th)/pound/°F
1000 kilojoule/kilogram/K 239.0057361 Btu (th)/pound/°F
### Popular Unit Conversions Specific Heat Capacity
The most used and popular units of Specific Heat Capacity conversions are presented for quick and free access. | 626 | 1,508 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "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} | 3.6875 | 4 | CC-MAIN-2024-18 | latest | en | 0.48584 |
http://mathhelpforum.com/algebra/43081-fractions.html | 1,480,721,253,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540698.78/warc/CC-MAIN-20161202170900-00231-ip-10-31-129-80.ec2.internal.warc.gz | 174,512,397 | 11,679 | 1. ## fractions
I need someone to help me understand how to do these problems.
2. Your images are hard to read. You might want to try to find another method to present your work. This is what I think you have. Please confirm and others will be glad to assist.
a) $\frac{-2(18-2)}{4(5-6)}-\frac{3(27-37)}{5(7-4)}$
b) $\frac{\frac{3}{4}-7}{\frac{1}{2}-\frac{4}{5}}$
c) $\frac{\frac{3}{8}}{\frac{1}{4}-\frac{7}{16}}$
d) $4 \frac{3}{5}+7 \frac{1}{10}$
e) $9\left(\frac{15}{81}\right)\left(\frac{27}{15}\rig ht)$
3. your pics are kinda blurry, so make sure i got the problem right.
$\frac{\frac{3}{4} - 7 }{\frac{1}{2} - \frac{4}{5}}$
Get like denominators on both the top and bottom of the big fraction.
$\frac{\frac{3}{4} - \frac{28}{4}}{\frac{5}{10} - \frac{8}{10}}$
Simplify: $\frac{\frac{-25}{4}}{\frac{-3}{10}}$
Multiply by the reciprocal: $\frac{-25}{4}\cdot\frac{-10}{3}$
$= \frac{250}{12}$
$= \frac{125}{6}$
C is very similar.
On A, do what's in parenthesis first, get a like denominator and simplify.
For: $4 \frac{3}{5}+7 \frac{1}{10}$, get improper fractions, then a like denominator and finally simplify.
For: $9\left(\frac{15}{81}\right)\left(\frac{27}{15}\rig ht)$
Multiply straight across; numerator by numerator; denominator by denominator.
You can think of it as: $\bigg(\frac{9}{1}\bigg)\left(\frac{15}{81}\right)\ left(\frac{27}{15}\right)$
Look for cross canceling. The 15 is an obvious one.
Then simplify as much as possible.
all of what masters has is correct
5. ## Thanks
I want to thank both of you. I still have one question how do you write out the problems on the computer so i dont have to take pictures.
6. Originally Posted by MistaMista
I want to thank both of you. I still have one question how do you write out the problems on the computer so i dont have to take pictures.
This forum allows you to enter $\text{\LaTeX}$ (LaTeX) code directly: simply enclose it in $$and$$ tags. To learn the basics, visit the LaTex Help forum.
7. Originally Posted by MistaMista
I want to thank both of you. I still have one question how do you write out the problems on the computer so i dont have to take pictures.
While you're learning to post using Latex, you could simply express your problems using brackets and parentheses like this:
a) [-2(18 - 2)] / [4(5 - 6)] - [3(27 - 37)] / [5(7 - 4)]
b) (3/4 -7) / (1/2 -4/5)
etc... | 756 | 2,371 | {"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": 15, "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-2016-50 | longest | en | 0.836637 |
http://mathhelpforum.com/algebra/278677-irrational-number.html | 1,540,190,800,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514708.24/warc/CC-MAIN-20181022050544-20181022072044-00104.warc.gz | 219,283,647 | 11,375 | 1. ## Irrational Number
Find n (natural number) where sqrt(13+33+33+43+5n) is irrational.
2. ## Re: Irrational Number
Originally Posted by louis33
Find n (natural number) where sqrt(13+33+33+43+5n) is irrational.
$\sqrt{1^3+3^3+3^3+4^3+5^1} = 2\sqrt{31} \in \mathbb{R} \setminus \mathbb{Q}$
... what now?
3. ## Re: Irrational Number
Originally Posted by louis33
Find n (natural number) where sqrt(13+33+33+43+5n) is irrational.
Here are some comments. Is there a typo: $\sqrt{1^3+{\color{red}2^3}+3^3+4^3+5^n}~?$
Even if it is, nothing really changes in the question.
Also: if $k$ is a non-square positive integer then $\large\sqrt{k}$ is irrational.
4. ## Irrational number
Find all the natural numbers for which sqrt(13+23+33+43+5n) is irrational.
6. ## Re: Irrational number
I still don't know how to solve it. Could you give me the complet explanation?
7. ## Re: Irrational number
in your other thread with the same problem, I just tried $n=1$ ... once again, on your edited expression I tried $n=1$ ...
$\sqrt{1^3+2^3+3^3+4^3+5^1} = \sqrt{105}$
As stated by Plato in your previous thread, note that 105 is not a perfect square, therefore its square root is irrational.
8. ## Re: Irrational Number
Yeah, there was a typo.
9. ## Re: Irrational number
Originally Posted by louis33
I still don't know how to solve it. Could you give me the complet explanation?
Originally Posted by louis33
Find all the natural numbers for which sqrt(13+23+33+43+5n) is irrational.
So the question reduces to: Find all $n\in\mathbb{N}^+$ such that $\Large5^n+100~$ is not a perfect square.
10. ## Re: Irrational Number
Let us find all $\displaystyle n$ for which $\displaystyle 5^n+100$ is a perfect square
$\displaystyle 5^n+100=m^2$ for some $\displaystyle m$
assume $\displaystyle n\geq 3$. Now m must be divisible by 5 so we can write m=5t.
replacing we get
$\displaystyle 5^{n-2}=(t-2)(t+2)$
so $\displaystyle t-2$ and $\displaystyle t+2$ are both powers of 5 but cannot both be divisible by 5 since otherwise their difference = 4 would be divisible by 5
therefore $\displaystyle t-2=1$ and $\displaystyle t+2=5^{n-2}$
this gives t=3 and n=3 | 689 | 2,158 | {"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 | 4 | CC-MAIN-2018-43 | latest | en | 0.82482 |
http://oeis.org/A160099 | 1,569,118,689,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574765.55/warc/CC-MAIN-20190922012344-20190922034344-00030.warc.gz | 148,957,464 | 3,873 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A160099 Decimal expansion of (843 + 418*sqrt(2))/601. 4
2, 3, 8, 6, 2, 5, 8, 3, 5, 1, 2, 0, 1, 2, 5, 4, 1, 2, 7, 1, 1, 9, 9, 7, 6, 5, 1, 7, 0, 0, 4, 4, 1, 5, 6, 3, 5, 3, 4, 4, 6, 3, 0, 3, 5, 5, 9, 1, 9, 4, 0, 8, 1, 4, 4, 0, 7, 2, 9, 6, 5, 2, 7, 5, 3, 4, 1, 3, 0, 8, 1, 1, 3, 1, 4, 0, 0, 3, 4, 5, 1, 2, 8, 7, 7, 6, 4, 0, 5, 8, 1, 2, 7, 9, 4, 5, 7, 8, 6, 5, 8, 9, 5, 5, 6 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Equals Lim_{n -> infinity} b(n)/b(n-1) for n mod 3 = {1, 2}, b = A111258. Equals Lim_{n -> infinity} b(n)/b(n-1) for n mod 3 = {0, 2}, b = A160098. LINKS G. C. Greubel, Table of n, a(n) for n = 1..10000 FORMULA Equals (38 + 11*sqrt(2))/(38 - 11*sqrt(2)). EXAMPLE (843 + 418*sqrt(2))/601 = 2.38625835120125412711... MATHEMATICA RealDigits[(843 +418*Sqrt[2])/601, 10, 100][[1]] (* G. C. Greubel, Apr 21 2018 *) PROG (PARI) (843+418*sqrt(2))/601 \\ G. C. Greubel, Apr 21 2018 (MAGMA) (843+418*Sqrt(2))/601; // G. C. Greubel, Apr 21 2018 CROSSREFS Cf. A111258, A160098, A002193 (decimal expansion of sqrt(2)), A160100 (decimal expansion of (361299+5950*sqrt(2))/601^2). Sequence in context: A047930 A073875 A220395 * A309967 A321336 A093098 Adjacent sequences: A160096 A160097 A160098 * A160100 A160101 A160102 KEYWORD cons,nonn AUTHOR Klaus Brockhaus, May 18 2009 STATUS approved
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Last modified September 21 22:17 EDT 2019. Contains 327284 sequences. (Running on oeis4.) | 828 | 1,807 | {"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-2019-39 | latest | en | 0.567681 |
http://mathhelpforum.com/pre-calculus/145223-solve-x.html | 1,526,904,186,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864063.9/warc/CC-MAIN-20180521102758-20180521122758-00533.warc.gz | 185,198,331 | 9,303 | 1. ## Solve for X
9 = 40(2)^2x+1
9 = 40(2)^2x+1
3. Originally Posted by harish21
$\displaystyle 9 = 40(2)^{2x+1 }$
$\displaystyle ln ( \frac{ 9}{40} ) = ln (2^{2x+1}) = (2x+1) ln2$
$\displaystyle \frac { ln ( \frac{ 9}{40} ) }{ ln2} = 2x + 1$
$\displaystyle \frac { ln ( \frac{ 9}{40} ) }{2 ln2} - \frac{1}{2} = x$ | 157 | 317 | {"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} | 3.890625 | 4 | CC-MAIN-2018-22 | latest | en | 0.410955 |
https://gmatclub.com/forum/if-the-salary-cap-of-the-national-basketball-association-17519.html?fl=similar | 1,511,122,098,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805761.46/warc/CC-MAIN-20171119191646-20171119211646-00227.warc.gz | 626,320,707 | 44,267 | It is currently 19 Nov 2017, 13:08
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# If the salary cap of the National Basketball Association
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21 Jun 2005, 17:20
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If the salary cap of the National Basketball Association were to be removed, the cost of running a championship team would increase by at least two million dollars or more.
If the salary cap of the National Basketball Association were to be removed, the cost of running a championship team would increase by at least two million dollars or more
Were the salary cap of the National Basketball Association to be removed, the cost of running a championship team would increase by at least two million dollars
Was the salary cap of the National Basketball Association to be removed, the cost of running a championship team would increase by at least two million dollars or more
If the salary cap of the National Basketball Association was removed, the cost of running a championship team would increase by at least two million dollars
Should the salary cap of the National Basketball Association be removed, the cost of running a championship team would increase by at least two million dollars or more
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21 Jun 2005, 21:09
I choose B.
... increase by at least two million dollars or more sounds redundant.
Last edited by kdhong on 21 Jun 2005, 21:27, edited 1 time in total.
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21 Jun 2005, 21:26
Agree with B. E is also likely but has redundent: ... \$2bn and more.
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21 Jun 2005, 23:19
A, C and E can be eliminated because 'or more' is redundant after 'at least'.
B & D both sound correct to me...B sounds better...but it'd be great if someone could give a reason for why one of the two is better than the other
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22 Jun 2005, 05:08
Yes it looks like B....Subjunctive is what is happening here
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22 Jun 2005, 06:30
the sentence consists of a main and a conditional clause. the conditional clause must contain "were" because it is the 2nd form of conditional, which describes unlikely situations. the main clause must contain "would". A) is indeed wrong because of redundancy.
_________________
If your mind can conceive it and your heart can believe it, have faith that you can achieve it.
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22 Jun 2005, 06:52
Which one is better? Be careful, i have modified E.
If the salary cap of the National Basketball Association were to be removed, the cost of running a championship team would increase by at least two million dollars or more.
B. Were the salary cap of the National Basketball Association to be removed, the cost of running a championship team would increase by at least two million dollars
E. Should the salary cap of the National Basketball Association be removed, the cost of running a championship team would increase by at least two million dollars
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22 Jun 2005, 10:46
I will go with D on this.
I am not very sure but here is what I read.
If things described in subjunctive mood are impossible and hypothetical, then "were" or "had" will be used.
But if things described in subjunctive mood are possible, then "was" can be used and is grammatically correct.
If you look at the answer choices "were and to be" do not go along well.
E looks impressive but I dont think it is grammtically correct because of atleast and more...
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22 Jun 2005, 13:50
OA is B
the rule of subjunctive applies...
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25 Jun 2005, 05:45
I am quite confused with sujunctive moods and if. Can someone please explain when are they applied & when are they not. Eg. in the following why is it not applied?
A wildlife expert predicts that the reintroduction of the caribou into northern Minnesota will fail if the density of the timber wolf population in that region is greater than one wolf for every 39 square miles.
Shouldn't it be "if the density ... were greater than"??
Kindly explain.
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25 Jun 2005, 22:01
"at least ... or more" is redundant.
D is wrong because of tense mix-up.
B for me
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27 Jun 2005, 06:47
I disagree with the OA.
B sounds awkward.
What is the source of this quesiton pls.
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27 Jun 2005, 06:47
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# If the salary cap of the National Basketball Association
Moderators: GMATNinjaTwo, GMATNinja
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®. | 1,832 | 6,556 | {"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-2017-47 | latest | en | 0.928291 |
https://web2.0calc.com/questions/a-math-question_12 | 1,585,498,875,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370494349.3/warc/CC-MAIN-20200329140021-20200329170021-00090.warc.gz | 795,532,105 | 6,106 | +0
# a math question
0
80
1
what is 90 x 73 / 34 + 73 - 50
Jan 7, 2020
#1
+1
Here is a hint:
Multiply first then divide then add and subtract. (ORDER MATTER!!) (From left to right)
Another hint:
BODMAS Is the rule you should follow , (We solve brackets (B) then orders (O) which is powers,exponents etc, then Division (D) then multiplication (M) then addition (A) then subtraction (S)) Most importantly, from left to right, if M is first on the left then you should start with M
M have the same value as D
A and S have the same value
In both order matters tho! From left to right again!!
Solution obtained by this website calculator:
90*73/34+73-50 = 216.2352941176470588
Without a calculator:
$$90*\frac{73}{2*17}+23$$
$$45*\frac{73}{17}+23$$
now you need to multiply 45*73
$$\frac{3285}{17}+23$$
Long division..
$$193.24+23=216.24$$
.
Jan 7, 2020 | 280 | 870 | {"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 | 4 | CC-MAIN-2020-16 | latest | en | 0.832815 |
https://math.wikia.org/wiki/Inscribed_sphere | 1,623,705,607,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487613453.9/warc/CC-MAIN-20210614201339-20210614231339-00280.warc.gz | 357,161,421 | 32,766 | 1,196 Pages
In geometry, the inscribed sphere or insphere of a convex polyhedron is a sphere that is contained within the polyhedron and tangent to each of the polyhedron's faces. It is the largest sphere that is contained wholly within the polyhedron, and is dual to the dual polyhedron's circumsphere.
All regular polyhedra have inscribed spheres, but some irregular polyhedra do not have all facets tangent to a common sphere, although it is still possible to define the largest contained sphere for such shapes. For such cases, the notion of an insphere does not seem to have been properly defined and various interpretations of an insphere are to be found:
• The sphere tangent to all faces (if one exists).
• The sphere tangent to all face planes (if one exists).
• The sphere tangent to a given set of faces (if one exists).
• The largest sphere that can fit inside the polyhedron.
Often these spheres coincide, leading to confusion as to exactly what properties define the insphere for polyhedra where they do not coincide.
For example the regular small stellated dodecahedron has a sphere tangent to all faces, while a larger sphere can still be fitted inside the polyhedron. Which is the insphere? Important authorities such as Coxeter or Cundy & Rollett are clear enough that the face-tangent sphere is the insphere. Again, such authorities agree that the Archimedean polyhedra (having regular faces and equivalent vertices) have no inspheres while the Archimedean dual or Catalan polyhedra do have inspheres. But many authors fail to respect such distinctions and assume other definitions for the 'inspheres' of their polyhedra.
The radius of the sphere inscribed in a polyhedron P is called the inradius of P.
## Volume
The volume of the sphere inscribed in a polyhedron is therefore:
## Surface area
The surface area of the sphere inscribed in a polyhedron is therefore:
## Inscribed hemisphere
### Volume
The volume of the hemisphere inscribed in a polyhedron is therefore:
### Surface area
The surface area of the hemisphere inscribed in a polyhedron is therefore:
## References
• Coxeter, H.S.M. Regular polytopes 3rd Edn. Dover (1973).
• Cundy, H.M. and Rollett, A.P. Mathematical Models, 2nd Edn. OUP (1961). | 515 | 2,244 | {"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} | 3.546875 | 4 | CC-MAIN-2021-25 | latest | en | 0.917289 |
https://metanumbers.com/67031 | 1,638,469,839,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362287.26/warc/CC-MAIN-20211202175510-20211202205510-00454.warc.gz | 430,963,862 | 7,324 | # 67031 (number)
67,031 (sixty-seven thousand thirty-one) is an odd five-digits composite number following 67030 and preceding 67032. In scientific notation, it is written as 6.7031 × 104. The sum of its digits is 17. It has a total of 2 prime factors and 4 positive divisors. There are 63,072 positive integers (up to 67031) that are relatively prime to 67031.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 17
• Digital Root 8
## Name
Short name 67 thousand 31 sixty-seven thousand thirty-one
## Notation
Scientific notation 6.7031 × 104 67.031 × 103
## Prime Factorization of 67031
Prime Factorization 17 × 3943
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 67031 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 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 67,031 is 17 × 3943. Since it has a total of 2 prime factors, 67,031 is a composite number.
## Divisors of 67031
1, 17, 3943, 67031
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 70992 Sum of all the positive divisors of n s(n) 3961 Sum of the proper positive divisors of n A(n) 17748 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 258.903 Returns the nth root of the product of n divisors H(n) 3.77682 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 67,031 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 67,031) is 70,992, the average is 17,748.
## Other Arithmetic Functions (n = 67031)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 63072 Total number of positive integers not greater than n that are coprime to n λ(n) 31536 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 6677 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 63,072 positive integers (less than 67,031) that are coprime with 67,031. And there are approximately 6,677 prime numbers less than or equal to 67,031.
## Divisibility of 67031
m n mod m 2 3 4 5 6 7 8 9 1 2 3 1 5 6 7 8
67,031 is not divisible by any number less than or equal to 9.
## Classification of 67031
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (67031)
Base System Value
2 Binary 10000010111010111
3 Ternary 10101221122
4 Quaternary 100113113
5 Quinary 4121111
6 Senary 1234155
8 Octal 202727
10 Decimal 67031
12 Duodecimal 3295b
16 Hexadecimal 105d7
20 Vigesimal 87bb
36 Base36 1fpz
## Basic calculations (n = 67031)
### Multiplication
n×y
n×2 134062 201093 268124 335155
### Division
n÷y
n÷2 33515.5 22343.7 16757.8 13406.2
### Exponentiation
ny
n2 4493154961 301180670190791 20188441503558911521 1353251422425057398164151
### Nth Root
y√n
2√n 258.903 40.6217 16.0905 9.23114
## 67031 as geometric shapes
### Circle
Radius = n
Diameter 134062 421168 1.41157e+10
### Sphere
Radius = n
Volume 1.26158e+15 5.64627e+10 421168
### Square
Length = n
Perimeter 268124 4.49315e+09 94796.1
### Cube
Length = n
Surface area 2.69589e+10 3.01181e+14 116101
### Equilateral Triangle
Length = n
Perimeter 201093 1.94559e+09 58050.5
### Triangular Pyramid
Length = n
Surface area 7.78237e+09 3.54945e+13 54730.6
## Cryptographic Hash Functions
md5 f55b61d6b3bf265db7d801774062598a e1c39c18db1995dae23fd3521d9ddbd06859d3ba d36d52f2fb9f41015dd6042da1b946b36fa9cabe60f584865bb4a959e2d91ae9 05aa8b2af3151fed78ace816c915aa291f0c5ea7bc80b340b4ced8e650d0c41c87f50188eadf129b4e937cd546efceac83d2c538731a7de3df0e6c23aa37f3c6 3dc87cde43b84fc173a5d93c05ac6966cc26d07e | 1,477 | 4,200 | {"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-2021-49 | latest | en | 0.815752 |
https://www.cuemath.com/ncert-solutions/exercise-5-6-continuity-and-differentiability-class-12-maths/ | 1,623,591,829,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487608856.6/warc/CC-MAIN-20210613131257-20210613161257-00257.warc.gz | 677,802,217 | 16,759 | # NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.6
## Chapter 5 Ex.5.6 Question 1
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = 2a{t^2},\;y = a{t^4}$$, without eliminating the parameter, find $$\frac{{dy}}{{dx}}$$
### Solution
Given, $$x = 2a{t^2},\; y = a{t^4}$$
Then,
$$\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {2a{t^2}} \right) = 2a.\frac{d}{{dt}}\left( {{t^2}} \right) = 2a.2t = 4at$$
\begin{align} & \frac{{dy}}{{dt}}= \frac{d}{{dt}}\left( {a{t^4}} \right)= a.\frac{d}{{dt}}\left( {{t^4}} \right) = a.4.{t^3} = 4a{t^3}\\&\therefore \frac{{dy}}{{dt}} = \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{4a{t^3}}}{{4at}} = {t^2}\end{align}
## Chapter 5 Ex.5.6 Question 2
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = a\cos \theta,\;y = b\cos \theta$$, without eliminating the parameter, find $$\frac{{dy}}{{dx}}$$
### Solution
Given, $$x = a\cos \theta,\;y = b\cos \theta$$
Then,
\begin{align}&\frac{{dx}}{{d\theta }} = \frac{d}{{d\theta }}\left( {a\cos \theta } \right) = a\left( { - \sin \theta } \right) = - a\sin \theta \\&\frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left( {b\cos \theta } \right) = b\left( { - \sin \theta } \right) = - b\sin \theta \\&\therefore \frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}} = \frac{{ - b\sin \theta }}{{ - a\sin \theta }} = \frac{b}{a}\end{align}
## Chapter 5 Ex.5.6 Question 3
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = \sin t,\;y = \cos 2t$$, without eliminating the parameter, find $$\frac{{dy}}{{dx}}$$
### Solution
Given, $$x = \sin t,\;y = \cos 2t$$
Then, $$\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {\sin t} \right) = \cos t$$
\begin{align}&\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {\cos 2t} \right) = - \sin 2t.\frac{d}{{dt}}\left( {2t} \right) = - 2\sin 2t\\&\therefore \frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{ - 2\sin 2t}}{{\cos t}} = \frac{{ - 2.2\sin t\cos t}}{{\cos t}} = - 4\sin t\end{align}
## Chapter 5 Ex.5.6 Question 4
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = 4t,\;y = \frac{4}{t}$$, without eliminating the parameter, find $$\frac{{dy}}{{dx}}$$
### Solution
Given, $$x = 4t,\;y = \frac{4}{t}$$
\begin{align}&\frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {4t} \right) = 4\\&\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {\frac{4}{t}} \right) = 4.\frac{d}{{dt}}\left( {\frac{1}{t}} \right) = 4.\left( {\frac{{ - 1}}{{{t^2}}}} \right) = \frac{{ - 4}}{{{t^2}}}\\&\therefore \frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{\left( {\frac{{ - 4}}{{{t^2}}}} \right)}}{4} = \frac{{ - 1}}{{{t^2}}}\end{align}
## Chapter 5 Ex.5.6 Question 5
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = \cos \theta - \cos 2\theta,\;y = \sin \theta - \sin 2\theta$$, without eliminating the parameter, find
### Solution
Given, $$x = \cos \theta - \cos 2\theta ,\;y = \sin \theta - \sin 2\theta$$
Then,
\begin{align}\frac{{dx}}{{d\theta }} &= \frac{d}{{d\theta }}\left( {\cos \theta - \cos 2\theta } \right) = \frac{d}{{d\theta }}\left( {\cos \theta } \right) - \frac{d}{{d\theta }}\left( {\cos 2\theta } \right)\\[5pt]&= - \sin \theta - \left( { - 2\sin 2\theta } \right) \\[5pt]&= 2\sin 2\theta - \sin \theta \\[10pt] \frac{{dy}}{{d\theta }}& = \frac{d}{{d\theta }}\left( {\sin \theta - \sin 2\theta } \right) = \frac{d}{{d\theta }}\left( {\sin \theta } \right) - \frac{d}{{d\theta }}\left( {\sin 2\theta } \right)\\[5pt]&= \cos \theta - 2\cos 2\theta \\\therefore \frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}} &= \frac{{\cos \theta - 2\cos 2\theta }}{{2\sin 2\theta - \sin \theta }}\end{align}
## Chapter 5 Ex.5.6 Question 6
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = a\left( {\theta - \sin \theta } \right),\;y = a\left( {1 + \cos \theta } \right)$$, without eliminating the parameter, find $$\frac{{dy}}{{dx}}$$
### Solution
Given, $$x = a\left( {\theta - \sin \theta } \right),\;y = a\left( {1 + \cos \theta } \right)$$
Then, $$\frac{{dx}}{{d\theta }} = a\left[ {\frac{d}{{d\theta }}\left( \theta \right) - \frac{d}{{d\theta }}\left( {\sin \theta } \right)} \right] = a\left( {1 - \cos \theta } \right)$$
\begin{align}\frac{{dy}}{{d\theta }} &= a\left[ {\frac{d}{{d\theta }}\left( 1 \right) + \frac{d}{{d\theta }}\left( {\cos \theta } \right)} \right] = a\left[ {0 + \left( { - \sin \theta } \right)} \right] = - a\sin \theta \\\therefore \frac{{dy}}{{dx}} &= \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}} = \frac{{ - a\sin \theta }}{{a\left( {1 - \cos \theta } \right)}} = \frac{{ - 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}{{2{{\;\sin }^2}\frac{\theta }{2}}} = \frac{{ - \cos \frac{\theta }{2}}}{{\sin \frac{\theta }{2}}} = - \cot \frac{\theta }{2}\end{align}
## Chapter 5 Ex.5.6 Question 7
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = \frac{{{{\sin }^3}t}}{{\sqrt {\cos 2t} }},\;y = \frac{{{{\cos }^3}t}}{{\sqrt {\cos 2t} }}$$, without eliminating the parameter, find $$\frac{{dy}}{{dx}}$$
### Solution
Given, $$x = \frac{{{{\sin }^3}t}}{{\sqrt {\cos 2t} }},\;y = \frac{{{{\cos }^3}t}}{{\sqrt {\cos 2t} }}$$
Then,
\begin{align}\frac{{dx}}{{dt}} &= \frac{d}{{dt}}\left[ {\frac{{{{\sin }^3}t}}{{\sqrt {\cos 2t} }}} \right]\\[5pt]&= \frac{{\sqrt {\cos 2t} .\frac{d}{{dt}}\left( {{{\sin }^3}t} \right) - {{\sin }^3}t.\frac{d}{{dt}}\sqrt {\cos 2t} }}{{\cos 2t}}\\[5pt]&= \frac{{\sqrt {\cos 2t} .3{{\sin }^2}t.\frac{d}{{dt}}\left( {\sin t} \right) - {{\sin }^3}t \times \frac{1}{{2\sqrt {\cos 2t} }}.\frac{d}{{dt}}\left( {\cos 2t} \right)}}{{\cos 2t}}\\[5pt]&= \frac{{3\sqrt {\cos 2t} .{{\sin }^2}t.\cos t - \frac{{{{\sin }^3}t}}{{2\sqrt {\cos 2t} }}.\left( { - 2\sin 2t} \right)}}{{\cos 2t}}\\[5pt]&= \frac{{3\cos 2t.{{\sin }^2}t\cos t + {{\sin }^3}t.\sin 2t}}{{\cos 2t\sqrt {\cos 2t} }}\\[5pt]\frac{{dy}}{{dt}}& = \frac{d}{{dt}}\left[ {\frac{{{{\cos }^3}t}}{{\sqrt {\cos 2t} }}} \right]\\&= \frac{{\sqrt {\cos 2t} .\frac{d}{{dt}}\left( {{{\cos }^3}t} \right) - {{\cos }^3}t.\frac{d}{{dt}}\left( {\sqrt {\cos 2t} } \right)}}{{\cos 2t}}\\[5pt]&= \frac{{\sqrt {\cos 2t} .3{{\cos }^2}t.\frac{d}{{dt}}\left( {\cos t} \right) - {{\cos }^3}t.\frac{1}{{2\sqrt {\cos 2t} }}.\frac{d}{{dt}}\left( {\cos 2t} \right)}}{{\cos 2t}}\\[5pt]&= \frac{{3\sqrt {\cos 2t} .{{\cos }^2}t\left( { - \sin t} \right) - {{\cos }^3}t.\frac{1}{{\sqrt {\cos 2t} }}.\left( { - 2\sin 2t} \right)}}{{\cos 2t}}\\[5pt]&= \frac{{ - 3\cos 2t.{{\cos }^2}t.\sin t + {{\cos }^3}t.\sin 2t}}{{\cos 2t.\sqrt {\cos 2t} }}\end{align}
\begin{align}\therefore \frac{{dy}}{{dx}} &= \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{\frac{{ - 3\cos 2t.{{\cos }^2}t.\sin t + {{\cos }^3}t\sin 2t}}{{\cos 2t.\sqrt {\cos 2t} }}}}{{\frac{{3\cos 2t.{{\sin }^2}t.\cos t + {{\sin }^3}t\sin 2t}}{{\cos 2t.\sqrt {\cos 2t} }}}}\\[5pt]&= \frac{{ - 3\cos 2t.{{\cos }^2}t.\sin t + {{\cos }^3}t\sin 2t}}{{3\cos 2t.{{\sin }^2}t.\cos t + {{\sin }^3}t\sin 2t}}\\[5pt]&= \frac{{ - 3\cos 2t.{{\cos }^2}t.\sin t + {{\cos }^3}t\left( {2\sin t\cos t} \right)}}{{3\cos 2t.{{\sin }^2}t.\cos t + {{\sin }^3}t\left( {2\sin t\cos t} \right)}}\\[5pt]&= \frac{{\sin t\cos t\left[ { - 3\cos 2t.\cos t + 2{{\cos }^3}t} \right]}}{{\sin t\cos t\left[ {3\cos 2t\sin t + 2{{\sin }^3}t} \right]}}\\[5pt]&= \frac{{\left[ { - 3\left( {2{{\cos }^2}t - 1} \right)\cos t + 2{{\cos }^3}t} \right]}}{{\left[ {3\left( {1 - 2{{\sin }^2}t} \right)\sin t + 2{{\sin }^3}t} \right]}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \begin{array}{l}\cos 2t = \left( {2{{\cos }^2}t - 1} \right)\\\cos 2t = \left( {1 - 2{{\sin }^2}t} \right)\end{array} \right]\\[5pt]&= \frac{{ - 4{{\cos }^3}t + 3\cos t}}{{3\sin t - 4{{\sin }^3}t}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \begin{array}{l}\cos 3t = 4{\cos ^3}t - 3\cos t\\\sin 3t = 3\sin t - 4{\sin ^2}t\end{array} \right]\\[5pt]&= \frac{{ - \cos 3t}}{{\sin 3t}} = - \cot 3t\end{align}
## Chapter 5 Ex.5.6 Question 8
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = a\left( {\cos t + \log \tan \frac{t}{2}} \right),\;y = a\sin t$$, without eliminating the parameter, find $$\frac{{dy}}{{dx}}$$
### Solution
Given, $$x = a\left( {\cos t + \log \tan \frac{t}{2}} \right),\;y = a\sin t$$
Then,
\begin{align}\frac{{dx}}{{dt}} &= a.\left[ {\frac{d}{{dt}}\left( {\cos t} \right) + \frac{d}{{dt}}\left( {\log \tan \frac{t}{2}} \right)} \right]\\&= a\left[ { - \sin t + \frac{1}{{\tan \frac{t}{2}}}.\frac{d}{{dt}}\left( {\tan \frac{t}{2}} \right)} \right]\\&= a\left[ { - \sin t + \cot \frac{t}{2}.{{\sec }^2}\frac{t}{2}.\frac{d}{{dt}}\left( {\frac{t}{2}} \right)} \right]\\&= a\left[ { - \sin t + \frac{{\cos \frac{t}{2}}}{{\sin \frac{t}{2}}} \times \frac{1}{{{{\cos }^2}\frac{t}{2}}} \times \frac{1}{2}} \right]\\&= a\left[ { - \sin t + \frac{1}{{2\sin \frac{t}{2}\cos \frac{t}{2}}}} \right]\\&= a\left( { - \sin t + \frac{1}{{\sin t}}} \right)\\&= a\left( {\frac{{ - {{\sin }^2}t + 1}}{{\sin t}}} \right)\\&= a\left( {\frac{{{{\cos }^2}t}}{{\sin t}}} \right)\\\frac{{dy}}{{dt}} &= a\frac{d}{{dt}}\left( {\sin t} \right) = a\cos t\end{align}
Therefore,
$\frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{a\cos t}}{{\left( {a\frac{{{{\cos }^2}t}}{{\sin t}}} \right)}} = \frac{{\sin t}}{{\cos t}} = \tan t$
## Chapter 5 Ex.5.6 Question 9
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = a\;\sec \theta,\;y = b\;\tan \theta$$, without eliminating the parameter, find $$\frac{{dy}}{{dx}}$$
### Solution
Given, $$x = a\;\sec \theta,\;y = b\;\tan \theta$$
Then,
$$\frac{{dx}}{{d\theta }} = a.\frac{d}{{d\theta }}\left( {\sec \theta } \right) = a\sec \theta \tan \theta$$
$$\frac{{dy}}{{d\theta }} = b.\frac{d}{{d\theta }}\left( {\tan \theta } \right) = b{\;\sec ^2}\theta$$
Therefore,
\begin{align}\frac{{dy}}{{dx}} &= \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}}\\&= \frac{{b{{\;\sec }^2}\theta }}{{a\sec \theta \tan \theta }}\\&= \frac{b}{a}\sec \theta \cot \theta \\&= \frac{{b\cos \theta }}{{a\cos \theta \sin \theta }}\\&= \frac{b}{a} \times \frac{1}{{\sin \theta }}\\&= \frac{b}{a}{\mathop{\text cosec\;}\nolimits} \theta \end{align}
## Chapter 5 Ex.5.6 Question 10
If $$x$$ and $$y$$ are connected parametrically by the equations $$x = a\left( {\cos \theta + \theta \sin \theta } \right),\;y = a\left( {\sin \theta - \theta \cos \theta } \right)$$, without eliminating the parameter, find $$\frac{{dy}}{{dx}}$$
### Solution
Given, $$x = a\left( {\cos \theta + \theta \sin \theta } \right),\;y = a\left( {\sin \theta - \theta \cos \theta } \right)$$
Then,
\begin{align}\frac{{dx}}{{d\theta }} &= a\left[ {\frac{d}{{d\theta }}\cos \theta + \frac{d}{{d\theta }}\left( {\theta \sin \theta } \right)} \right]\\&= a\left[ { - \sin \theta + \theta \frac{d}{{d\theta }}\left( {\sin \theta } \right) + \sin \theta \frac{d}{{d\theta }}\left( \theta \right)} \right]\\[5pt]&= a\left[ { - \sin \theta + \theta \cos \theta + \sin \theta } \right]\\[5pt]&= a\;\theta \cos \theta \end{align}
\begin{align}\frac{{dy}}{{d\theta }} &= a\left[ {\frac{d}{{d\theta }}\left( {\sin \theta } \right) - \frac{d}{{d\theta }}\left( {\theta \cos \theta } \right)} \right] \\&= a\left[ {\cos \theta - \left\{ {\theta \frac{d}{{d\theta }}\left( {\cos \theta } \right) + \cos \theta .\frac{d}{{d\theta }}\left( \theta \right)} \right\}} \right]\\[5pt]&= a\left[ {\cos \theta + \theta \sin \theta - \cos \theta } \right]\\[5pt]&= a\; \theta \sin \theta \end{align}
Therefore,
\begin{align}\frac{{dy}}{{dx}} &= \frac{{\left( {\frac{{dy}}{{d\theta }}} \right)}}{{\left( {\frac{{dx}}{{d\theta }}} \right)}}\\[5pt]&= \frac{{a\;\theta \sin \theta }}{{a\;\theta \cos \theta }}\\[5pt]&= \tan \theta \end{align}
## Chapter 5 Ex.5.6 Question 11
If $$x = \sqrt {{a^{{{\sin }^{ - 1}}t}}},\;y = \sqrt {{a^{{{\cos }^{ - 1}}t}}}$$, show that $$\frac{{dy}}{{dx}} = - \frac{y}{x}$$
### Solution
Given, $$x = \sqrt {{a^{{{\sin }^{ - 1}}t}}}$$ and $$y = \sqrt {{a^{{{\cos }^{ - 1}}t}}}$$
Hence,
$$x = \sqrt {{a^{{{\sin }^{ - 1}}t}}} = {\left( {{a^{{{\sin }^{ - 1}}t}}} \right)^{\frac{1}{2}}} = {a^{\frac{1}{2}{{\sin }^{ - 1}}t}}$$ and $$y = \sqrt {{a^{{{\cos }^{ - 1}}t}}} = {\left( {{a^{{{\cos }^{ - 1}}t}}} \right)^{\frac{1}{2}}} = {a^{\frac{1}{2}{{\cos }^{ - 1}}t}}$$
Consider $$x = {a^{\frac{1}{2}{{\sin }^{ - 1}}t}}$$
Taking log on both sides, we get
$$\log x = \frac{1}{2}{\sin ^{ - 1}}t\log a$$
Therefore,
\begin{align}&\Rightarrow \; \; \frac{1}{x}.\frac{{dx}}{{dt}} = \frac{1}{2}\log a.\frac{d}{{dt}}\left( {{{\sin }^{ - 1}}t} \right)\\&\Rightarrow \; \; \frac{{dx}}{{dt}} = \frac{x}{2}\log a.\frac{1}{{\sqrt {1 - {t^2}} }}\\&\Rightarrow \; \; \frac{{dx}}{{dt}} = \frac{{x\log a}}{{2\sqrt {1 - {t^2}} }}\end{align}
Now, $$y = {a^{\frac{1}{2}{{\cos }^{ - 1}}t}}$$
Taking log on both sides, we get
$$\log x = \frac{1}{2}{\cos ^{ - 1}}t\log a$$
Therefore,
\begin{align}&\Rightarrow \; \; \frac{1}{y}.\frac{{dy}}{{dt}} = \frac{1}{2}\log a.\frac{d}{{dt}}\left( {{{\cos }^{ - 1}}t} \right)\\&\Rightarrow \; \; \frac{{dy}}{{dt}} = \frac{y}{2}\log a.\frac{{ - 1}}{{\sqrt {1 - {t^2}} }}\\&\Rightarrow \; \; \frac{{dy}}{{dt}} = \frac{{ - y\log a}}{{2\sqrt {1 - {t^2}} }}\end{align}
Hence,
$\frac{{dy}}{{dx}} = \frac{{\left( {\frac{{dy}}{{dt}}} \right)}}{{\left( {\frac{{dx}}{{dt}}} \right)}} = \frac{{\left( {\frac{{ - y\log a}}{{2\sqrt {1 - {t^2}} }}} \right)}}{{\left( {\frac{{x\log a}}{{2\sqrt {1 - {t^2}} }}} \right)}} = - \frac{y}{x}$
Instant doubt clearing with Cuemath Advanced Math Program | 6,073 | 13,903 | {"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": 30, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2021-25 | latest | en | 0.403828 |
https://sportsbizusa.com/20-number-3-preschool-worksheet/ | 1,627,998,352,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154459.22/warc/CC-MAIN-20210803124251-20210803154251-00195.warc.gz | 533,568,538 | 11,369 | HomeTemplate ➟ 20 20 Number 3 Preschool Worksheet
# 20 Number 3 Preschool Worksheet
Learn to Count and Write Number 3 printable number 3 worksheet preschool, number 3 tracing worksheets preschool, free printable number 3 worksheets for preschoolers, via: pinterest.com
Numbering Worksheets for Kids. Kids are usually introduced to this topic matter during their math education. The main reason behind this is that learning math can be done with the worksheets. With an organized worksheet, kids will be able to describe and explain the correct answer to any mathematical problem. But before we talk about how to create a math worksheet for kids, let’s have a look at how children learn math.
In elementary school, children are exposed to a number of different ways of teaching them how to do a number of different subjects. Learning these subjects is important because it would help them develop logical reasoning skills. It is also an advantage for them to understand the concept behind all mathematical concepts.
To make the learning process easy for children, the educational methods used in their learning should be easy. For example, if the method is to simply count, it is not advisable to use only numbers for the students. Instead, the learning process should also be based on counting and dividing numbers in a meaningful way.
The main purpose of using a worksheet for kids is to provide a systematic way of teaching them how to count and multiply. Children would love to learn in a systematic manner. In addition, there are a few benefits associated with creating a worksheet. Here are some of them:
Children have a clear idea about the number of objects that they are going to add up. A good worksheet is one which shows the addition of different objects. This helps to give children a clear picture about the actual process. This helps children to easily identify the objects and the quantities that are associated with it.
This worksheet helps the child’s learning. It also provides children a platform to learn about the subject matter. They can easily compare and contrast the values of various objects. They can easily identify the objects and compare it with each other. By comparing and contrasting, children will be able to come out with a clearer idea.
He or she will also be able to solve a number of problems by simply using a few cells. He or she will learn to organize a worksheet and manipulate the cells. to arrive at the right answer to any question.
This worksheet is a vital part of a child’s development. When he or she comes across an incorrect answer, he or she can easily find the right solution by using the help of the worksheets. He or she will also be able to work on a problem without having to refer to the teacher. And most importantly, he or she will be taught the proper way of doing the mathematical problem.
Math skills are the most important part of learning and developing. Using the worksheet for kids will improve his or her math skills.
Many teachers are not very impressed when they see the number of worksheets that are being used by their children. This is actually very much true in the case of elementary schools. In this age group, the teachers often feel that the child’s performance is not good enough and they cannot just give out worksheets.
However, what most parents and educators do not realize is that there are several ways through which you can improve the child’s performance. You just need to make use of a worksheet for kids. elementary schools.
As a matter of fact, there is a very good option for your children to improve their performance in math. You just need to look into it.
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https://mathhelpboards.com/threads/fourier-series-showing-converges-to-pi-16.6539/ | 1,610,732,719,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00113.warc.gz | 503,936,812 | 15,244 | # Fourier series--showing converges to pi/16
#### dwsmith
##### Well-known member
When a Fourier series contains only sine and cosine terms, evaluating the series isn't too difficult.
However, I want to show a Fourier series with sine and sinh converges to $$\frac{\pi}{16}$$.
$T(50, 50) = \sum_{n = 1}^{\infty} \frac{\sin\left(\frac{\pi(2n - 1)}{2}\right) \sinh\left(\frac{\pi(2n - 1)}{2}\right)} {(2n - 1)\sinh[(2n - 1)\pi]} = \frac{\pi}{16}$
Since this series contains sinh, I am not sure how to evaluate it.
#### Deveno
##### Well-known member
MHB Math Scholar
My thought: use
$\text{sinh}(x) = -i \sin(ix)$
#### dwsmith
##### Well-known member
My thought: use
$\text{sinh}(x) = -i \sin(ix)$
I think I will still have some trouble though. I end up with:
$\sum_{n = 1}^{\infty}\frac{(-1)^{n+1}}{2n - 1}\frac{\sin\left[\frac{i\pi}{2}(2n-1)\right]}{\sin\left[i\pi(2n-1)\right]}$
#### Opalg
##### MHB Oldtimer
Staff member
When a Fourier series contains only sine and cosine terms, evaluating the series isn't too difficult.
However, I want to show a Fourier series with sine and sinh converges to $$\frac{\pi}{16}$$.
$T(50, 50) = \sum_{n = 1}^{\infty} \frac{\sin\left(\frac{\pi(2n - 1)}{2}\right) \sinh\left(\frac{\pi(2n - 1)}{2}\right)} {(2n - 1)\sinh[(2n - 1)\pi]} = \frac{\pi}{16}$
Since this series contains sinh, I am not sure how to evaluate it.
The sin function isn't really there at all, because $\sin\left(\frac{\pi(2n - 1)}{2}\right) = \sin\left(\bigl(n-\frac12\bigr)\pi\right) = (-1)^{n-1}$. Also, you can use the identity $\sinh(2x) = 2\sinh x\cosh x$, to rewrite the sum as $$\displaystyle \sum_{n = 1}^{\infty}\frac{(-1)^n}{2(2n-1)\cosh\left(\bigl(n-\frac12\bigr)\pi\right)}.$$ That looks a bit less complicated than the original series, but I still don't see how to sum it. It clearly converges very fast, because the cosh term in the denominator will get very large after the first few terms. A numerical check shows that the sum of the first three terms is $0.1968518...$, which is very close to $\pi/16$. But that isn't a proof! | 713 | 2,059 | {"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": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-04 | longest | en | 0.838143 |
https://kinerjamall.com/checker-404 | 1,669,526,032,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710192.90/warc/CC-MAIN-20221127041342-20221127071342-00620.warc.gz | 391,028,242 | 5,167 | # How to solve with square roots
These can be very helpful when you're stuck on a problem and don't know How to solve with square roots. We will give you answers to homework.
## How can we solve with square roots
Read on for some helpful advice on How to solve with square roots easily and effectively. Piecewise functions can be tricky to solve, but there are a few steps you can follow to make it easier. First, identify the different parts of the function and what values they apply to. Second, graph the function to help you visualize what's happening. Finally, solve each part of the function separately and piece the solution together. Following these steps should help you solve most piecewise functions.
There are many methods for solving systems of linear equations. The most common method is using elimination. This involves adding or subtracting equations in order to cancel out variables. Another common method is substitution, which involves solving for one variable in terms of the others and then plugging this back into the other equations. These methods can be used for systems with any number of variables.
To solve a piecewise function, you need to carefully examine the given equations and determine which one to use for which x-values. You also need to be careful when graphing piecewise functions, as the different equations can produce different shapes for the graph.
Scanning math problems can be a quick and easy way to check your work for mistakes. By scan reading, you can quickly identify where you made a mistake and correct it. This can save you time and frustration in the long run.
In this case, we can subtract 4 from both sides of the equation to get: y - 4 = 2x Then, we need to divide both sides of the equation by 2 to get: y/2 = x Finally, we can multiply both sides of the equation by 2 to get: y = 2x
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Math problem search Geometry helper app Three equations three unknowns solver Problem solver com Completing the square solver Expression solver | 542 | 2,583 | {"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} | 4.375 | 4 | CC-MAIN-2022-49 | latest | en | 0.950607 |
http://mathhelpforum.com/calculus/15437-max-mins-without-numerical-values-graph-print.html | 1,519,447,633,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815318.53/warc/CC-MAIN-20180224033332-20180224053332-00258.warc.gz | 219,726,072 | 3,520 | # Max/Mins Without Numerical Values (On Graph)
• May 28th 2007, 06:24 PM
SportfreundeKeaneKent
Max/Mins Without Numerical Values (On Graph)
These are two different types of max min problems but I understand that they go on a graph instead of being the usual rectangle with a given value question.
http://img516.imageshack.us/img516/4438/45lc1.png
• May 28th 2007, 06:31 PM
Jhevon
Quote:
Originally Posted by SportfreundeKeaneKent
These are two different types of max min problems but I understand that they go on a graph instead of being the usual rectangle with a given value question.
http://img516.imageshack.us/img516/4438/45lc1.png
For the first:
two lines are parrallel if they have the same slope. the slope of the line is 9. so we simply want to find the $x$'s that make the tangent lines to the curve have a slope of 9. the derivative gives the formula for the slope of the tangent line.
$y = x^3 - 3x^2$
$\Rightarrow y' = 3x^2 - 6x$
Now we simply solve:
$3x^2 - 6x = 9$
i leave that to you
• May 28th 2007, 06:35 PM
Jhevon
Quote:
Originally Posted by SportfreundeKeaneKent
These are two different types of max min problems but I understand that they go on a graph instead of being the usual rectangle with a given value question.
http://img516.imageshack.us/img516/4438/45lc1.png
see http://www.mathhelpforum.com/math-he...ma-minima.html as a hint for the second
i assume you remember how to maximize a function
• Jun 1st 2007, 04:09 PM
SportfreundeKeaneKent
A question about the one with the tangent. Would you sub x=-1,3 (the two values I obtain after factoring and finding x) into y=x^3-3x^2 or into y=9x+7?
I'm guessing that you'd sub it into y=9x+7 to obtain the two points to be (3,34) and (-1,-2)
• Jun 1st 2007, 04:11 PM
Jhevon
Quote:
Originally Posted by SportfreundeKeaneKent
A question about the one with the tangent. Would you sub x=-1,3 (the two values I obtain after factoring and finding x) into y=x^3-3x^2 or into y=9x+7?
I'm guessing that you'd sub it into y=9x+7 to obtain the two points to be (3,34) and (-1,-2)
we want two points ON THE CURVE, therefore you would plug them into the original function to obtain the y-values for the points. we wouldn't sub them into the line, unless we knew that the line intersected with our curve exactly at the points where the slope of the curve is 9. so just plug them into $y = x^3 - 3x^2$
why did you think you would plug it into the line?
• Jun 1st 2007, 04:35 PM
SportfreundeKeaneKent
Oh wait, I wrote it the wrong way around. So the points are just (-1,-4) and (3,0)
• Jun 1st 2007, 04:39 PM
Jhevon
Quote:
Originally Posted by SportfreundeKeaneKent
Oh wait, I wrote it the wrong way around. So the points are just (-1,-4) and (3,0)
correct :D
• Jun 2nd 2007, 09:43 AM
curvature
tangent lines
tangent lines | 846 | 2,800 | {"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": 5, "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-2018-09 | longest | en | 0.904676 |
https://socratic.org/questions/59e7439a11ef6b2d324ff210 | 1,597,077,363,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736057.87/warc/CC-MAIN-20200810145103-20200810175103-00323.warc.gz | 459,378,350 | 6,485 | # Question ff210
Oct 20, 2017
$\text{185.0 u}$
#### Explanation:
The first thing that you need to do here is to use the percent abundance of rhenium-187 to calculate the percent abundance of rhenium-185.
To do that, use the fact that rhenium has only two naturally occurring isotopes. This implies that their percent abundances must add up to give 100%.
Alternatively, you can say that their decimal abundances must add up to $1$. This is the case because a decimal abundance is simply a percent abundance divided by 100%.
This means that you have
$\text{decimal abundance"color(white)(.)""^185"Re} = 1 - 0.6260$
Now, the average atomic mass of the element is calculated by taking the weighted average of the atomic masses of its naturally occurring isotopes.
This means that you can write
"186.207" color(red)(cancel(color(black)("u"))) = overbrace(186.956 color(red)(cancel(color(black)("u"))) * 0.6260)^(color(blue)("contribution from"color(white)(.)""^187"Re")) + overbrace(? color(red)(cancel(color(black)("u"))) * (1 - 0.6260))^(color(blue)("contribution from"color(white)(.)""^185"Re"))
Here ? represents the value of the atomic mass of rhenium-185. Rearrange the equation to solve for ?
? = (186.207 - 186.956 * 0.6260 )/(1 - 0.06260) = 184.953
Therefore, you can say that the atomic mass of rhenium-185 is equal to
"atomic mass"color(white)(.)""^185"Re" = color(darkgreen)(ul(color(black)("185.0 u")))#
The answer must be rounded to four sig figs, the number of sig figs you have for the percent abundance of rhenium-187. | 436 | 1,546 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "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} | 4.25 | 4 | CC-MAIN-2020-34 | latest | en | 0.837953 |
http://yojih.net/standard-error/guide-what-does-standard-error-mean-in-regression-analysis.php | 1,591,200,690,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347435238.60/warc/CC-MAIN-20200603144014-20200603174014-00058.warc.gz | 219,934,734 | 4,591 | Home > Standard Error > What Does Standard Error Mean In Regression Analysis
# What Does Standard Error Mean In Regression Analysis
in contiguous columns (here columns B and C). They will be subsumed S is 3.53399, which tells us that the average distance ofSee also unbiased estimation of standard deviation for more discussion.gung 74.7k19163312 Excellent and very clear answer!
WHY are you looking at numeral for 1980 to 1989? Confidence intervals for what http://yojih.net/standard-error/tutorial-what-does-the-standard-error-mean-in-regression-analysis.php )*(k-1)/(n-k) = .8025 - .1975*2/2 = 0.6050. mean Standard Error Of Estimate Calculator Thus, larger SEs them can be expressed exactly as a linear combination of the others. JSTOR2340569. (Equation 1) what
There's not much I can conclude without understanding standard Deer in German: Hirsch, Reh What grid should be equal to the population mean.
Of theorem of calculus be proved in just two lines? DM. Standard Error Of Regression Formula The log transformation is also in the better performing school was ‘significantly' better than the other.It is possible to compute confidence intervals for either means or predictions arounddo with the sampling distributions of your slopes.
The Standard Error of the estimate is the The Standard Error of the estimate is the You can see that in Graph A, the points are sample, plotted on the distribution of ages for all 9,732 runners.
in the values fall outside the range plus-or-minus 2.Therefore, the predictions in Graph A Standard Error Of Estimate Interpretation test whether HH SIZE has coefficient β2 = 1.0. the amount of uncertainty in that distribution. The sample mean x ¯ {\displaystyle {\bar {x}}} = 37.25 is greater
I actually haven't read does the dependent variable is affected multiplicatively by the independent variables.many cases, I prefer the standard error of the regression over R-squared. does email [email protected] *No appts.The estimated coefficients for the two dummy variables would exactly equal the difference http://yojih.net/standard-error/repairing-what-is-the-meaning-of-standard-error-in-regression-analysis.php
Here Feb 6-May 5Walk-in, 1-5 pm* May the answer to that question.3 (3): 113–116. There’s no http://onlinestatbook.com/lms/regression/accuracy.html significance, and you generally don't scrutinize its t-statistic too closely.Sokal and Rohlf (1981)[7] give an equation analysis
the following table of coefficients and associated output: Coefficient St. in companion page Introduction to Regression first.Then t = (b2 - H0 value of β2) / (standard error of standard error of the mean describes bounds on a random sampling process.
Use of the standard error statistic presupposes the user is familiar with the central mean the statistic, the statistic will typically be non-significant. x1 and y in the population, but you only have access to your sample. Read more about how to obtain and use Standard Error Of Regression Coefficient if the number of degrees of freedom is more than about 30.Scatterplots involving such variables will be very strange looking: the points will
It is particularly important to use the standard error to estimate an see this our model needs to be more precise.How to Fill Between two Curves Dealing with a nasty a model, depending on the amount of "leverage" that it has.The numerator is the sum of squared differences error recruiter Was user-agent identification used for some scripting attack techique?The residual standard deviation has nothing to mean
doi:10.2307/2340569. Therefore, which is the Linear Regression Standard Error Other standard errors Every inferential statistic has an associated standard error.Excel does not provide alternaties, such asheteroskedastic-robustwho have had open heart surgery that lasted more than 4 hours. that the data points fall from the fitted values.
does the usual estimator of a population mean.The graph shows the ages for the 16 runners in theIn an example above, n=16 runners wereto calculate confidence intervals.
http://yojih.net/standard-error/repair-unstandardized-regression-coefficient-standard-error-of-the-regression-coefficient-t-value.php judged by its t-statistic, then there is really no need to look at the F-ratio.However, you can’t use R-squared to assessIn most cases, the effect size statistic It concludes, "Until a better case can Standard Error Of Prediction hard-and-fast rule, just an arbitrary threshold that indicates the possibility of a problem.
So do not reject null hypothesis at advice may be correct.However, many statistical results obtained from a computer statistical package (such as Hence, you can think of the standard error of the estimated coefficient of Xapproximately the same expected value; i.e., the F-ratio should be roughly equal to 1.
between the actual scores and the predicted scores. The mean age The Standard Error Of The Estimate Is A Measure Of Quizlet error. error A quantitative measure of uncertainty is reported: a margin ofslope and the intercept) were estimated in order to estimate the sum of squares.
is represented by the symbol σ x ¯ {\displaystyle \sigma _{\bar {x}}} . For example, to find 99% confidence intervals: in the Regression dialog box (in thethen Y is expected to change by b1 + b2 units. in Because your independent variables may be correlated, a condition known as multicollinearity, the coefficients What Is A Good Standard Error The reason N-2 is used rather than N-1 is that two parameters (thegood one!
I.e., the five variables Q1, Q2, Q3, Q4, and CONSTANT are not linearly independent: mean deviation of the coefficient, the amount it varies across cases. Column "P-value" gives the p-value for test ofa measure of the accuracy of predictions. does Pearson-Prentice Hall, 2006. 3. Standard error.
Standard Error of the that you give, and that people usually have in mind when they ask this question. There is no contradiction, regressor variables be in adjoining columns. Is there a name for the (anti- ) pattern of passing parameters columns need to be copied to get the regressors in contiguous columns.
In "classical" statistical methods such as linear regression, information about the precision is likely that the population mean is zero or near zero.
The margin of error and the confidence interval are There is little extra to know I use the graph for simple sales will fall within a given distance--say, $5M or$10M--of the predicted value of \$83.421M. | 1,424 | 6,495 | {"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} | 3.578125 | 4 | CC-MAIN-2020-24 | latest | en | 0.872475 |
http://www.statisticslectures.com/topics/addingintegers/ | 1,685,994,035,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652161.52/warc/CC-MAIN-20230605185809-20230605215809-00650.warc.gz | 94,575,410 | 2,535 | Remember that any number that can be found on a number line is an integer.
The number line can be used to help with addition of integers that have different signs. Let's use the number line to solve the following equation:
(-2) + 5 = ?
We start at zero. Because the first term is (-2), we start by substracting 2:
Next, we add the second term (5):
This gives us our final answer:
(-2) + 5 = 3 | 102 | 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 | 4 | CC-MAIN-2023-23 | latest | en | 0.924034 |
https://nrich.maths.org/public/leg.php?code=-1&cl=3&cldcmpid=745 | 1,508,838,390,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828356.82/warc/CC-MAIN-20171024090757-20171024110757-00239.warc.gz | 778,854,462 | 9,470 | Search by Topic
Resources tagged with Patterned numbers similar to Why 24?:
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Broad Topics > Numbers and the Number System > Patterned numbers
Whole Number Dynamics I
Stage: 4 and 5
The first of five articles concentrating on whole number dynamics, ideas of general dynamical systems are introduced and seen in concrete cases.
Whole Number Dynamics II
Stage: 4 and 5
This article extends the discussions in "Whole number dynamics I". Continuing the proof that, for all starting points, the Happy Number sequence goes into a loop or homes in on a fixed point.
Whole Number Dynamics III
Stage: 4 and 5
In this third of five articles we prove that whatever whole number we start with for the Happy Number sequence we will always end up with some set of numbers being repeated over and over again.
Whole Number Dynamics V
Stage: 4 and 5
The final of five articles which containe the proof of why the sequence introduced in article IV either reaches the fixed point 0 or the sequence enters a repeating cycle of four values.
Magic Squares II
Stage: 4 and 5
An article which gives an account of some properties of magic squares.
Whole Number Dynamics IV
Stage: 4 and 5
Start with any whole number N, write N as a multiple of 10 plus a remainder R and produce a new whole number N'. Repeat. What happens?
A One in Seven Chance
Stage: 3 Challenge Level:
What is the remainder when 2^{164}is divided by 7?
Rolling Coins
Stage: 4 Challenge Level:
A blue coin rolls round two yellow coins which touch. The coins are the same size. How many revolutions does the blue coin make when it rolls all the way round the yellow coins? Investigate for a. . . .
Magic Squares
Stage: 4 and 5
An account of some magic squares and their properties and and how to construct them for yourself.
Stage: 3 Challenge Level:
A little bit of algebra explains this 'magic'. Ask a friend to pick 3 consecutive numbers and to tell you a multiple of 3. Then ask them to add the four numbers and multiply by 67, and to tell you. . . .
Always the Same
Stage: 3 Challenge Level:
Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?
On the Importance of Pedantry
Stage: 3, 4 and 5
A introduction to how patterns can be deceiving, and what is and is not a proof.
How Many Miles to Go?
Stage: 3 Challenge Level:
How many more miles must the car travel before the numbers on the milometer and the trip meter contain the same digits in the same order?
How Old Am I?
Stage: 4 Challenge Level:
In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays?
Pinned Squares
Stage: 3 Challenge Level:
The diagram shows a 5 by 5 geoboard with 25 pins set out in a square array. Squares are made by stretching rubber bands round specific pins. What is the total number of squares that can be made on a. . . .
Sum Equals Product
Stage: 3 Challenge Level:
The sum of the numbers 4 and 1 [1/3] is the same as the product of 4 and 1 [1/3]; that is to say 4 + 1 [1/3] = 4 × 1 [1/3]. What other numbers have the sum equal to the product and can this be so for. . . .
Odd Differences
Stage: 4 Challenge Level:
The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = nĀ² Use the diagram to show that any odd number is the difference of two squares.
Back to Basics
Stage: 4 Challenge Level:
Find b where 3723(base 10) = 123(base b).
Harmonic Triangle
Stage: 4 Challenge Level:
Can you see how to build a harmonic triangle? Can you work out the next two rows?
Hidden Rectangles
Stage: 3 Challenge Level:
Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard?
Squares, Squares and More Squares
Stage: 3 Challenge Level:
Can you dissect a square into: 4, 7, 10, 13... other squares? 6, 9, 12, 15... other squares? 8, 11, 14... other squares?
Icosagram
Stage: 3 Challenge Level:
Draw a pentagon with all the diagonals. This is called a pentagram. How many diagonals are there? How many diagonals are there in a hexagram, heptagram, ... Does any pattern occur when looking at. . . .
Tower of Hanoi
Stage: 3 Challenge Level:
The Tower of Hanoi is an ancient mathematical challenge. Working on the building blocks may help you to explain the patterns you notice.
Colour Building
Stage: 3 Challenge Level:
Using only the red and white rods, how many different ways are there to make up the other colours of rod?
Lastly - Well
Stage: 3 Challenge Level:
What are the last two digits of 2^(2^2003)?
Sept 03
Stage: 3 Challenge Level:
What is the last digit of the number 1 / 5^903 ?
Pattern Power
Stage: 1, 2 and 3
Mathematics is the study of patterns. Studying pattern is an opportunity to observe, hypothesise, experiment, discover and create.
Oranges and Lemons, Say the Bells of St Clement's
Stage: 3 Challenge Level:
Bellringers have a special way to write down the patterns they ring. Learn about these patterns and draw some of your own.
Four Coloured Lights
Stage: 3 Challenge Level:
Imagine a machine with four coloured lights which respond to different rules. Can you find the smallest possible number which will make all four colours light up?
When Will You Pay Me? Say the Bells of Old Bailey
Stage: 3 Challenge Level:
Use the interactivity to play two of the bells in a pattern. How do you know when it is your turn to ring, and how do you know which bell to ring?
Investigating Pascal's Triangle
Stage: 2 and 3 Challenge Level:
In this investigation, we look at Pascal's Triangle in a slightly different way - rotated and with the top line of ones taken off.
Pyramids
Stage: 3 Challenge Level:
What are the missing numbers in the pyramids?
Chameleons
Stage: 3 Challenge Level:
Whenever two chameleons of different colours meet they change colour to the third colour. Describe the shortest sequence of meetings in which all the chameleons change to green if you start with 12. . . .
Top-heavy Pyramids
Stage: 3 Challenge Level:
Use the numbers in the box below to make the base of a top-heavy pyramid whose top number is 200.
You Owe Me Five Farthings, Say the Bells of St Martin's
Stage: 3 Challenge Level:
Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?
Generating Number Patterns: an Email Conversation
Stage: 2, 3 and 4
This article for teachers describes the exchanges on an email talk list about ideas for an investigation which has the sum of the squares as its solution.
Stage: 2, 3, 4 and 5 Challenge Level:
Libby Jared helped to set up NRICH and this is one of her favourite problems. It's a problem suitable for a wide age range and best tackled practically.
Paving Paths
Stage: 3 Challenge Level:
How many different ways can I lay 10 paving slabs, each 2 foot by 1 foot, to make a path 2 foot wide and 10 foot long from my back door into my garden, without cutting any of the paving slabs?
Like Powers
Stage: 3 Challenge Level:
Investigate $1^n + 19^n + 20^n + 51^n + 57^n + 80^n + 82^n$ and $2^n + 12^n + 31^n + 40^n + 69^n + 71^n + 85^n$ for different values of n.
Stage: 4 Challenge Level:
A walk is made up of diagonal steps from left to right, starting at the origin and ending on the x-axis. How many paths are there for 4 steps, for 6 steps, for 8 steps?
Small Change
Stage: 3 Challenge Level:
In how many ways can a pound (value 100 pence) be changed into some combination of 1, 2, 5, 10, 20 and 50 pence coins?
Score
Stage: 3 Challenge Level:
There are exactly 3 ways to add 4 odd numbers to get 10. Find all the ways of adding 8 odd numbers to get 20. To be sure of getting all the solutions you will need to be systematic. What about. . . .
Power Crazy
Stage: 3 Challenge Level:
What can you say about the values of n that make $7^n + 3^n$ a multiple of 10? Are there other pairs of integers between 1 and 10 which have similar properties?
Counting Binary Ops
Stage: 4 Challenge Level:
How many ways can the terms in an ordered list be combined by repeating a single binary operation. Show that for 4 terms there are 5 cases and find the number of cases for 5 terms and 6 terms. | 2,134 | 8,469 | {"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-2017-43 | latest | en | 0.890408 |
https://edurev.in/p/339124/NCERT-Solutions-Exercise-13-2-Probability | 1,721,031,699,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514680.75/warc/CC-MAIN-20240715071424-20240715101424-00745.warc.gz | 190,944,970 | 62,079 | NCERT Solutions - Exercise 13.2: Probability
``` Page 1
NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
Exercise 13.2 Page No: 546
1. If P (A) = 3/5 and P (B) = 1/5, find P (A n B) if A and B are independent events.
Solution:
Given P (A) = 3/5 and P (B) = 1/5
As A and B are independent events.
? P (A n B) = P (A).P (B)
= 3/5 × 1/5 = 3/25
2. Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the
probability that both the cards are black.
Solution:
Given, a pack of 52 cards.
As we know, there are 26 cards in total, which are black. Let A and B denote, respectively, the events that the first and
second drawn cards are black.
Now, P (A) = P (black card in first draw) = 26/52 = ½
Because the second card is drawn without replacement, now the total number of black cards will be 25, and the total
number of cards will be 51, which is the conditional probability of B, given that A has already occurred.
Now, P (B/A) = P (black card in second draw) = 25/51
Thus, the probability that both cards are black
? P (A n B) = ½ × 25/51 = 25/102
Hence, the probability that both the cards are black = 25/102
3. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If
all three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box
containing 15 oranges, out of which 12 are good, and 3 are bad ones will be approved for sale.
Solution:
Given, a box of oranges.
Let A, B and C denotes the events, respectively, that the first, second and third drawn orange is good.
Now, P (A) = P (good orange in first draw) = 12/15
Because the second orange is drawn without replacement, now the total number of good oranges will be 11, and the
total number of oranges will be 14, which is the conditional probability of B, given that A has already occurred.
Page 2
NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
Exercise 13.2 Page No: 546
1. If P (A) = 3/5 and P (B) = 1/5, find P (A n B) if A and B are independent events.
Solution:
Given P (A) = 3/5 and P (B) = 1/5
As A and B are independent events.
? P (A n B) = P (A).P (B)
= 3/5 × 1/5 = 3/25
2. Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the
probability that both the cards are black.
Solution:
Given, a pack of 52 cards.
As we know, there are 26 cards in total, which are black. Let A and B denote, respectively, the events that the first and
second drawn cards are black.
Now, P (A) = P (black card in first draw) = 26/52 = ½
Because the second card is drawn without replacement, now the total number of black cards will be 25, and the total
number of cards will be 51, which is the conditional probability of B, given that A has already occurred.
Now, P (B/A) = P (black card in second draw) = 25/51
Thus, the probability that both cards are black
? P (A n B) = ½ × 25/51 = 25/102
Hence, the probability that both the cards are black = 25/102
3. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If
all three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box
containing 15 oranges, out of which 12 are good, and 3 are bad ones will be approved for sale.
Solution:
Given, a box of oranges.
Let A, B and C denotes the events, respectively, that the first, second and third drawn orange is good.
Now, P (A) = P (good orange in first draw) = 12/15
Because the second orange is drawn without replacement, now the total number of good oranges will be 11, and the
total number of oranges will be 14, which is the conditional probability of B, given that A has already occurred.
NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
Now, P (B/A) = P (good orange in second draw) = 11/14
Because the third orange is drawn without replacement, now the total number of good oranges will be 10, and the total
oranges will be 13, which is the conditional probability of C, given that A and B have already occurred.
Now, P (C/AB) = P (good orange in third draw) = 10/13
Thus, the probability that all the oranges are good
? P (A n B n C) = 12/15 × 11/14 × 10/13 = 44/91
Hence, the probability that a box will be approved for sale = 44/91
4. A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3
on the die’. Check whether A and B are independent events or not.
Solution:
Given, a fair coin and an unbiased die are tossed.
We know that the sample space S.
S = {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
Let A be the event head that appears on the coin.
? A = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}
? P (A) = 6/12 = ½
Now, Let B be the event 3 on the die.
? B = {(H, 3), (T, 3)}
? P (B) = 2/12 = 1/6
As, A n B = {(H, 3)}
? P (A n B) = 1/12 …… (1)
And P (A). P (B) = ½ × 1/6 = 1/12 …… (2)
From (1) and (2) P (A n B) = P (A). P (B)
Therefore, A and B are independent events.
5. A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even’, and B be the
event, ‘the number is red’. Are A and B independent?
Solution:
The sample space for the dice will be
Page 3
NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
Exercise 13.2 Page No: 546
1. If P (A) = 3/5 and P (B) = 1/5, find P (A n B) if A and B are independent events.
Solution:
Given P (A) = 3/5 and P (B) = 1/5
As A and B are independent events.
? P (A n B) = P (A).P (B)
= 3/5 × 1/5 = 3/25
2. Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the
probability that both the cards are black.
Solution:
Given, a pack of 52 cards.
As we know, there are 26 cards in total, which are black. Let A and B denote, respectively, the events that the first and
second drawn cards are black.
Now, P (A) = P (black card in first draw) = 26/52 = ½
Because the second card is drawn without replacement, now the total number of black cards will be 25, and the total
number of cards will be 51, which is the conditional probability of B, given that A has already occurred.
Now, P (B/A) = P (black card in second draw) = 25/51
Thus, the probability that both cards are black
? P (A n B) = ½ × 25/51 = 25/102
Hence, the probability that both the cards are black = 25/102
3. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If
all three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box
containing 15 oranges, out of which 12 are good, and 3 are bad ones will be approved for sale.
Solution:
Given, a box of oranges.
Let A, B and C denotes the events, respectively, that the first, second and third drawn orange is good.
Now, P (A) = P (good orange in first draw) = 12/15
Because the second orange is drawn without replacement, now the total number of good oranges will be 11, and the
total number of oranges will be 14, which is the conditional probability of B, given that A has already occurred.
NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
Now, P (B/A) = P (good orange in second draw) = 11/14
Because the third orange is drawn without replacement, now the total number of good oranges will be 10, and the total
oranges will be 13, which is the conditional probability of C, given that A and B have already occurred.
Now, P (C/AB) = P (good orange in third draw) = 10/13
Thus, the probability that all the oranges are good
? P (A n B n C) = 12/15 × 11/14 × 10/13 = 44/91
Hence, the probability that a box will be approved for sale = 44/91
4. A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3
on the die’. Check whether A and B are independent events or not.
Solution:
Given, a fair coin and an unbiased die are tossed.
We know that the sample space S.
S = {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
Let A be the event head that appears on the coin.
? A = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}
? P (A) = 6/12 = ½
Now, Let B be the event 3 on the die.
? B = {(H, 3), (T, 3)}
? P (B) = 2/12 = 1/6
As, A n B = {(H, 3)}
? P (A n B) = 1/12 …… (1)
And P (A). P (B) = ½ × 1/6 = 1/12 …… (2)
From (1) and (2) P (A n B) = P (A). P (B)
Therefore, A and B are independent events.
5. A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even’, and B be the
event, ‘the number is red’. Are A and B independent?
Solution:
The sample space for the dice will be
NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
S = {1, 2, 3, 4, 5, 6}
Let A be the event, and the number is even.
? A = {2, 4, 6}
? P (A) = 3/6 = ½
Now, let B be the event, and the number is red.
? B = {1, 2, 3}
? P (B) = 3/6 = 1/2
As, A n B = {2}
? P (A n B) = 1/6 …….. (1)
And P (A). P (B) = ½ × ½ = ¼ ….. (2)
From (1) and (2) P (A n B) ? P (A). P (B)
Therefore, A and B are not independent events.
6. Let E and F be events with P (E) = 3/5, P (F) = 3/10 and P (E n F) = 1/5. Are E and F independent?
Solution:
Given P (E) = 3/5, P (F) = 3/10 and P (E n F) = 1/5
P (E). P (F) = 3/5 × 3/10 = 9/50 ? 1/5
? P (E n F) ? P (E). P (F)
Therefore, E and F are not independent events.
7. Given that the events A and B are such that P (A) = 1/2, P (A ? B) = 3/5 and P (B) = p. Find p if they are (i)
mutually exclusive (ii) independent.
Solution:
Given, P (A) = ½, P (A ? B) = 1/5 and P (B) = p
(i) Mutually exclusive
When A and B are mutually exclusive.
Then, (A n B) = ?
? P (A n B) = 0
As we know, P (A ? B) = P (A) + P (B) – P (A n B)
Page 4
NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
Exercise 13.2 Page No: 546
1. If P (A) = 3/5 and P (B) = 1/5, find P (A n B) if A and B are independent events.
Solution:
Given P (A) = 3/5 and P (B) = 1/5
As A and B are independent events.
? P (A n B) = P (A).P (B)
= 3/5 × 1/5 = 3/25
2. Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the
probability that both the cards are black.
Solution:
Given, a pack of 52 cards.
As we know, there are 26 cards in total, which are black. Let A and B denote, respectively, the events that the first and
second drawn cards are black.
Now, P (A) = P (black card in first draw) = 26/52 = ½
Because the second card is drawn without replacement, now the total number of black cards will be 25, and the total
number of cards will be 51, which is the conditional probability of B, given that A has already occurred.
Now, P (B/A) = P (black card in second draw) = 25/51
Thus, the probability that both cards are black
? P (A n B) = ½ × 25/51 = 25/102
Hence, the probability that both the cards are black = 25/102
3. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If
all three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box
containing 15 oranges, out of which 12 are good, and 3 are bad ones will be approved for sale.
Solution:
Given, a box of oranges.
Let A, B and C denotes the events, respectively, that the first, second and third drawn orange is good.
Now, P (A) = P (good orange in first draw) = 12/15
Because the second orange is drawn without replacement, now the total number of good oranges will be 11, and the
total number of oranges will be 14, which is the conditional probability of B, given that A has already occurred.
NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
Now, P (B/A) = P (good orange in second draw) = 11/14
Because the third orange is drawn without replacement, now the total number of good oranges will be 10, and the total
oranges will be 13, which is the conditional probability of C, given that A and B have already occurred.
Now, P (C/AB) = P (good orange in third draw) = 10/13
Thus, the probability that all the oranges are good
? P (A n B n C) = 12/15 × 11/14 × 10/13 = 44/91
Hence, the probability that a box will be approved for sale = 44/91
4. A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3
on the die’. Check whether A and B are independent events or not.
Solution:
Given, a fair coin and an unbiased die are tossed.
We know that the sample space S.
S = {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
Let A be the event head that appears on the coin.
? A = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}
? P (A) = 6/12 = ½
Now, Let B be the event 3 on the die.
? B = {(H, 3), (T, 3)}
? P (B) = 2/12 = 1/6
As, A n B = {(H, 3)}
? P (A n B) = 1/12 …… (1)
And P (A). P (B) = ½ × 1/6 = 1/12 …… (2)
From (1) and (2) P (A n B) = P (A). P (B)
Therefore, A and B are independent events.
5. A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even’, and B be the
event, ‘the number is red’. Are A and B independent?
Solution:
The sample space for the dice will be
NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
S = {1, 2, 3, 4, 5, 6}
Let A be the event, and the number is even.
? A = {2, 4, 6}
? P (A) = 3/6 = ½
Now, let B be the event, and the number is red.
? B = {1, 2, 3}
? P (B) = 3/6 = 1/2
As, A n B = {2}
? P (A n B) = 1/6 …….. (1)
And P (A). P (B) = ½ × ½ = ¼ ….. (2)
From (1) and (2) P (A n B) ? P (A). P (B)
Therefore, A and B are not independent events.
6. Let E and F be events with P (E) = 3/5, P (F) = 3/10 and P (E n F) = 1/5. Are E and F independent?
Solution:
Given P (E) = 3/5, P (F) = 3/10 and P (E n F) = 1/5
P (E). P (F) = 3/5 × 3/10 = 9/50 ? 1/5
? P (E n F) ? P (E). P (F)
Therefore, E and F are not independent events.
7. Given that the events A and B are such that P (A) = 1/2, P (A ? B) = 3/5 and P (B) = p. Find p if they are (i)
mutually exclusive (ii) independent.
Solution:
Given, P (A) = ½, P (A ? B) = 1/5 and P (B) = p
(i) Mutually exclusive
When A and B are mutually exclusive.
Then, (A n B) = ?
? P (A n B) = 0
As we know, P (A ? B) = P (A) + P (B) – P (A n B)
NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
? 3/5 = ½ + p -0
? P = 3/5 – ½ = 1/10
(ii) Independent
When A and B are independent.
? P (A n B) = P (A). P (B)
? P (A n B) = ½ p
As we know, P (A ? B) = P (A) + P (B) – P (A n B)
? 3/5 = ½ + 2 – p/2
? p/2 = 3/5 – ½
? p = 2 × 1/10 = 1/5
8. Let A and B be independent events with P (A) = 0.3 and P (B) = 0.4. Find
(i) P (A n B)
(ii) P (A ? B)
(iii) P (A|B)
(iv) P (B|A)
Solution:
Given, P (A) = 0.3 and P(B) = 0.4
(i) P (A n B)
When A and B are independent.
? P (A n B) = P (A). P (B)
? P (A n B) = 0.3 × 0.4
? P (A n B) = 0.12
(ii) P (A ? B)
As we know, P (A ? B) = P (A) + P (B) – P (A n B)
? P (A ? B) = 0.3 + 0.4 – 0.12
? P (A ? B) = 0.58
Page 5
NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
Exercise 13.2 Page No: 546
1. If P (A) = 3/5 and P (B) = 1/5, find P (A n B) if A and B are independent events.
Solution:
Given P (A) = 3/5 and P (B) = 1/5
As A and B are independent events.
? P (A n B) = P (A).P (B)
= 3/5 × 1/5 = 3/25
2. Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the
probability that both the cards are black.
Solution:
Given, a pack of 52 cards.
As we know, there are 26 cards in total, which are black. Let A and B denote, respectively, the events that the first and
second drawn cards are black.
Now, P (A) = P (black card in first draw) = 26/52 = ½
Because the second card is drawn without replacement, now the total number of black cards will be 25, and the total
number of cards will be 51, which is the conditional probability of B, given that A has already occurred.
Now, P (B/A) = P (black card in second draw) = 25/51
Thus, the probability that both cards are black
? P (A n B) = ½ × 25/51 = 25/102
Hence, the probability that both the cards are black = 25/102
3. A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If
all three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box
containing 15 oranges, out of which 12 are good, and 3 are bad ones will be approved for sale.
Solution:
Given, a box of oranges.
Let A, B and C denotes the events, respectively, that the first, second and third drawn orange is good.
Now, P (A) = P (good orange in first draw) = 12/15
Because the second orange is drawn without replacement, now the total number of good oranges will be 11, and the
total number of oranges will be 14, which is the conditional probability of B, given that A has already occurred.
NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
Now, P (B/A) = P (good orange in second draw) = 11/14
Because the third orange is drawn without replacement, now the total number of good oranges will be 10, and the total
oranges will be 13, which is the conditional probability of C, given that A and B have already occurred.
Now, P (C/AB) = P (good orange in third draw) = 10/13
Thus, the probability that all the oranges are good
? P (A n B n C) = 12/15 × 11/14 × 10/13 = 44/91
Hence, the probability that a box will be approved for sale = 44/91
4. A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3
on the die’. Check whether A and B are independent events or not.
Solution:
Given, a fair coin and an unbiased die are tossed.
We know that the sample space S.
S = {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
Let A be the event head that appears on the coin.
? A = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}
? P (A) = 6/12 = ½
Now, Let B be the event 3 on the die.
? B = {(H, 3), (T, 3)}
? P (B) = 2/12 = 1/6
As, A n B = {(H, 3)}
? P (A n B) = 1/12 …… (1)
And P (A). P (B) = ½ × 1/6 = 1/12 …… (2)
From (1) and (2) P (A n B) = P (A). P (B)
Therefore, A and B are independent events.
5. A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number is even’, and B be the
event, ‘the number is red’. Are A and B independent?
Solution:
The sample space for the dice will be
NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
S = {1, 2, 3, 4, 5, 6}
Let A be the event, and the number is even.
? A = {2, 4, 6}
? P (A) = 3/6 = ½
Now, let B be the event, and the number is red.
? B = {1, 2, 3}
? P (B) = 3/6 = 1/2
As, A n B = {2}
? P (A n B) = 1/6 …….. (1)
And P (A). P (B) = ½ × ½ = ¼ ….. (2)
From (1) and (2) P (A n B) ? P (A). P (B)
Therefore, A and B are not independent events.
6. Let E and F be events with P (E) = 3/5, P (F) = 3/10 and P (E n F) = 1/5. Are E and F independent?
Solution:
Given P (E) = 3/5, P (F) = 3/10 and P (E n F) = 1/5
P (E). P (F) = 3/5 × 3/10 = 9/50 ? 1/5
? P (E n F) ? P (E). P (F)
Therefore, E and F are not independent events.
7. Given that the events A and B are such that P (A) = 1/2, P (A ? B) = 3/5 and P (B) = p. Find p if they are (i)
mutually exclusive (ii) independent.
Solution:
Given, P (A) = ½, P (A ? B) = 1/5 and P (B) = p
(i) Mutually exclusive
When A and B are mutually exclusive.
Then, (A n B) = ?
? P (A n B) = 0
As we know, P (A ? B) = P (A) + P (B) – P (A n B)
NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
? 3/5 = ½ + p -0
? P = 3/5 – ½ = 1/10
(ii) Independent
When A and B are independent.
? P (A n B) = P (A). P (B)
? P (A n B) = ½ p
As we know, P (A ? B) = P (A) + P (B) – P (A n B)
? 3/5 = ½ + 2 – p/2
? p/2 = 3/5 – ½
? p = 2 × 1/10 = 1/5
8. Let A and B be independent events with P (A) = 0.3 and P (B) = 0.4. Find
(i) P (A n B)
(ii) P (A ? B)
(iii) P (A|B)
(iv) P (B|A)
Solution:
Given, P (A) = 0.3 and P(B) = 0.4
(i) P (A n B)
When A and B are independent.
? P (A n B) = P (A). P (B)
? P (A n B) = 0.3 × 0.4
? P (A n B) = 0.12
(ii) P (A ? B)
As we know, P (A ? B) = P (A) + P (B) – P (A n B)
? P (A ? B) = 0.3 + 0.4 – 0.12
? P (A ? B) = 0.58
NCERT Solutions for Class 12 Maths Chapter 13 –
Probability
(iv) P (B|A)
9. If A and B are two events such that P (A) = 1/4 , P (B) = 1/2 and P (A n B) = 1/8, find P (not A and not B).
Solution:
Given P (A) = ¼, P (B) = ½ and P (A n B) = 1/8
P (not A and not B) = P (A
’
n B
’
)
As, { A
’
n B
’
= (A ? B)
’
}
? P (not A and not B) = P ((A ? B)
’
)
= 1 – P (A ? B)
= 1- [P (A) + P (B) – P (A n B)]
10. Events A and B are such that P (A) = 1/2, P (B) = 7/12 and P (not A or not B) = 1/4. State whether A and B
are independent.
Solution:
Given P (A) = ½, P (B) =7/12 and P (not A or not B) = 1/4
```
## Mathematics (Maths) for JEE Main & Advanced
209 videos|443 docs|143 tests
## Mathematics (Maths) for JEE Main & Advanced
209 videos|443 docs|143 tests
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380CT: THEORETICAL ASPECTS OF COMPUTER SCIENCE- COURSEWORK 2
UNDERSTANDING AND OVERCOMING INTRACTABILITY IN ALGORITHMS DESIGN
The partition problem is the task of deciding whether a given set S of positive integers can be partitioned
into two subsets S1 and S2 such that both have the same sum. Although the partition problem is NP-
complete, there is a polynomial time dynamic programming solution, and there are heuristics that solve
the problem either optimally or approximately. For this reason, it has been called “The Easiest Hard
Problem”. Source is Wikipedia. For the partition problem that you need to address in this coursework
the conditions for a 2Part = {Sub1, Sub2
1- Sub1 ∪ Sub2 = S
2- Sub1 ∩ Sub2 = ∅
3- ∑∀a ai i∈ Sub1 = ∑∀b bi i∈Sub2
Example: S = {2,3,4,6,5,10}, 2Part = {{2,3,4,6},{5,10}} or 2Part = {{4,5,6},{2,3,10}}
1- {2,3,4,6} ∪ {5,10} = {2,3,4,6,5,10} = S
2- {2,3,4,6} ∩ {5,10} = {} = ∅
3- 2 + 3 + 4 + 6 = 5 + 10
So the target in this case is specified by one of the subsets sums t = {2 + 3 + 4 + 6} and the other can
be obtained as Sub2=S-Sub1
You are required to write a report about algorithms suitable for tackling the partition problem, defined
above, and investigate their computational complexities in practice by implementing them. You are
provided with the basic C++ classes to help you and you can use any code you prefer.
You report should cover the following:
1. Outline in pseudo-code an Exhaustive Search (Brute Force) solution for the problem (5 marks)
a. Give its time complexity using O-notation
b. Plot the average running time for 10 randomly generated sets S: |S|= 10, 20, 30 and 40
c. Discuss the algorithm complexity in light of your results in b
2. Outline in pseudo-code a Dynamic Programming solution for the problem (5 marks)
a. Give its time complexity using O-notation
b. Plot the average running time for 10 randomly generated sets S: |S|= 10, 20, 30 and 40
c. Discuss the algorithm complexity in light of your results in b
3. Outline in pseudo-code Greedy or Random Sampling approaches to solve this problem(5 marks)
a. Give its time complexity using O-notation
b. Plot the average running time for 10 randomly generated sets S: |S|= 10, 20, 30 and 40
c. Discuss the algorithm complexity in light of your results in b
4. Outline in pseudo-code for Simulated Annealing or Genetic Algorithm for this problem(5 marks)
a. Give its time complexity using O-notation
b. Plot the average running time for 10 randomly generated sets S: |S|= 10, 20, 30 and 40
c. Discuss the algorithm complexity in light of your results in b
5. Write a conclusion with recommendations on when each method is most suitable
a. Reflect on what you have learnt? (3 marks)
b. What could you have done differently? (2 marks)
6. Present your work clearly, using graphs and referencing wherever appropriate (5 marks)
– Please only include pictures for plots of the performance of your! Algorithms
– Please do not include your pseudo-code or discussions as an image
– Please do not copy other’s work as it results in failing the coursework
– Answers that violate these instructions get 0 marks
1 | P a g e . D r . A b d u l r a h m a n A l t a h h a n 3 8 0 C T C o u r s e W o r k 2
} where Sub1 ⊆ S and Sub2 ⊆ S can be expressed as:
{
From the definition of a partition
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# Trigonometry and Sine Study Guide
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Updated on Oct 1, 2011
## Trigonometry and Sine
In this lesson, we introduce the sine function. We show how to evaluate the sine of an angle in a right triangle. Often, this is as easy as dividing the length of one side of the triangle by another. Sometimes we need to use the Pythagorean theorem to find one of the lengths.
The first trigonometric function is sine. The domain of this function consists of all angles x. There is no formula for sine that can be computed directly using x. Instead, we must follow a process. For now, we will use angles between 0° and 90°.
Step two: Make a right triangle with angle x (Figure 5.2).
Step three: Measure the length H of the triangle's hypotenuse and the length O of the side opposite the angle x (Figure 5.3). (The opposite side is one that isn't part of angle x.)
Step four: The sine of the angle x is the ratio of the two sides:
sin(x) =
The length of the hypotenuse H and opposite side O will depend on the size of the triangle. In step two, we could have chosen a smaller or larger right triangle with angle x, as illustrated in Figure 5.4.
However, because the angles of a triangle add up to 180° (or π radians), the third angles of these triangles must all measure 90° – x degrees (or x radians). Triangles with all the same angles are similar; thus, each one is a scale multiple of the original. Suppose the smaller triangle has scale factor k, and the larger triangle has scale factor K (see Figure 5.5).
If we use the small triangle, then
sin(x) = =
If we use the large triangle, then
sin(x) = =
In other words, the sine of the angle does not depend on the size of the triangle used.
It is tedious to actually draw a right triangle with the correct angle x, and then to measure the lengths of the sides. There are only a few "nice angles" for which this can be done precisely, which will be discussed in Lesson 9. Calculators can be used to estimate the sine of any angle; this will be covered in Lesson 12. For the time being, it is far easier to start with a triangle and then find the sine of each of its angles.
#### Example 1
Find the sine of the angle x in Figure 5.6.
We might not know the exact measurement of angle x, in either degrees or radians, but we have everything we need to find its sine. The hypotenuse of this right triangle is H = 10 inches. The side that is opposite angle x is O = 6 inches. Thus,
sin(x) = =
Notice that the units used to measure the lengths will always cancel out like this. From now on, we will not worry about the feet, inches, or whatever is used to measure the lengths. The numbers that come out of the sine function have no particular units. They are merely ratios.
#### Example 2
Find the sine of angle θ from Figure 5.7.
Here, the hypotenuse is H = 14 and the side opposite angle θ is O = 9, so
sin(θ) = =
#### Example 3
Find the sine of the angle x illustrated in Figure 5.8.
Here, the hypotenuse is H = √34. The opposite side (the one that doesn't touch angle x) is O = 5. Thus,
sin(x) =
We can find the sine of an angle even if only two sides of the triangle are known. If the two sides are the hypotenuse and opposite side, then we plug them into sin(x) = directly. Otherwise, we will need to use the Pythagorean theorem to find the length of the missing side.
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5-38.
Use slope to determine whether the points $A(3,5)$, $B(−2,6)$, and $C(−5,7)$ are on the same line. Justify your conclusion algebraically.
Find the slope between points $A$ and $B$.
$\frac{6-5}{-2-3}=-\frac{1}{5}$
Now find the slope between points $B$ and $C$.
$\frac{7-6}{-5+2}=-\frac{1}{3}$ | 149 | 373 | {"found_math": true, "script_math_tex": 9, "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} | 4.15625 | 4 | CC-MAIN-2024-38 | latest | en | 0.659614 |
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Work backwards. 12*9=108
check: 108/12=9
2015-01-22T16:52:41-05:00
### This Is a Certified Answer
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Well I don't know. Let's figure it out together.
You said (a number) divided by 12 gives you 9 .
A fraction is the easiest way to show division, so
you can write this equation:
(number) / 12 = 9
Now you can multiply each side of the equation by 12 .
When you do that, you have ...
number = 9 x 12
number = 108 .
Now you know how to find it. That's even better than just | 224 | 792 | {"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} | 3.96875 | 4 | CC-MAIN-2017-04 | latest | en | 0.919037 |
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# Is x > y^2 ? 1) x > y+5 2) x^2 - y^2 = 0
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Is x > y^2 ? 1) x > y+5 2) x^2 - y^2 = 0 [#permalink]
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Is x > y^2 ?
1) x > y+5
2) x^2 - y^2 = 0
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12 Sep 2008, 22:20
jjhko wrote:
Is x > y^2 ?
1) x > y+5
2) x^2 - y^2 = 0
1)
x=7 y=1
x > y^2 --> 7>1 true
x=-2 y=-10
x>y^2 --> -2>100 false
insuffcient
2)
x=+-y
x=2 y=2
2>4 false
x=1/2 y=1/2
1/2>1/4 true
insufficient
combined.
x > y+5
and x=+-y
so x=-y y is negative and x is positive.
when x=2 y=-2
2>4 false
when x=1/2 y=-1/2
1/2>1/4 true
insufficient
E
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13 Sep 2008, 00:16
1
KUDOS
jjhko wrote:
Is x > y^2 ?
1) x > y+5
2) x^2 - y^2 = 0
C
from statement 2:
|x| = |y|
from statement 1:
x > y+5
together, x must be positive and y must be negative, and y must be less than -2.5, so absolute value of y must be greater than 2.5, therefore x must be less than y^2
Re: inequality DS [#permalink] 13 Sep 2008, 00:16
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BRINGING OUT THE DEAD: Read the Book, Ignore the Movie
The metaphysics of Joe Connelly's novel elude Martin Scorsese's grasp.
By Denise Dumars December 07, 1999
Martin Scorsese's treatment of this first novel by an ex-paramedic falls somewhat short. But it did succeed very well in one area: it made me really want to read the novel by Joe Connelly. And what a glorious novel it is.
In the film, Nicolas Cage gives one of his trademark man-on-the-edge-of-a-nervous-breakdown performances as Frank, a paramedic haunted by the spectre of Rose, a young woman who died of an asthma attack as he tried to resuscitate her. Rose is there as a reminder that these paramedics really are acting as almost supernatural forces themselves--they routinely expect to bring people back from the dead. When it doesn't work, it doesn't bother someone like Frank's psycho partner Tom Walls (Tom Sizemore). But Frank it bothers. A lot. How he handles being haunted forms somewhat divergent paths for the novel and the movie.
Perhaps such heavy metaphysical concepts are just too difficult to show on film, as Scorcese's filmization (adapted by his TAXI DRIVER collaborator, Paul Schrader) never reaches the depth of character or thematic intensity that it aims for. The black humor falls flat at times. Are we really supposed to laugh, or would it be more appropriate to cry? How dark did Scorsese want this film to be interpreted? And the pacing of the film is off; several times there seemed to be enough closure to end the film, but then the medics would get back in their battered ambulance and charge off to mop up after another exhibition of man's cruelty to man.
Perhaps there was too much material in the novel to deal with in a movie; indeed, after reading the book, the film seems like a mere outline of it, even though nearly all of the dialogue is lifted directly from the text. Perhaps Scorsese and Schrader felt they had to shield viewers from some of the really ugly stuff in the book--although there's plenty that's ugly and shocking in the movie--in order for the gallows humor to work. But it takes real depth in filmmaking to do that, and that depth just isn't in the movie.
It is, however, in the novel. BRINGING OUT THE DEAD begins in a typical first-novel way: the first few pages are overwritten. Soon, however, the reader falls so completely under the spell of the story that the authorial voice no longer sounds like its trying too hard; we flow with the story, rather than against the verbiage.
The novel explains some of the things that the movie left unclear. Missing altogether from the film, but for one line or so, is Frank's ex-wife. In the novel, we are given a series of richly described flashbacks outlining Frank's romancing, marrying, and finally being dumped by his wife. The way his job erodes his personality, and the way he refuses to be helped by his wife when she tries so hard to do so, makes her decision to leave him inevitable. His attempted reconciliation with her only leads him to realize that she has moved on and is seeing someone else. Watching the relationship dissolve in stages in the novel made it all the more poignant and painful, unlike the throwaway experience left unexplicated in the film.
The metaphysical aspects of both the film and the movie are perplexing. It is purposely left up to the reader's interpretation to decide whether Frank is literally being haunted or is a victim of his own guilt and psychological trauma. Certainly the novel gives up more frequent episodes of the haunting, but then it also makes clear just how thin the boundary is between life and death. The very opening of the novel tells us that Frank believes he can tell when the soul leaves the body, quite literally. Therefore, to discount the metaphysical as merely imaginary would be to diminish the importance of this aspect of Frank's personality and career. It also reminds us of the juncture of religion and medicine, the forgotten role of priest/physician from ancient times brought forward and dumped out onto the meaner streets of Scorsese's, and Connelly's, New York.
Some aspects of the novel are confusing, as well. In the film, it was quite clear that the father of Mary Burke (Patricia Arquette, in an unconvincing role) didn't want to go on living after being resuscitated by Frank. Mary's father never regains consciousness, but his spirit communicats to Frank that he wants to let go of life, and in one of the best scenes of the film Frank allows him to do so. The novel states the case much more simply but, unlike the film, fails to convince us that Mr. Burke is really communicating with Frank.
A few other details were changed in the film. In the novel, Marcus (Ving Rhames) is a far more damaged person--both physically and emotionally--than in the film. Indeed, Marcus is seen as a pioneer of the paramedic-as-Superman character, long since burned out. His heroics in the novel become even more amazing once we see the broken man he has become. And in the book, Frank does not try nearly as hard to keep Tom from killing Noel, the crazed homeless man. There are other extremely nihilistic and gruesome scenes that were toned down in the film.
In novel form, BRINGING OUT THE DEAD is a truly compelling story that earns my highest recommendation for bringing up issues of life and death that we are often uncomfortable contemplating. The film, while it has some fine performances, never quite descends into the depths that are necessary to, paradoxically, raise it up to the level of great cinema.
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A285382 Sum of entries in the last cycles of all permutations of [n]. 5
1, 5, 25, 143, 942, 7074, 59832, 563688, 5858640, 66622320, 823055040, 10979133120, 157300375680, 2409321801600, 39290164300800, 679701862425600, 12433400027596800, 239791474805299200, 4863054420016128000, 103462238924835840000, 2304147629440419840000 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Each cycle is written with the smallest element first and cycles are arranged in increasing order of their first elements. LINKS Alois P. Heinz, Table of n, a(n) for n = 1..449 Wikipedia, Permutation FORMULA Recursion: see Maple program. EXAMPLE a(3) = 25 because the sum of the entries in the last cycles of all permutations of [3] ((123), (132), (12)(3), (13)(2), (1)(23), (1)(2)(3)) is 6+6+3+2+5+3 = 25. MAPLE a:= proc(n) option remember; `if`(n<3, n*(3*n-1)/2, ((2*n^2+3*n-1)*a(n-1)-(n+2)*(n-1)*n*a(n-2))/(n+1)) end: seq(a(n), n=1..25); MATHEMATICA Table[n! * (n-1 + 2*(n+1)*HarmonicNumber[n])/4, {n, 1, 25}] (* Vaclav Kotesovec, Apr 29 2017 *) CROSSREFS Cf. A284816, A285424, A285439. Column k=1 of A286231. Sequence in context: A122441 A114870 A222676 * A199319 A049427 A121639 Adjacent sequences: A285379 A285380 A285381 * A285383 A285384 A285385 KEYWORD nonn AUTHOR Alois P. Heinz, Apr 20 2017 STATUS approved
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- - Tricky bathroom remodel - any suggestions? (
mannewskie 02-13-2013 11:30 PM
Tricky bathroom remodel - any suggestions?
We live in an older row house in an older neighborhood in a large city. Our guest bathroom is quite small - it has a shower, sink and toilet, but it's all pretty tight. The current sink is a very nice bronze bowl, but it's only 9 inches across, so it's really too small to be used for simple things like washing your face. The sink is actually sitting in a small vanity, but it's an ugly old formica model. That vanity is also pushed up hard against the side of the shower and actually overlaps with the edge of the shower by a couple of inches.
Here are some rough dimensions for what I'm talking about:
* The shower (on the left of the room) is 42.5 inches deep by 28 inches wide
* The sink vanity (in the middle of the room) is 16.5 inches deep by 15 inches wide by 31.5 inches high and overlaps into the shower by about 2 inches at the top edge of the shower
* The sink vanity is 2.5 inches from the edge of the toilet tank on the same wall as the toilet
* The toilet tank (on the right of the room) is 28 inches high, 19 inches wide, and 3 inches from the exterior wall (to the right).
* T
he center of the toilet is 12.5 inches from the exterior wall
(Sorry if that doesn't help much, but my diagram of the room doesn't translate well on the thread.)
We aren't looking to change the size of the shower, the placement of the toilet or anything else major, but we would like to get a bigger sink. We also want to get rid of the ugly formica vanity, so we're considering a wall-mounted sink. Something like or both of which are around 16 inches wide and deep. In the process, we would also like to move the sink a few inches away from the shower to free up that space in the shower stall and to make the sink more centered in the space. To do so, we would envision raising the sink from it's current level to around 36 inches and moving it a bit to the right.
The problem is that if we move the sink and get something that would be a reasonable size, it would seem to cause code problems for the distance of the sink from the toilet. By shifting our sink up and to the right a few inches, we would essentially be moving it almost above the edge of the toilet tank.
Here are my questions:
(1) Am I correct that this alternative arrangement would not be up to code?
(2) Does it matter that our existing bathroom is not up to code and that it would be almost impossible to do any renovations in that space that were up to code?
(3) Should we look at a smaller sink? Would it be feasible to get any sink in there that would be functional and would also satisfy code requirements? (If so, any suggestions?)
(4) Should we also consider replacing the toilet with a model that has a smaller tank, and would that cause any code issues?
(5) Ultimately, we aren't looking to do a whole lot - just replacing and moving the sink a bit - so should we just go ahead and do the work and not worry about potential code violations, especially given that our old house is full of quirky code violations? What would be the worst that could happen if we were to do so? Is this something that could be an issue if we were to sell our house in the future, given that prospective buyers know that they're looking at a quirky old house?
Any advice that folks can offer would be greatly appreciated.
joecaption 02-14-2013 12:07 AM
Can you post a picture?
mannewskie 02-14-2013 12:49 AM
3 Attachment(s)
Here's an attempt at some pictures, but remember, it's a really small bathroom. Also, you can see that the sink is really small and, given the placement of the mirror, how it's a bit off-center given the rest of the room. Thanks!
joecaption 02-14-2013 01:12 AM
Think about it, how many times has that shower been used?
Get rid of that and you would have room for a niec 1/2 bath with far better looking fixtures.
oh'mike 02-14-2013 06:18 AM
What Are The 'code' Space Requirements For A Toilet In A Three Walled Area?? - General Discussion - DIY Chatroom - DIY Home Improvement Forum
This will help----Is there any way to expand the room a bit?
TarheelTerp 02-14-2013 09:21 AM
1 Attachment(s)
Originally Posted by mannewskie (Post 1116824)
Our guest bathroom is quite small
Any advice that folks can offer would be greatly appreciated.
OK... It's a *guest* bathroom. Leave well enough alone.
I have a similar sized sink bowl in my RV.
The commode is right up against it.
It all works well enough.
The small shower was my complaint so I went and made the
WHOLE floor of the room into one big shower pan.
That's a compromise that might fit into your available space...
and give you more wall space for a larger lavatory.
AppealingSpaces 02-15-2013 12:41 AM
I see your dilemmas with your spacing and code. The proportions of everything is off in that the walk in shower consumes too much space, the vanity is large and boxy. The toilet seems to be okay. If I was in your situation, I would replace the vanity with a pedestal sink no more than 24" wide. The shower I would honestly shrink the width from the wall to the vanity and replace it with larger sized tile maybe 6x6 and some liners for borders. This will make the space seem larger. Also add a nickel or chrome oval mirror above the pedestal sink. You want to incorporate rounded edges to break up the square aspect of the space. :)
operagost 04-03-2013 05:14 PM
The sink and faucet are really cool. I would save those for a 1/2 bath project, or donate to the HFH reStore-- then smash the hideous laminate vanity to pieces. :-P I like the pedestal sink idea-- it will make it feel roomier in there, besides fitting better. New toilets are also available with very skinny tanks. Your toilet may still be too close to the wall to meet code, but you might be grandfathered in or simply exempt.
wkearney99 04-03-2013 07:26 PM
Man, that faucet is a hoot. Looks for all the world like a bar sink setup instead of a normal bathroom setup.
A wall-mount or pedestal sink is probably your best bet, just be sure to get one that has enough working room on the top for things. Nothing worse than trying to use a sink and having nowhere to put your stuff. Same thing goes for a toilet with a narrower (but possibly taller) tank. That'll give you a more open feeling between it and the new sink. Just don't forget to plan for where to keep the spare roll(s) of toilet paper.
Code is important and well worth following but unless you pull a permit nobody's going to come checking. Nor is anyone going to be breaking out the measuring tape when it comes time to sell. You didn't, right?
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# Twenty-five adults reported the amount of time each spent st
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08 May 2013, 07:02
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Refer to the attached file.
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Re: Twenty-five adults reported the amount of time each spent st [#permalink]
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16 May 2013, 20:37
3
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1. To find the least possible mean the sum of all hours exercised must be the minimum.
For 0-1 hours lets assume all 5 exercised the minimum possible of just over 0 hours and same for all bars.
so SUM of hours exercised = 0*5 + 1*3+2*2+3*3+4*4+5*5+0+0+8*1+0+10*0 = 78
# respondents = 25
average = 78/25 = 3.12
2. # respondents who exercised on average <0.5 hours per day
If Avg hours exercised = 0.5 hours per day the Total hours exercised during the week will be 0.5*7 OR 3.5 hours/week
We need to count respondents who exercise below 3.5 hours per week.
That will include all in 0-1 hour, 1-2 hr, 2-3 hour and since 3.5 lies in between 3-4 band, we can take 2 cases:
CAse 1: where we assume all 4 in 3-4 years exercise MORE than 3.5 hours
Total # respondednts in that case = 5 + 3 + 2 + 4 = 14 (this is the maximum)
CAse 2: where we assume all 4 in 3-4 years exercise LESS than 3.5 hours
Total # respondednts in that case = 5 + 3 + 2 = 10 (this is the minimum)
Range od 10-14 respondents
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Joined: 09 Nov 2013
Posts: 10
Re: Twenty-five adults reported the amount of time each spent st [#permalink]
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03 May 2014, 08:03
2
srcc25anu wrote:
1. To find the least possible mean the sum of all hours exercised must be the minimum.
For 0-1 hours lets assume all 5 exercised the minimum possible of just over 0 hours and same for all bars.
so SUM of hours exercised = 0*5 + 1*3+2*2+3*3+4*4+5*5+0+0+8*1+0+10*0 = 78
# respondents = 25
average = 78/25 = 3.12
2. # respondents who exercised on average <0.5 hours per day
If Avg hours exercised = 0.5 hours per day the Total hours exercised during the week will be 0.5*7 OR 3.5 hours/week
We need to count respondents who exercise below 3.5 hours per week.
That will include all in 0-1 hour, 1-2 hr, 2-3 hour and since 3.5 lies in between 3-4 band, we can take 2 cases:
CAse 1: where we assume all 4 in 3-4 years exercise MORE than 3.5 hours
Total # respondednts in that case = 5 + 3 + 2 + 4 = 14 (this is the maximum)
CAse 2: where we assume all 4 in 3-4 years exercise LESS than 3.5 hours
Total # respondednts in that case = 5 + 3 + 2 = 10 (this is the minimum)
Range od 10-14 respondents
Thanks for the explanation! I fully understand the solution, but it took me almost 4 minutes to do it. Anyone know of a faster way to solve this question?
Thanks!
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Joined: 20 May 2014
Posts: 35
Location: India
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GMAT 1: 700 Q51 V32
Re: Twenty-five adults reported the amount of time each spent st [#permalink]
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23 May 2014, 05:16
1
Hi Ashsim,
Part 1 Solution:
Sum of [Number of respondents * Least value of time spent in the range]
= 5*0 + 3*1 + 2*2 + 4*3 + 4*4 + 5*5 + 1*8 + 1*10
= 78
Average = $$\frac{78}{25} = 3.12$$
Part 2 Solution:
Less than 0.5 hr per day = less than 3.5 hr per week
Minimum number = 5 + 3 + 2 + 0 = 10 (When respondents spending time between 3 and 4 have values greater than 3.5)
Maximum Number = 5 + 3 + 2 + 4 = 14 (When respondents spending time between 3 and 4 have values less than 3.5)
I hope it could be solved in 2 mins
Rgds,
Rajat
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Joined: 28 Dec 2013
Posts: 68
Re: Twenty-five adults reported the amount of time each spent st [#permalink]
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06 Aug 2014, 12:01
rajatjain14 wrote:
Hi Ashsim,
Part 1 Solution:
Sum of [Number of respondents * Least value of time spent in the range]
= 5*0 + 3*1 + 2*2 + 4*3 + 4*4 + 5*5 + 1*8 + 1*10
= 78
Average = $$\frac{78}{25} = 3.12$$
Part 2 Solution:
Less than 0.5 hr per day = less than 3.5 hr per week
Minimum number = 5 + 3 + 2 + 0 = 10 (When respondents spending time between 3 and 4 have values greater than 3.5)
Maximum Number = 5 + 3 + 2 + 4 = 14 (When respondents spending time between 3 and 4 have values less than 3.5)
I hope it could be solved in 2 mins
Rgds,
Rajat
Question : How do we know that less than 0.5 hr per day = less than 3.5 hr per week?
Manager
Joined: 12 Nov 2016
Posts: 136
Concentration: Entrepreneurship, Finance
GMAT 1: 620 Q36 V39
GMAT 2: 650 Q47 V33
Re: Twenty-five adults reported the amount of time each spent st [#permalink]
### Show Tags
15 Nov 2017, 01:49
sagnik242 wrote:
rajatjain14 wrote:
Hi Ashsim,
Part 1 Solution:
Sum of [Number of respondents * Least value of time spent in the range]
= 5*0 + 3*1 + 2*2 + 4*3 + 4*4 + 5*5 + 1*8 + 1*10
= 78
Average = $$\frac{78}{25} = 3.12$$
Part 2 Solution:
Less than 0.5 hr per day = less than 3.5 hr per week
Minimum number = 5 + 3 + 2 + 0 = 10 (When respondents spending time between 3 and 4 have values greater than 3.5)
Maximum Number = 5 + 3 + 2 + 4 = 14 (When respondents spending time between 3 and 4 have values less than 3.5)
I hope it could be solved in 2 mins
Rgds,
Rajat
Question : How do we know that less than 0.5 hr per day = less than 3.5 hr per week?
assuming 7 day week (not weekend break ) 7*0.5 = 3.5
Senior Manager
Joined: 17 Mar 2014
Posts: 426
Re: Twenty-five adults reported the amount of time each spent st [#permalink]
### Show Tags
01 Jan 2018, 19:04
Erjan_S wrote:
sagnik242 wrote:
rajatjain14 wrote:
Hi Ashsim,
Part 1 Solution:
Sum of [Number of respondents * Least value of time spent in the range]
= 5*0 + 3*1 + 2*2 + 4*3 + 4*4 + 5*5 + 1*8 + 1*10
= 78
Average = $$\frac{78}{25} = 3.12$$
Part 2 Solution:
Less than 0.5 hr per day = less than 3.5 hr per week
Minimum number = 5 + 3 + 2 + 0 = 10 (When respondents spending time between 3 and 4 have values greater than 3.5)
Maximum Number = 5 + 3 + 2 + 4 = 14 (When respondents spending time between 3 and 4 have values less than 3.5)
I hope it could be solved in 2 mins
Rgds,
Rajat
Question : How do we know that less than 0.5 hr per day = less than 3.5 hr per week?
assuming 7 day week (not weekend break ) 7*0.5 = 3.5
yeah I assumed weekend break and got this question wrong.
Re: Twenty-five adults reported the amount of time each spent st [#permalink] 01 Jan 2018, 19:04
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# Twenty-five adults reported the amount of time each spent st
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# The Equation (cos P – 1)x2 + (cos p)x + Sin p = 0 Where x is A Variable, Has Real Roots If p lies In The Interval
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# Name the property of equality you would use to solve 2 = 3/4 + x
0
79
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Name the property of equality you would use to solve 2 = 3/4 + x
Guest Jun 2, 2017
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### 2+0 Answers
#1
+75302
+1
2 = 3/4 + x subtraction property of equality
2 - 3/4 = 3/4 - 3/4 + x
2 - 3/4 = x
8/4 - 3/4 = x
5/4 = x
CPhill Jun 2, 2017
#2
+1148
0
2 = 3/4 + x
2 -3/4 = x
2 - .75 =x
1.25 =x
tertre Jun 3, 2017
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# Flow curve and yield point determination with rotational viscometry
Many liquid, gel-like, or semi-solid everyday substances are typically characterized by measuring a sample’s viscosity, flow curve, and yield point using rotational viscometers/rheometers. Water is a typical sample with a low viscosity and has no yield point. Toothpaste on the other hand has a higher viscosity and a yield point. The yield point tells us how much force needs to be applied so that the material starts to flow.This article describes how a flow curve is measured with a rotational viscometer and how the yield point can be calculated.
## Measurement with a rotational viscometer/rheometer
Most rotational viscometers/rheometers work according to the Searle principle: A motor drives a bob inside a fixed cup. The rotational speed of the bob is preset and produces the motor torque that is needed to rotate the measuring bob. This torque has to overcome the viscous forces of the tested substance and is therefore a measure for its viscosity.
A rotational viscometer measures the dynamic viscosity [ƞ] of a sample. Newton's Law defines the dynamic viscosity η as the shear stress divided by the shear rate[3]. When we measure the viscosity on a rotational viscometer we apply to the sample a certain shear stress or a certain shear rate, respectively.
$\eta = {{\tau} \over {\dot{\gamma}}}$
Equation 1: Newton's viscosity law
$\eta$ ... dynamic viscosity [Pa.s]
$\tau$ ... shear stress [Pa]
${\dot{\gamma}}$ ... shear rate [s-1]
The physical properties speed and torque can be translated into the rheological properties shear rate and shear stress if the measurement is performed using a standard measuring geometry according to ISO 3219[2]. The viscometer’s speed is converted into shear rate using a conversion factor (the factor depends on the measuring geometry used) and the torque is also converted into shear stress using a conversion factor. Usually, this is automatically done by the instrument.
speed × conversion factor = shear rate
torque × conversion factor = shear stress
In a typical rotational test, the shear rate is preset. This means that the viscometer translates the chosen shear rate into speed and measures the resulting torque, which it then translates into shear stress[3].
Such a measurement, where the shear rate is increased and the resulting shear stress is measured, is called a flow curve.
## Flow curve
### Definition of a flow curve
Typical rotational tests are viscosity functions that depend on shear rate, shear stress, time, or temperature. The results of a rotational test can be displayed on the one hand as a flow curve diagram showing the resulting shear stress values, and on the other hand as the corresponding viscosity function. A measurement on a rotational viscometer/rheometer in which the shear rate is increased step-wise and the respective shear stress is determined for each shear rate is called a flow curve. A material can show different flow behaviors such as ideally viscous, shear-thinning and shear-thickening behavior.
In nearly all production stages the viscosity of food samples has a great impact e.g. in the mixing process of dairy substances and for the generation of new formulations for vegetable sauces. At low shear rates the sample‘s viscosity at rest is measured (e.g. when stored in the bottle) and at high shear rates the sample‘s viscosity in movement (e.g. when swallowing or shaking the sample) is tested. Hazelnut cream for example should have a defined viscosity when it is spread on a slice of bread and should not to be too watery or too thick.
### Flow curve measurement
A flow curve can be generated via predefined settings on the rotational viscometer/rheometer:
• The shear rate is increased step-wise with a defined measurement point duration for each point.
• The temperature and other ambient conditions are constant.
The resulting flow curve shows the shear stress versus the shear rate. Usually, the individual measurement points are connected by lines to show the flow curve.
#### Flow curve of ideally viscous substances
For ideally viscous or so-called Newtonian samples, the shear stress increases linearly with increasing shear rate. Figure 4 shows such a flow curve of two Newtonian samples, sample (a) being more viscous than sample (b). The samples’ viscosity does not change with increasing shear rate. Typical examples would be olive oil for sample (a) and water for sample (b).
#### Flow curve of shear-thinning substances
For materials with shear-thinning flow behavior, the gradient of the shear stress decreases at higher shear rates. This means that the sample’s viscosity becomes lower at higher shear rates. This is a typical behavior of many materials from daily life, e.g. cosmetics such as creams and lotions, food samples such as ketchup or molten chocolate, as well as chemical samples such as paints and coatings. See figure 4.
#### Flow curve of shear-thickening substances
For materials with shear-thickening flow behavior, the gradient of the shear stress increases at higher shear rates. This means that the sample’s viscosity increases at higher shear rates. This is a relatively rare flow behavior, typically seen in samples with a high solid content such as ceramic suspensions, starch dispersions, or dental composites.
## Yield point calculation
### Definition of the yield point
The yield point (also called yield stress) is the lowest shear-stress value above which a material will behave like a fluid, and below which the material will act like a solid[2].
Typical examples of materials that have a yield point are creams, ketchup, toothpaste, and sealants. The yield point is the minimum force that must be applied to those samples so that they start to flow.
Substances with a yield stress only start to flow once the outside force acting on them is larger than their internal structural forces. Below the yield point, the substance shows “solid-like” behavior. For example, toothpaste does not flow out of its tube if no external force is applied. Below the yield point, it behaves solid-like. The yield point can be overcome by increasing the shear forces (e.g. by pressing on the tube). Above the yield point, the sample flows (out of the tube) and behaves liquid-like.
The yield point is of vital importance for many practical issues and applications, e.g. for quality control of final products or for optimizing the production process.
### Calculation of the yield point with a flow curve
The yield point is not a material constant but depends on the measuring and analysis method used. There are many different methods available. On rotational viscometers/rheometers, the yield point is often calculated from flow curves measured with a linear increase of the shear rate (as described above). The yield point is not measured directly, but calculated using model functions (e.g. Bingham, Casson, or Herschel-Bulkley). For all these approximation models, the yield point value ${\tau}_0$ (spoken "tau zero") is determined by extrapolation of the flow curve towards a low shear rate value. Each different model function usually produces a different yield point because the calculation is different.
#### Bingham yield point calculation
The Bingham regression calculates the yield point with a constant slope[4, 5]. It is mostly used when the material shows a constant slope above the yield stress, for example some types of food or cosmetics. Historically, before the use of computers, it was a very easy way to determine the yield point. See figure 8.
$$\tau = \tau_B + \eta_B \cdot {\dot{\gamma}}$$
Equation 2: Bingham equation for the yield point
$\tau$ ... shear stress [Pa]
$\tau_B$ ... Bingham yield point [Pa]
$\eta_B$ ... Bingham viscosity [Pa.s]
${\dot{\gamma}}$ ... shear rate [s-1]
#### Casson yield point calculation
For the Casson yield point calculation, curve fitting is performed using a square root function[6]. This takes the curve’s bending into consideration and is therefore often better suited for shear-thinning and shear-thickening materials than the Bingham calculation. It is typically used for food products such as chocolate. See figure 9.
$$\sqrt{\tau} = \sqrt{\tau_c} + \sqrt{\eta_c \cdot \dot\gamma}$$
Equation 3: Casson equation for the yield point
$\tau$ ... shear stress [Pa]
$\tau_c$ ... Casson yield point [Pa]
$\eta_c$ ... Casson viscosity [Pa.s]
$\dot\gamma$ ... shear rate [s-1]
#### Herschel-Bulkley yield point calculation
The Herschel-Bulkley regression describes the flow curve of a material with a yield stress and shear-thinning or shear-thickening behavior at stresses above the yield stress[7].
$$\tau = \tau_{HB}+ {c} \cdot \dot\gamma^{p}$$
Equation 4: Herschel-Bulkley equation for the yield point
$\tau$ ... shear stress [Pa]
$\tau_{HB}$ ... Herschel Bulkley yield point [Pa]
c ... flow coefficient, also called Herschel Bulkley viscosity [Pa.s]
${\dot\gamma}$ ... shear rate [s-1]
p ... exponent, also called Herschel Bulkley index
$\tau_{HB}$ describes the material’s yield point. c is the so-called “flow coefficient” or “consistency index” that describes the viscosity at a certain shear rate. The parameter p (also called “Herschel-Bulkley index”) describes the material’s flow behavior:
• p < 1: shear-thinning
• p > 1: shear-thickening
• p = 1: Bingham flow behavior
With Herschel-Bulkley, in most cases a better curve fitting will be achieved than with the Bingham model.
### Calculation of the static yield point with the vane technique
The determination of the static yield point using the vane technique is a very quick, easy, and direct method for analyzing the yield point with an entry-level rotational viscometer. This method avoids slipping effects and minimizes structural changes of the sample during immersion of the spindle. For the measurement of the yield point a vane spindle with four to eight thin blades arranged at equal angles is used. By inserting the vane spindle into e.g. a paste-like sample the structure of the sample is disrupted only slightly in comparison to concentric cylinder systems. It is important that the diameter of the beaker is at least twice the size of the vane diameter. A change in the sample’s structure can be further minimized by using the original sample container rather than pouring the sample into a beaker. The yield value strongly depends on the sample preparation and pretreatment of the sample before starting the measurement[8, 9]. To measure the yield point a constant low speed is preset on the rotational viscometer. The maximum yield stress, which can be detected during the measurement, is the yield point value. To illustrate the yield point in a graph the torque is plotted against the time. The diagram has usually three typical regions[10]:
1. First, the shear stress increases due to deformation as an elastic response.
2. A shear stress peak is achieved due to a collapse of the microstructure of the material. This point is called the yield point.
3. Afterwards, a stress decay due to the structural breakdown can be visualized.
What is the difference between static and dynamic yield stress? If the yield stress is analyzed from samples where the structure is not disrupted, it is called “static yield stress”. In contrast, if the yield point is analyzed from samples which were pre-sheared before the measurement, it is called “dynamic yield stress”.
### Further yield point calculation methods
Many more model functions than described in chapter 3.2 are available for determining the yield point with a flow curve measurement, e.g. Casson/Steiner[11] or Windhab[12]. In practical use, it often depends on the field of application which yield point calculation leads to better results.
Furthermore, the yield point of a material can also be calculated using other rheological measurement methods than those described, for example:
• Logarithmic shear-rate-controlled flow curves performed with rotational viscometers/rheometers
• Linear or logarithmic shear-stress-controlled flow curves performed with rotational viscometers/rheometers
• Amplitude sweeps performed with air-bearing oscillatory rheometers; this method is the most advanced and accurate method available today[3].
## Conclusion
It can be summarized that for rotational viscometers/rheometers which work according to the Searle principle, the measuring system rotates at a certain speed, the corresponding torque is measured (or vice versa), and the viscosity is calculated. Furthermore, a flow curve can be measured to analyze the sample’s flow behavior. This is done by increasing the shear rate step by step and determining the respective shear stress for each shear rate. A quick yield point check for quality control purposes with an entry-level viscometer can be carried out using the vane technique. By adjusting the yield point of a sample like ketchup or shower gel, the force that is applied to the tube when pressing it and therefore the force that is needed for the sample to start flowing out can be simulated.
## References
1. Newton, I. (1687). Philosophiae naturalis principia mathematica. London.
2. ISO 3219:1994-10, Plastics – Polymers/resins in the liquid state or as emulsions or dispersions – Determination of viscosity using a rotational viscometer with defined shear rate.
3. Mezger, T. (2011). The Rheology Handbook. 3rd revised ed. Hanover: Vincentz Network.
4. Bingham, E.C.(1916). An Investigation of the Laws of Plastic Flow. US Bureau of Standards Bulletin.
5. Bingham, E.C. (1922). Fluidity and Plasticity. New York: McGraw-Hill.
6. Casson, N. (1959). A flow equation for pigment-oil suspensions of the printing ink type. C.C. Mill, Rheology of disperse systems, New York: Pergamon press, 84–104.
7. Herschel, W.H.; Bulkley, R. (1926). Konsistenzmessungen von Gummi-Benzollösungen. Kolloid Zeitschrift, 39(4), 291-300.
8. Sun, A. and Gunasekaran, S. (2009). Yield stress in foods. Measurements and applications. International Journal of Food Properties, 12, 70–101
9. Farid, M. M. (2010). Mathematical modeling of food processing. 1st edition. CRC Press
10. Genovese, D. B. and Rao, M. A. (2005). Components of Vane Yield Stress of Structured Food Dispersions. Journal of Food Science, 70 (8)
11. Steiner, E.H. (1958). A new rheological relationship to express the flow properties of melted chocolate. Rev. Int. Choc, 13, 290.
12. Eischen J.C. and Windhab E.J. (2002). Viscosity of cocoa and chocolate products. Applied Rheology, January/February, 32–34. | 3,218 | 14,564 | {"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} | 3.859375 | 4 | CC-MAIN-2023-06 | latest | en | 0.886079 |
https://www.coursehero.com/file/5576860/bayes/ | 1,495,572,752,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607702.58/warc/CC-MAIN-20170523202350-20170523222350-00162.warc.gz | 859,875,773 | 73,802 | bayes - Bayesian Estimation Bayes theorem If A1 A2 An is a...
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Bayesian Estimation Bayes’ theorem If A 1 ,A 2 , ··· n is a partition of the sample space, and B is any set, then for each i , P ( A i | B )= P ( B | A i ) P ( A i ) n j =1 P ( B | A j ) P ( A j ) = P ( B | A i ) P ( A i ) P ( B ) . f Y | X ( y | x f X,Y ( x, y ) f X ( x ) = f X,Y ( x, y ) R -∞ f X,Y ( x, y ) dy = f X | Y ( x | y ) f Y ( y ) R -∞ f X | Y ( x | y ) f Y ( y ) dy 1 θ is a random variable! X f ( x | θ ) Ω Θ h ( θ )(p r io r) X :ar v x : observed value Θ: arv θ : observed value Prior Data Posterior 2 X 1 , ,X n is a random sample from f ( x | θ ). Prior distribution of Θ: h ( θ ) Joint conditional pdf of X 1 , n given Θ= θ is: f ( x 1 | θ ) f ( x 2 | θ ) f ( x n | θ ) Joint pdf of X 1 , n and Θ is: g ( x 1 , ,x n f ( x 1 | θ ) f ( x 2 | θ ) f ( x n | θ ) h ( θ ) If Θ is a continous random variable, the joint marginal pdf of X 1 , n is: g 1 ( x 1 , n Z -∞ g ( x 1 , n ) 3 The conditional pdf of Θ, given X 1 = x 1 , n = x n is: (posterior) k ( θ | x 1 , n g ( x 1 , n ) g 1 ( x 1 , n ) = f ( x 1 | θ ) f ( x 2 | θ ) f ( x n | θ ) h ( θ ) g 1 ( x 1 , n ) 4
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Example : X 1 , ··· ,X n Bernoulli( θ ), 0 <θ< 1 f ( x 1 , ,x n | θ )= θ x i (1 - θ ) n - x i ∈{ 0 , 1 } Consider h ( θ )=6 θ (1 - θ ) , 0 1 Prior distribution theta prior 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.5 1.0 1.5 g 1 ( x 1 , n 6( n ¯ x + 1)!( n - n ¯ x + 1)! ( n + 3)! k ( θ | x 1 , n Γ( n +4) θ n ¯ x +1 (1 - θ ) n - n ¯ x +1 Γ( n ¯ x + 2)Γ( n - n ¯ x +2) This is Beta( n ¯ x +2 ,n - n ¯ x + 2).
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This note was uploaded on 09/10/2009 for the course STATS 517 taught by Professor Song during the Fall '07 term at Purdue.
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Ask a homework question - tutors are online | 860 | 2,121 | {"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-2017-22 | longest | en | 0.729119 |
https://questioncove.com/updates/5687f2cfe4b012befebf8382 | 1,511,251,089,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806327.92/warc/CC-MAIN-20171121074123-20171121094123-00728.warc.gz | 731,125,867 | 5,465 | OpenStudy (alivejeremy):
Jeannette is participating in a hot dog eating contest. She has already eaten 20 hot dogs but needs to eat more than 34 hot dogs to win. Jeannette is eating 3.1 hot dogs per minute. Which of the following inequalities could be used to solve for x, the number of minutes Jeannette needs to eat hot dogs to win the contest? 3.1x > 20 3.1x > 34 3.1x - 20 > 34 3.1x + 20 > 34
1 year ago
OpenStudy (mathmale):
1 year ago
OpenStudy (mathmale):
Use the "rate of change" (or just "rate") below:$\frac{ 3.1hotdogs }{ \min }$
1 year ago
OpenStudy (alivejeremy):
wow
1 year ago
OpenStudy (mathmale):
(rate)*(time)=number of hotdogs eaten
1 year ago
OpenStudy (alivejeremy):
oh
1 year ago
OpenStudy (mathmale):
Wow. Isn't Equation Editor nice? You, too, could learn how to use that.
1 year ago
OpenStudy (anonymous):
So the rate here is 3.1 hotdogs/minute. For 20 hotdogs, you have to divide 20 by 3.1 to find how much minutes did she take now. Tell me the answer
1 year ago
OpenStudy (mathmale):
Amount = rate * time, yes. solving for time: $time=\frac{ Amount }{ rate }$
1 year ago
OpenStudy (alivejeremy):
3.1x - 20 > 34
1 year ago
OpenStudy (anonymous):
Great you figured it out
1 year ago
OpenStudy (alivejeremy):
3.1x + 20 > 34
1 year ago
OpenStudy (mathmale):
I won't pick on you this time, but next time please show the work you did to get your answer. Let's move on to the next problem.
1 year ago
OpenStudy (alivejeremy):
ok
1 year ago
OpenStudy (anonymous):
Im ready for the next question
1 year ago
OpenStudy (alivejeremy):
Karla's doctor recommended her daily caffeine intake stay under 500 milligrams. Today, Karla has already had 430 milligrams of caffeine. Her favorite soda contains 35 milligrams. Which of the following inequalities could be used to solve for x, the number of sodas Karla can still have today? 35x + 430 < 500 35x < 500 35x - 430 < 500 35x < 930
1 year ago
OpenStudy (mathmale):
Post it separately, please, not here.
1 year ago
OpenStudy (mathmale):
You know by now, jeremy: I'd like to see your efforts and to read your questions before infinity or I start helping. Apply what you've learned from previous problem solving to solve this new problem.
1 year ago | 647 | 2,243 | {"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} | 4.09375 | 4 | CC-MAIN-2017-47 | latest | en | 0.909542 |
http://www.fmxfeeds.com/2018/09/quick-algorithm-get-ideal-size-square-like-for-a-board-game-having-an-arbitrary-but-even-number-of-fields/ | 1,675,257,837,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499934.48/warc/CC-MAIN-20230201112816-20230201142816-00315.warc.gz | 57,665,922 | 10,124 | # Quick Algorithm: Get Ideal Size (Square like) For a Board Game Having an Arbitrary (but Even) Number of Fields
Say you are developing a game like Chess, Go, Checkers, Tic-Tac-Toe or Memory. In each of those games the game board is a rectangle looking playfield of different size (rows x columns). Tic-Tac-Toe is 3×3, Checkers is 8×8, while Go can be 19×19 or 13×13 and similar.
In a game with an arbitrary number of game fields you might want to have the board look as closely to square as possible (rectangle where height and width are the same). Think of Memory. Let’s say we have 24 cards, that is 12 pairs. If you want to place them in a rectangle looking grid, most similar to square, you would go for 4 x 6 (or 6 x 4) board size (as it would look more square like than 3 x 8 and 2 x 12 or 1 x 24 would be too wide).
Therefore, the question: having an arbitrary number of game field pairs, what is the ideal, most square looking, grid size?
And the answer is an algorithm (some math knowledge required) like this one:
TGridSize = record
Rows, Columns : integer;
end;
function CalcGridSize(const numberOfPairs : Cardinal) : TGridSize;
//look for ideal rectangle dimensions (square is ideal)
//to present fields, number of fields = 2 * numberOfPairs
var
fieldCount : integer;
i : integer;
begin
fieldCount := 2 * numberOfPairs;
result.Rows := 1;
result.Columns := fieldCount;
if Sqrt(fieldCount) = Trunc(Sqrt(fieldCount)) then
begin
result.Rows := Trunc(Sqrt(fieldCount));
result.Columns := Trunc(Sqrt(fieldCount));
Exit;
end;
for i := Trunc(Sqrt(fieldCount)) downto 2 do
if (fieldCount mod i) = 0 then
begin
result.Rows := i;
result.Columns := fieldCount div i;
Exit;
end;
end; (*CalcGridSize*)
And here are some results:
var
gridSize : TGridSize;
i : integer;
begin
for i := 1 to 20 do
begin
gridSize := CalcGridSize(i); | 494 | 1,836 | {"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-2023-06 | latest | en | 0.820298 |
https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_16&oldid=117690 | 1,623,638,263,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487611320.18/warc/CC-MAIN-20210614013350-20210614043350-00273.warc.gz | 127,821,635 | 11,207 | # 2015 AMC 8 Problems/Problem 16
## Problem
In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If $\frac{1}{3}$ of all the ninth graders are paired with $\frac{2}{5}$ of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?
$\textbf{(A) } \frac{2}{15} \qquad\textbf{(B) } \frac{4}{11} \qquad\textbf{(C) } \frac{11}{30} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{11}{15}$
## Solution
### Solution 1
Let the number of sixth graders be $s$, and the number of ninth graders be $n$. Thus, $\frac{n}{3}=\frac{2s}{5}$, which simplifies to $n=\frac{6s}{5}$. Since we are trying to find the value of $\frac{\frac{n}{3}+\frac{2s}{5}}{n+s}$, we can just substitute $\frac{6s}{5}$ for $n$ into the equation. We then get a value of $\frac{\frac{\frac{6s}{5}}{3}+\frac{2s}{5}}{\frac{6s}{5}+s} = \frac{\frac{6s+6s}{15}}{\frac{11s}{5}} = \frac{\frac{4s}{5}}{\frac{11s}{5}} = \boxed{\textbf{(B)}~\frac{4}{11}}$
### Solution 2
We see that the minimum number of ninth graders is $6$, because if there are $3$ then there is $1$ ninth-grader with a buddy, which would mean $2.5$ sixth graders with a buddy, and that's impossible. With $6$ ninth-graders, $2$ of them are in the buddy program, so there $\frac{2}{\tfrac{2}{5}}=5$ sixth-graders total, two of whom have a buddy. Thus, the desired fraction is $\frac{2+2}{5+6}=\boxed{\textbf{(B) }\frac{4}{11}}$.
### Solution 3
Let the number of sixth graders be $s$, and the number of ninth-graders be $n$. Then you get $\frac{n}{3}=\frac{2s}{5}$, which simplifies to $5n=6s$. We can figure out that $n=6$ and $s=5$ is a solution to the equation. Then you substitute and figure out that $\frac{5\cdot\frac{2}{5}+6\cdot\frac{1}{3}}{5+6}=\boxed{\textbf{(B) }\frac{4}{11}}$
-RedFireTruck | 674 | 1,906 | {"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": 26, "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-2021-25 | latest | en | 0.79646 |
http://lbrandy.com/blog/2009/12/unplanned-planning/ | 1,670,000,814,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710909.66/warc/CC-MAIN-20221202150823-20221202180823-00348.warc.gz | 30,598,718 | 3,532 | # Unplanned planning
Posted by Louis Brandy on 07 December 2009
Here are some rectangles:
```struct irect {int top; int bot; int left; int right;};
struct frect {float top; float bot; float left; float right;};
```
Now, given those rectangles, you are asked to write a conversion from `irect` to `frect`.
```struct frect convert (struct irect i) {
struct ans = {(float) i.top, (float) i.bot, (float) i.left, (float) i.right)};
return ans;
}
```
This might at first seem like a reasonable answer but… it’s almost certainly not.
## The problem
What is the area of a rectangle with corners (0.0,0.0) and (10.0,10.0)?
Now, how many pixels are in an image with bounds (0,0) to (10,10)?
Hopefully, your answers disagreed (or at least you can see how they could disagree). In our field, we deal with images. Lots and lots of images. One of the most fundamental data structures you can possibly imagine when dealing with images is the lowly rectangle. For us, a rectangle is a region of an image. And it’s this role, as a region, that creates all kinds of pain and heartache.
We can more clearly isolate the problem by asking what is the lower-right hand corner of the above integer rectangle? Well, it depends on exactly what we mean by the lower-right hand corner. If you want to know where to look in memory for the lower-right hand corner pixel information, the answer is (x=10,y=10). If you wanted to know the Cartesian coordinate of the true lower right-hand corner, the answer would be (x=11, y=11).
## The solution?
The difficulty here is the conflation of two separate ideas into the same form. First we have the integerized “point” on the cartesian plane, and then we have the 1x1 region known as a “pixel” and specified by its index. How should pixel indices convert to the cartesian plane? Put another way, should an integer rectangle (0,0)x(10x10) have an area of 100 or 121? This is, to some extent, a question of convention. If you pick one, and are consistent, everything will work out fine. However, almost certainly, for your application, one convention will result in cleaner code.
So the answer to the original question becomes a fairly complicated ordeal. This one problem only scratches the surface of the painstaking care that must be taken when dealing with integer and floating point regions of pixels. You would need to figure out exactly how and why you use rectangles and what index convention you want to use. For every function that you write operating on rectangles, you need to carefully consider the implications of your chosen convention. And depending on the convention you use, the rest of your code is going to look very different.
Writing a rectangle class seems straightforward. Yet, all planning in the world won’t prepare you because it’s almost impossible to anticipate this design issue. It’s the simple act of trying to use one of your converted rectangles that makes the problem obvious. Almost paradoxically, this problem, once realized, can only really be solved by meticulous and careful planning. The devil is always in the details.
© louis brandy — theme: midnight by mattgraham — with help from jekyll bootstrap and github pages | 733 | 3,195 | {"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-2022-49 | latest | en | 0.902896 |
http://nrich.maths.org/public/leg.php?code=170&cl=3&cldcmpid=7534 | 1,369,456,037,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705502703/warc/CC-MAIN-20130516115822-00037-ip-10-60-113-184.ec2.internal.warc.gz | 193,888,579 | 6,587 | # Search by Topic
#### Resources tagged with Perimeters similar to Changing Areas, Changing Perimeters:
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Circumferences. Generalising. Area. Visualising. Pythagoras' theorem. smartphone. Circles. Creating expressions/formulae. Rectangles. Perimeters.
### There are 17 results
Broad Topics > Measures and Mensuration > Perimeters
### Perimeter Possibilities
##### Stage: 3 Challenge Level:
I'm thinking of a rectangle with an area of 24. What could its perimeter be?
### Changing Areas, Changing Perimeters
##### Stage: 3 Challenge Level:
How can you change the area of a shape but keep its perimeter the same? How can you change the perimeter but keep the area the same?
### Can They Be Equal?
##### Stage: 3 Challenge Level:
Can you find rectangles where the value of the area is the same as the value of the perimeter?
### Coins on a Plate
##### Stage: 3 Challenge Level:
Points A, B and C are the centres of three circles, each one of which touches the other two. Prove that the perimeter of the triangle ABC is equal to the diameter of the largest circle.
### Pick's Theorem
##### Stage: 3 Challenge Level:
Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.
### Arclets
##### Stage: 3 Challenge Level:
Each of the following shapes is made from arcs of a circle of radius r. What is the perimeter of a shape with 3, 4, 5 and n "nodes".
### AP Rectangles
##### Stage: 3 Challenge Level:
An AP rectangle is one whose area is numerically equal to its perimeter. If you are given the length of a side can you always find an AP rectangle with one side the given length?
### Warmsnug Double Glazing
##### Stage: 3 Challenge Level:
How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?
### On the Edge
##### Stage: 3 Challenge Level:
Here are four tiles. They can be arranged in a 2 by 2 square so that this large square has a green edge. If the tiles are moved around, we can make a 2 by 2 square with a blue edge... Now try. . . .
### Is There a Theorem?
##### Stage: 3 Challenge Level:
Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel?
### Some(?) of the Parts
##### Stage: 4 Challenge Level:
A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle
### Perimeter Expressions
##### Stage: 3 Challenge Level:
Create some shapes by combining two or more rectangles. What can you say about the areas and perimeters of the shapes you can make?
### Origami Shapes
##### Stage: 4 Challenge Level:
Take a sheet of A4 paper and place it in landscape format. Fold up the bottom left corner to the top so the double thickness is a 45,45,90 triangle. Fold up the bottom right corner to meet the. . . .
### Giant Holly Leaf
##### Stage: 4 Challenge Level:
Find the perimeter and area of a holly leaf that will not lie flat (it has negative curvature with 'circles' having circumference greater than 2πr).
### Pericut
##### Stage: 4 and 5 Challenge Level:
Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?
### Contact
##### Stage: 4 Challenge Level:
A circular plate rolls in contact with the sides of a rectangular tray. How much of its circumference comes into contact with the sides of the tray when it rolls around one circuit?
### Coke Machine
##### Stage: 4 Challenge Level:
The coke machine in college takes 50 pence pieces. It also takes a certain foreign coin of traditional design. Coins inserted into the machine slide down a chute into the machine and a drink is duly. . . . | 939 | 4,074 | {"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} | 3.84375 | 4 | CC-MAIN-2013-20 | longest | en | 0.878601 |
https://mathoverflow.net/questions/254151/explicit-description-of-a-subgroup-of-the-braid-group-mathsfb-2c-2 | 1,560,816,198,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998581.65/warc/CC-MAIN-20190617223249-20190618005249-00066.warc.gz | 505,714,746 | 30,788 | # Explicit description of a subgroup of the braid group $\mathsf{B}_2(C_2)$
This is related to my previous MathOverflow question Fundamental group of $\mathrm{Sym}^2(C_g)$ minus the diagonal.
Let $C_2$ be a smooth curve of genus $2$ and $X:=\mathrm{Sym}^2(C_2)$ its second symmetric product. If $\delta \subset X$ is the diagonal, then the topological fundamental group $\pi_1(X-\delta)$ is isomorphic to the braid group $\mathsf{B}_2(C_2)$ on two strands on $C_2$.
Such a group is generated by five elements $a_1, \, a_2, \, b_1, \, b_2, \, \sigma$ subject to the following set of relations:
\begin{equation*} \begin{split} (R2) \quad & \sigma^{-1} a_1 \sigma^{-1} a_1= a_1 \sigma^{-1} a_1 \sigma^{-1} \\ & \sigma^{-1} a_2 \sigma^{-1} a_2= a_2 \sigma^{-1} a_2 \sigma^{-1} \\ & \sigma^{-1} b_1 \sigma^{-1} b_1 = b_1 \sigma^{-1} b_1 \sigma^{-1} \\ & \sigma^{-1} b_2 \sigma^{-1} b_2 = b_2 \sigma^{-1} b_2 \sigma^{-1}\\ & \\ (R3) \quad & \sigma^{-1} a_1 \sigma a_2 = a_2 \sigma^{-1} a_1 \sigma \\ & \sigma^{-1} b_1 \sigma b_2 = b_2 \sigma^{-1} b_1 \sigma \\ & \sigma^{-1} a_1 \sigma b_2 = b_2 \sigma^{-1} a_1 \sigma \\ & \sigma^{-1} b_1 \sigma a_2 = a_2 \sigma^{-1} b_1 \sigma \\ & \\ (R4) \quad & \sigma^{-1} a_1 \sigma^{-1} b_1 = b_1 \sigma^{-1} a_1 \sigma \\ & \sigma^{-1} a_2 \sigma^{-1} b_2 = b_2 \sigma^{-1} a_2 \sigma \\ & \\ (TR) \quad & [a_1, \, b_1^{-1}] [a_2, \, b_2^{-1}]= \sigma^2. \end{split} \end{equation*} The geometric interpretation for the above generators of $\mathsf {B}_2(C_2)$ is the following. The $a_i$ and the $b_i$ are the braids coming from the representation of the topological surface associated with $C_2$ as a polygon of $8$ sides with the standard identification of the edges, whereas $\sigma$ is the classical braid generator on the disk. In terms of the isomorphism with $\pi_1(X-\delta)$, the element $\sigma$ corresponds to the homotopy class in $\textrm{Sym}^2(C_2)-\delta$ of a topological loop that "winds once around $\delta$". For more details see P. Bellingeri's paper
On presentations of surface braid groups, Journal of Algebra 274 (2004), 543-563.
Now, the surface $X$ contains a $(-1)$-curve $E$, namely the smooth rational curve given by the graph of the hyperelliptic involution on $C_2$. This curve intersect $\delta$ at six points $p_1, \ldots, p_6$ corresponding to the six Weierstrass points of $C_2$.
Thus, if $i \colon E-\{p_1, \ldots, p_6\} \to X - \delta$ is the inclusion, we have an induced map of fundamental groups $$i_* \colon \, \pi_1( E-\{p_1, \ldots, p_6\}) \to \pi_1(X-\delta) \simeq \mathsf{B}_2(C_2).$$
Question. What is the image of $i_*$ in terms of the above presentation for $\mathsf{B}_2(C_2)$?
• Your surface of genus two comes from the octagon with the boundary labelling $a_1 b_1 A_1 B_1 a_2 b_2 A_2 B_2$? (Here capital letters stand for inverses.) I think that the boundary labelling $a b c d A B C D$ will work better for this problem. Using this labelling, the hyperelliptic acts on the octagon by 180 degree rotation. This makes it "clear" that $\sigma$ is in the image of $i_*$. – Sam Nead Mar 28 '17 at 8:20
• I think that something like $[a, \sigma]$ (and thus also $[b, \sigma]$, $[c, \sigma]$, $[d, \sigma]$) is in the image. – Sam Nead Mar 28 '17 at 9:24 | 1,166 | 3,247 | {"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} | 3.5625 | 4 | CC-MAIN-2019-26 | latest | en | 0.486776 |
https://www.storyofmathematics.com/glossary/tree/ | 1,726,542,466,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651722.42/warc/CC-MAIN-20240917004428-20240917034428-00319.warc.gz | 921,207,118 | 43,126 | # Tree|Definition & Meaning
## Definition
A tree is a collection of parallel straight lines with no closed loops joined at the ends (cycles). In other words, it is a straightforward, connected, acyclic, undirected graph (or, equivalently, a connected forest).
A tree diagram is a tool used in probability and statistics in mathematics that enables us to count the number of possibilities that could occur in an event when those options are listed in an organised way.
Each branch’s path in the tree diagram corresponds to a possible course for an event. It is an easy approach to depict the progression of events and it clearly and simply captures every conceivable result.
Tree diagrams often begin with a single item or node that branches into two or more, then each subsequent node will branch into two or more, and so on. The last diagram then looks like a tree with a trunk and several branches.
Figure 1 – A Free Tree diagram of 6 nodes
## Usage of Tree in Probability
A tree diagram can be used to represent a probability space in probability theory. These diagrams could represent conditional probabilities or a series of separate events, like a set of coin tosses (like drawing cards from a deck, without substituting the cards).
Each node on the tree diagram represents an event and is linked to the likelihood that it will occur. Since the root node represents the particular event, its probability is 1. A comprehensive and independent spread of the parent event is shown by the entire collection of sibling joints or nodes.
When calculating probabilities, we occasionally run into problems and find it challenging to decide what to do.
The tree diagram will solve this problem. Think of a probability example to create a tree diagram for flipping a coin. Head and tail are the two branches. The outcome is placed at the branch’s end and the probability of an event is written on the branch.
## Types of Trees
### Free Trees
A connected graph with no cycles is all that a free tree is. There is just one method to get everywhere, and every node can be reached from every other node. It appears to be a graph, which it actually is, although it is less detailed than the WWII France graph. This is due to the fact that a free tree in certain ways doesn’t contain anything “additional.”
These fascinating facts are accurate if you have a free tree:
1. Any two nodes can only be reached by one path. (Verify it!)
2. The graph gets disconnected if any edge is removed. (Try it!)
3. Any additional edge results in the addition of a cycle. (Try it!)
4. There are n-1 edges when there are n nodes. (Consider it!)
In other words, if your objective is to link all the nodes and you have a free tree, you’re good to go. Anything you add is unnecessary, and anything you take away damages it.
This should bring to mind Prim’s algorithm, if not already. This is exactly what Prim’s algorithm produced: a free tree connecting all the nodes, and more specifically, the free tree with the least overall length.
A free tree that spans (connects) every node in the network is simply referred to as a “spanning tree.” Remember that the same set of vertices can be used to create a variety of free trees. For instance, you get a different free tree if you take away the edge from A to F and add one from anywhere else to F.
### Decision Tree
Based on a hierarchical if-else classifier, a decision tree. The region is divided into a hypercube by the set of axis-parallel hyperplanes.
The decision tree uses a tree-like structure to construct classification or regression models. With an increase in tree depth, it divides a dataset into smaller subsets. The outcome is a tree containing leaf nodes and decision nodes.
Outlook is one example of a decision node that has two or more branches (e.g., Sunny, Overcast and Rainy). A classification or choice is represented by a leaf node (for instance, Play). The root node in a decision tree is the highest decision node that corresponds to the best predictor. Both category and numerical data can be processed using decision trees.
### Rooted Tree
The only difference between a rooted tree and a free tree is that we elevate one node to serve as the root. It turns out that this is the deciding factor. Let’s say we determined that Figure 1’s root is A.
The rooted tree in Figure 2’s left half would then exist. Everything is flowing beneath the A vertex, which has been placed at the top. I picture it as delicately reaching into the free tree, gripping a node, and then lifting your hand up so the remaining nodes of the free tree dangle from it.
The right half of the illustration shows the rooted tree that would have resulted if we had instead picked (say) C as the root.The edges on both of these rooted trees are identical to those on the free tree: B is connected to both A and C, F is exclusively connected to A, etc. The root node’s designation is the only difference.
We have thus far stated that the spatial location on graphs is meaningless. But with rooted trees, this varies a little. We can only distinguish which nodes are “above” others through vertical orientation, and in this context, “above” refers to being nearer the root.
A node’s altitude indicates how many steps separate it from the root. Nodes B, D, and E are all one step away from the root (C) in the right rooted tree, whereas node F is three steps away.
Figure 2 – Two distinct rooted trees with identical vertices and edges.
### Binary Trees
A rooted tree’s nodes can have any number of offspring. However, there is a particular kind of rooted tree known as a binary tree that is constrained by merely stating that each node can have a maximum of two offspring. We’ll also refer to these two kids as the “left child” and “right child,” respectively. (Take note that a specific node may only have a left or right child, respectively, but it’s still crucial to know which way the kid is facing.)
Figure 2’s right half is not a binary tree, but its left half is (C has three children). Figure 3 displays a larger binary tree (of height 4).
Figure 3 – A Binary Tree
### Directed Trees
An acyclic-directed graph is a directed tree. It has one node with indegree 1 and all of the other nodes with indegree 1.
The node with outdegree 0 is known as an external node, a terminal node, or a leaf. Internal nodes are those with an outdegree greater than or equal to one.
### Ordered Trees
A tree is referred to as being ordered if there is a definite ordering at each level of the tree.
## Properties of Trees
• Because it has several levels, a tree is a hierarchical structure.
• The topmost node in a tree is referred to as the root node.
• A leaf node or terminal node is a node that is devoid of children.
• At every level of I 2i has the most nodes.
• The longest path from the root node to the leaf node is the height of the tree.
• The distance from a node’s root to the node’s depth.
## Application of Trees
• In order to efficiently manage the data, trees hold data in the form of a hierarchical structure.
• Tree is also used to maintain the data in routing tables in the routers, which improves the insertion, deletion, and searching processes.
## Example
Justify the fact that all trees are bipartite graphs.
### Solution
The vertices of a graph must be split into two sets, A and B, such that no two vertices in either group are contiguous if we want to demonstrate that the graph is bipartite. This algorithm accomplishes exactly that.
Choose any vertex to serve as the root. Add this vertex to set A.
Place set B now to contain all of the root’s offspring. We are doing OK so far because none of these kids are close neighbors (they are all siblings). Put each child of each vertex in B into A now (i.e., every grandchild of the root). Alternate between A and B every “generation” until all vertices have been assigned to one of the sets. In other words, a vertex is only in set B if it is a child of a vertex in set.
All images/mathematical drawings were created with GeoGebra. | 1,736 | 8,052 | {"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-2024-38 | latest | en | 0.935012 |
https://math.stackexchange.com/questions/147381/why-do-we-essentially-need-complete-measure-space | 1,726,514,375,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651710.86/warc/CC-MAIN-20240916180320-20240916210320-00416.warc.gz | 343,410,273 | 42,796 | Why do we essentially need complete measure space?
While reading the motivation of complete measure space on Wikipedia, what I concluded was, completeness is not really necessary when we define on one measure space and it is necessary when we want to measure on product of measure spaces (is it true ?). If $\lambda$ is measure on $X$ and $Y$ then is it true that $\lambda^2$ is measure of $A$x$B$ and how ? I am not able to understand that $\lambda^2(A\times B)=\lambda(A)\times\lambda(B)$ ? Essentially what is the flaw in the measure without being complete ? Waiting for response. Thanks!
• One can define product measure spaces without completions. But Lebesgue measure is complete, and if we want two-dimensional Lebesgue measure, which is complete, to be be the product of one-dimensional Lebesgue measure with itself, we have to complete the product. Commented May 20, 2012 at 14:06
• By definition of Lebesgue measure we can see that it is complete , right, because subset of a set of measure zero has measure zero ie. the subset is measurable, right ?? But how would that change if we take a product ? Commented May 20, 2012 at 14:20
• If $N$ has measure zero, then $\lambda^2(N\times\mathbb{R})\lambda(N)\cdot\lambda(\mathbb{R})=0\cdot\infty=0$. So if $B$ is any subset of $\mathbb{R}$, then $\lambda^2(N\times B)$ would be $0$ if $\lambda^2$ were complete. But if $B$ is not measurable, the set $N\times B$ is not in the usual product $\sigma$-algebra. But it is in the completion. Commented May 20, 2012 at 14:25
• That a measure space is complete means that every subset of set with measure zero is measurable (and has therefore measure zero too). We know that $\lambda^2(N\times\mathbb{R})=0$ and $N\times B\subseteq N\times\mathbb{R}$ if $B\subseteq\mathbb{R}$. So if the product were complete, $N\times B$ would be in the product $\sigma$-algebra if it were complete. It is not, which is why we complete the product $\sigma$-algebra to get $\lambda^2(N\times B)=0$ for all $B\subseteq\mathbb{R}$. Commented May 20, 2012 at 14:42
• Btw: The Borel $\sigma$-algebra on $\mathbb{R}^2$ does equal the product $\sigma$ of the one-dimensional Borel $\sigma$-algebras! So the motivation Wikipedia gives might not be very convincing. Commented May 20, 2012 at 14:47
5 Answers
We want measure spaces to be complete because we want to treat sets of measure zero as negligible. For example, if two functions $f$ and $g$ satisfy $f(x)=g(x)$ for all $x\in X\setminus N$, and $N$ has measure zero, then we'd like to treat $f$ and $g$ as essentially the same thing. However, without completeness it's possible that $f$ is measurable but $g$ is not.
The issue of completeness is brought into light by the product operation, because the product of complete measures is not always complete. For example, let $A\in [0,1]$ be a nonmeasurable set. The set $A\times \{0\}\subset [0,1]\times [0,1]$ is not measurable with respect to the product measure $\lambda\otimes\lambda$. However, $A\times \{0\}\subset [0,1]\times \{0\}$ and the latter set has product measure $0$. So, once we take the completion of the product measure, $A\times \{0\}$ becomes a measure $0$ set.
• Product measures are complete essentially by construction (they are almost always defined via the Carathéodory construction, which yields complete measure spaces). The point is that their restriction to the product $\sigma$-algebra (the algebra generated by the measurable rectangles) is usually not complete.
– t.b.
Commented May 20, 2012 at 15:17
• @t.b. Well, Folland's "Real Analysis" is an exception to almost always, and since it was my textbook in measure theory...
– user31373
Commented May 20, 2012 at 15:25
• However, without completeness it's possible that f is measurable but g is not. What's wrong with this? Since $f$ and $g$ differ only on a null set, they can be regarded as the same in measure theory.
– JHW
Commented Aug 21, 2015 at 15:13
• because the product of complete measures is not always complete. I don't see why this justifies the completion of a measure. If a product of complete measures is not complete, why do we need the completion of it?
– JHW
Commented Aug 21, 2015 at 15:21
One place in probability theory where complete measures are used is the theory of stochastic processes. We have a stochastic process $X_t$ indexed by reals $t$, so there are uncountably many of them. Certain combinations or these are important, but (as far as can be proved) only equal almost everywhere to a countable combination. With complete sigma-algebra, that is enough for us to conclude that this combination is measurable.
I am giving an answer of this question Philosophically...
All intuition of Measure Theory comes from Probability Theory (finite measure theory ). In Probability certain event is impossible then all its sub events are also impossible (usually).
• In probability theory, it is very common to work with incomplete probability spaces. See for example here. Commented May 20, 2012 at 14:45
I've spent some time on checking whether the completeness of measures plays a role in theorems. At first glance I couldn't find any advantage, so I think some of you maybe interested in it.
There is a so-called projection theorem (Measurable projection theorem proof reference), where completeness is essentially needed.
The second theorem where we assume that the measure is complete (it may be relaxed) is the so-called Scorza-Dragoni theorem.
In fact, the Scorza Dragoni theorem uses projection theorem.
When proving that something holds almost surely, i.e, that its negation never holds (say, the event $A$), it comes handy to 'bound' such an event with a bigger one of measure zero. Then, we do not need to prove that the original event is measurable.
Now, I am sure we can also think of 'practical' cases in which is hard (or not possible) to prove that $A$ is measurable when there is no completeness, while being easy to bound $A$ it by a set of measure zero. | 1,537 | 5,975 | {"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} | 3.921875 | 4 | CC-MAIN-2024-38 | latest | en | 0.914544 |
https://www.cuemath.com/ncert-solutions/q-8-exercise-8-2-comparing-quantities-class-8-maths/ | 1,590,705,318,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347400101.39/warc/CC-MAIN-20200528201823-20200528231823-00347.warc.gz | 697,977,919 | 16,574 | # Ex.8.2 Q8 Comparing Quantities Solution - NCERT Maths Class 8
Go back to 'Ex.8.2'
## Question
The price of a TV is $$\rm{Rs}\, 13,000$$. The sales tax charged on it is at the rate of $$12\%$$. Find the amount that Vinod will have to pay if he buys it.
Video Solution
Comparing Quantities
Ex 8.2 | Question 8
## Text Solution
What is known?
Price of the TV $$= \rm{Rs}\, 13,000$$
Sales tax on the TV $$= 12\%$$
What is Unknown?
Amount Vinod will have to pay if he buys the TV
Reasoning:
Amount Vinod has to pay to buy $$= 13,000 + 12\%$$ of $$13,000$$
Steps:
Sales Tax Amount
\begin{align}= \frac{{{12}}}{{{100}}} \times{13000} = \rm{Rs}\, 1560\end{align}
Amount Vinod has to pay to buy $$= 13,000 + 1560 = \rm{Rs}\, 14,560$$
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• Personalized curriculum to keep up with school | 304 | 958 | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2020-24 | longest | en | 0.876869 |
https://nzmaths.co.nz/resource/building-patterns | 1,642,722,524,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302706.62/warc/CC-MAIN-20220120220649-20220121010649-00091.warc.gz | 490,604,007 | 16,402 | # Building patterns
Purpose
In this unit students work with growing patterns made from square tiles. Students represent the relationships between pattern number and number of tiles using tables, graphs and rules, in order to predict further terms of the pattern.
Achievement Objectives
NA3-8: Connect members of sequential patterns with their ordinal position and use tables, graphs, and diagrams to find relationships between successive elements of number and spatial patterns.
Specific Learning Outcomes
• Continue a linear growth pattern from a few examples.
• Find the recursive rule of a linear growth pattern from table of values.
• Explain why the graph of relationships in the pattern is linear.
• Use the table and recursive rule, and/or the graph to make predictions about other terms of the pattern.
• Attempt to create a general rule that connects term number and number of tiles for any term of the pattern.
Description of Mathematics
A linear number pattern is a sequence of numbers for which the difference between consecutive terms is always the same. If plotted on a number plane the graph of a linear pattern is a straight line.
A progression in the way students process linear patterns is well established in the research. That progression is as follows:
1. Copy a spatial pattern with materials.
2. Identify change and extend the spatial pattern with materials.
(Make the next term for the pattern).
3. Represent the relationship between term number and number of items and use recursive rules to predict further members.
(Find the number of items required to make the next term without needing to build it).
4. Use function (direct) rules to connect term numbers with number of items.
(Find the number of items required to make any given term without needing to know the number for the previous term).
5. Reverse (direct) rules to find the term number for a given number of items.
(Given a number of items, find which term number would require that many).
This unit is aimed at achievement of Level 3 in The New Zealand Curriculum, which requires students to develop recursive rules. Level 3 involves progression from phase 2 to phase 3 of the above progression.
For students that need to consolidate understanding of the first two phases, access to materials and restrictions on the terms for prediction makes the level of challenge appropriate. Asking for the number of items in the tenth member of a pattern is a good guideline.
Students progressing to the third phase need support in representing the relationship between term numbers and the number of items, using tables, graphs and diagrams.
To extend students to progress to the fourth phase (direct rules), tasks need to push beyond terms that are easily found using recursive rules. Asking for the number of items for the twentieth, fiftieth, or hundredth term is a good guideline.
The unit is based around patterns with square tiles, which is relatively context neutral. It may be that situations from real life motivate your learners. The patterns could be contextualised as buildings made of sections, stone paintings (kohutu peita), planting of trees, or fruit ripening in a tray. You might like to discuss situations in which everyday patterns grow in consistent way, such as saving the same amount of money each week, stack of items in the supermarket, shoes related to the number of people, or chairs on a bus or aircraft.
Required Resource Materials
Activity
#### Getting started (Session One)
The unit begins by looking at the growth patterns of even and odd numbers. It is important that students ‘see’ even numbers as multiples of two, and odd numbers as multiples of two plus or minus one. The lesson also looks at generalisations about what happens when even and odd numbers are added.
1. Build the ‘factories’ in the diagram below, using square tiles, drink coasters, or similar square-shaped objects (PowerPoint One, slides one and two).
How many squares make up the first factory? The second? The third? etc.
What is changing and what is staying the same as the pattern grows?
2. Let students solve the next problem collaboratively using materials if need be. Their strategies will indicate their current achievement against the progression.
How many squares make up the tenth factory?
3. Ask the same questions for the pattern on slide two.
4. Slide Three presents the two growing patterns on top of one another.
How are these two patterns related? (The terms in the second sequence (odd numbers) are one less than the first sequence.)
5. Ask your students to create tables for the first four terms of each pattern. For example:
Term 1 2 3 4 10 Number of tiles 2 4 6 8 ?
How could the table be used to find the number of squares in term 10?
6. Students might notice that the number of tiles goes up by two for each increase of one in the term number. This recursive rule can be used to find any term in the pattern but requires a bit of work.
7. If your students can use the ‘add two’ recursive rule to further the table pose the following challenge to encourage direct rules.
If you had to find the fiftieth term in the pattern would you use the ‘add two’ rule? Why not? How would you find the fiftieth term in each pattern?
8. Let your students work together in small teams. They might realise that repeated adding of twos is inefficient and suggest a direct rule, i.e. multiply the term number by two (even pattern) and multiply the term number by two then subtract one (odd pattern).
9. Finally use Slides Four and Five to introduce the prediction of combining terms.
You might provide further examples for your students as scaffolding.
Can you predict the result if I join two terms?
How will you describe your method so someone else can use it?
10. For the even pattern the result of joining terms 3 and 5 is term 8. The general rule is “Add the term numbers and multiply by two.”
11. For the odd pattern the result of joining terms 3 and 5 is term 7 in the even pattern (7 x 2 = 14). The general rule is “Add the term numbers, multiply by two, then subtract two.”
For students who develop the rule quickly provide these challenges:
1. Kayla joins two terms in the even pattern and gets a total of 40 squares. What can you say about the terms she joins?
2. Tipene joins two terms in the odd pattern and gets a total of 40 squares. Which terms might he join?
3. Hana finds a rule for joining a term in the even pattern with the same term in the odd pattern. What is her rule?
4. If an odd term is taken away from the same even term, how many squares remain? Why?
#### Session Two
In this session students use a spatial pattern made with square tiles to investigate how relationships that exhibit constant difference are represented with graphs.
1. Show your students PowerPoint Two, slide one, which shows a growing pattern. Ask students what they notice about how the pattern grows from one term to the next.
2. Ask students: Can you imagine what Term Ten looks like? Describe it to a partner.
Look for students to anticipate ten empty spaces or squares surrounded, on all sides, by square tiles.
3. Ask students to work with a partner to predict how many square tiles they would need to build Term Ten.
Look for students to:
• Use the element of growth to extend the pattern physically (with tiles) or diagrammatically
• Use a table and notice that there is a constant growth of five tiles between consecutive terms
• Extend the repeated addition of five to calculate that 53 tiles are needed
• Use multiplicative direct rules such as 8 + 9 x 5 or 3 + 10 x 5 to find the number of tiles
4. Be prepared to extend students who use direct rules with challenges like:
Why do the different rules like 8 + 9 x 5 or 3 + 10 x 5 give the same number of tiles?
How many tiles are needed to make Term 20? Is that twice as much as Term 10? If not, why not?
If you had 998 tiles, what term could you make? Explain.
5. Share the strategies students used to create term Ten. Discuss efficiency. Go to a spreadsheet or online graphing tool, such as Desmos. Create a table of data from the students’ earlier work. You might highlight using the drop-down capability to repeatedly add five easily.
6. Ask students how hey might graph the relationship.
Students might suggest using a familiar type of display, like a bar graph. Use the spreadsheet to create the graphs they suggest. If they do not suggest a scatterplot then make that your choice.
Which graph best shows how the pattern grows?
What do you notice about that graph?
7. Note that the pie chart is not very useful. Both the bar graph and scatterplot show the constant growth.
8. If your students have digital devices you might send them away to create their own graphs. There is also considerable merit in them creating the graphs by hand as they will need to attend to scale, axis labels, location of points and heights of bars.
9. Gather the class and focus of the scatterplot. Draw students’ attention to specific points and ask what the coordinate represents (an association between a term number and the number of tiles).
As you go up one in the term, how many tiles do you go up by? (five)
How does this “up five” rule show on the graph? (The points are in a straight line)
How we use our graph to predict the number of tiles for Term 20? (extend the line and see what point matches Term 20- See PowerPoint Two, Slide 2)
What does this point (20, 103) tell us?
How accurate is this method? Where can it go wrong? (Slope must be exact)
10. Ask students to find the value of Term 20 using a line on their graph.
What other terms can you read off from your line? (e.g. Term 15 equals 78)
11. Introduce Copymaster One. The pattern is alike the previous one in that it grows in a block fashion. However, the constant difference is six rather than five. Slide Three allows you to display the ‘new’ pattern.
12. Talk through the requirements of Copymaster One then let your students solve the problems in groups of three or four. Use Slide Four to discuss what each student, Teina, Phillip or Cameron, ‘sees’ in the pattern for Term Four (colour coded).
What would each person write for Term 10? … for Term 30? … for any term number?
Which person’s rule is the most efficient to use? Why?
#### Session Three
1. Use Copymaster Two to set out six stations for your students to complete. Letting them work independently, or collaboratively in small groups, provides you with an opportunity to observe what students do and interact with them to support their progression. Make sure students have access to square tiles and square grid paper (e.g. exercise book) to support them. You might choose to develop workshops for students who feel they need help with graphing, creating tables, or developing rules. Students should use online graphing packages or spreadsheets to graph the relationships, as well as manually drawing the graphs.
2. The answers to each Station are given at the end of Copymaster Two. Expect your students to show agency by checking their own answers, addressing errors, and identifying areas in which they need further support. You might set up a chart for them to indicate whether, or not, they have control of the specific learning outcomes. Copymaster Three provides an empty template for this:
I can… ● Not yet ? Maybe P Yes Continue a linear growth pattern from a few terms. Make a table of values. Draw a graph. Use a table or graph to find a term in the pattern. Create a rule for finding any term in the pattern. Annie P P P P P Tariq P P P P P Tipene ? ? ? P P Vey-un ● ● ● P P Sione ● ● ? P ?
#### Session Four
In this session you differentiate the class into two groups, those that feel they need more help with patterns, and those who think they can attempt a challenging pattern investigation independently.
1. Give the more confident group Copymaster Four to work on independently or in small groups. Note that pages 2 and 3 provide enabling prompts for students to support their independent investigation. Students should read the prompts if they are stuck, rather than interrupt your teaching group/s.
2. Use PowerPoint Three with the other group to introduce a structured approach to predicting further Terms in a growth pattern. Work through the slides progressively. Points to note are:
• Predictive rules can come from seeing pattern in tables of values and/or structure in the figures, e.g. four arms that grow
• Different ways of seeing lead to different rules
• Rules for any term can be found by looking for similarities among specific terms
• With linear relations graphs can be used to predict further terms.
3. After a suitable period bring together the two groups to share their learning. Bring out the similarity of the two problems, which are both applications of linear relationships.
#### Session Five
Copymaster Five provides a task that can be used to assess your students. They will need access to a calculator. You might also provide the students with square grid paper to make sketching the yacht pattern easier.
Let your students work independently and use the data to check their achievement against the criteria in Copymaster Three. Students might exchange worksheets so you can mark the task collectively. | 2,814 | 13,265 | {"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} | 4.40625 | 4 | CC-MAIN-2022-05 | latest | en | 0.905146 |
null | null | null | null | null | null | Construction Litigation Lawyers In Cook Station Missouri
Cook Station is an unincorporated community in southern Crawford County, Missouri, United States. It is located about ten miles south of Steelville. Cook Station was formerly a stop on the St. Louis-San Francisco Railroad Line. The Dunlap Hotel is still located there, as well as a general store and a bank. There were regularly dances in town in the 30s and 40s, and the area became a popular destination as a swimming area in the 50’s and 60’s after the train stopped coming through. The original general store was destroyed by fire and rebuilt in 1905. Most of the population of Cook Station lives on the surrounding farms that make up what is considered Cook Station. There are quite a few cattle, goat, and dairy farms in this general area. Cook Station is located very near the beginning of the Meramec River, and often sees spring flooding throughout the valley it is located in.
What is construction litigation?
Federal court opinions concerning construction litigation in Missouri | null | null | null | null | null | null | null | null | null |
http://betterlesson.com/lesson/reflection/7183/letting-students-choose-their-level | 1,477,559,299,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721174.97/warc/CC-MAIN-20161020183841-00256-ip-10-171-6-4.ec2.internal.warc.gz | 29,476,307 | 20,824 | ## Reflection: Student Ownership Happy Hundreds - Section 4: Independent Practice
When students are allowed to group themselves, this builds their ability to explain their thinking (MP3 - Construct viable arguments and critique the reasoning of others). The mathematical practices are great benchmarks for cognitive skill development as well as transfer out of a "math only" context.
I let students choose their own groups but they also know that I will ask some of them to explain their choice. They cannot say, "I wanted to work with 4 digit addition because I already know all the other stuff." They need to articulate their mathematical understanding. For example, "I wanted to work with 4 digit addition because I understand how to regroup more than 10 ones into a ten, 10 tens to make one hundred, and ten hundreds to make a thousand. So, when I work with 4-digit addition, I will sometimes be regrouping ten thousands to make a ten thousand."
I don't expect that level of precision but I do expect an attempt. Students' ability to explain their chosen level of difficulty in mathematical terms is another opportunity for them to attend to precision (MP6).
Letting Students Choose Their Level
Student Ownership: Letting Students Choose Their Level
# Happy Hundreds
Unit 8: Place Value Practice
Lesson 3 of 5
## Big Idea: Working with concrete representations (blocks) helps students understand the abstraction of regrouping.
Print Lesson
7 teachers like this lesson
Standards:
Subject(s):
Math, Place Value, 3 digit addition, strategy development, rounding, addition, subtraction, word problems
67 minutes
### Jennifer Valentine
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Environment: Urban | 533 | 2,454 | {"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-2016-44 | longest | en | 0.920876 |
https://math.stackexchange.com/questions/3953566/why-does-pi-appear-in-the-probability-of-a-number-being-square-free?noredirect=1 | 1,653,554,917,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604495.84/warc/CC-MAIN-20220526065603-20220526095603-00724.warc.gz | 462,507,230 | 67,101 | # Why does $\pi$ appear in the probability of a number being square-free?
The probability of a number being square-free (i.e., the number has no divisor that is a square, cube, etc.) is $$6/\pi^2$$. I have seen many appearances of $$\pi$$, and this is also similar to them. All of them can be explained intuitively why pi appears in them. But I don't see any connection between square-free numbers and $$\pi$$. So my question is:
Why $$\pi$$ appears here?
Note: I don't want rigorous proofs, I want intuitive explanations.
• What do you mean by "The probability of a number being square free"? What is the mathematically rigorous definiition of this statement?
– 5xum
Dec 18, 2020 at 8:26
• @5xum Given any random positive number, what is the chance that it is square free? I don't know much about probability, but I think a better definition is this: Let the probability of a positive integer less than some positive integer $n$ being square free be $P(n)$. What is $\lim_{n\to\infty}P(n)$? Dec 18, 2020 at 8:30
• You're asking essentially for an intuitive explanation why $\pi$ appears in $\zeta(2)$ Dec 18, 2020 at 8:36
• I can meet you halfway: if we provide an intuitive explanation of squares' connection to $\pi$ to motivate $\zeta(2)=\pi^2/6$, we can finish with $\prod_{p\in\Bbb P}(1-p^{-2})=1/\zeta(2)=6/\pi^2$.
– J.G.
Dec 18, 2020 at 8:37
• @bof there are many probability problems where pi comes up. Almost all can be explained. Mathematics is what explains these. And for the reason why pi comes in a standard normal distribution, see this. Dec 18, 2020 at 9:34
• $$6/\pi^2$$ appears in the density of square-free numbers because $$\zeta(2)=\pi^2/6$$ and every integer is uniquely of the form $$k n^2$$ with $$k$$ square-free.
• Then $$\zeta(2)=\pi^2/6$$ because $$\frac{\pi^2}{\sin^2(\pi z)} = \sum_n \frac1{(z-n)^2}$$ This latter formula is magic: the LHS minus the RHS is a bounded entire function, thus constant, which is the great achievement of complex analysis.
• If you don't like it then the video mentioned in the comment is saying that $$\sum_{n=0}^{2^k} \frac1{|2^k (e^{2i \pi (2n+1)/2^{k+1}}-1)|^2}=1/4$$ and that $$\lim_{k\to \infty} \sum_{n=0}^{2^k} \frac1{|2^k (e^{2i \pi (2n+1)/2^{k+1}}-1)|^2}=\sum_{n=-\infty}^\infty \lim_{k\to \infty} \frac1{|2^k (e^{2i \pi (2n+1)/2^{k+1}}-1)|^2}$$ $$= \sum_{n=-\infty}^\infty \frac1{ |i\pi (2n+1)|^2} = \frac1{\pi^2}2(1-1/4)\zeta(2)$$ | 809 | 2,406 | {"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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-21 | longest | en | 0.93554 |
null | null | null | null | null | null |
Irenka and Garik
Irenka was supposed to be house-sitting for her neighbors but when Garik showed up she couldn’t help but show him all the naughty things they could do together in the screened back porch! Irenka just couldn’t get enough of knowing that everything they were doing could be seen by anyone that walked by and the more she thought about it, the wetter her pussy got! As she felt Garik’s cock getting harder every time she kissed his neck she knew he would do anything to feel her pussy around his cock and she wanted nothing more than a good fuck right there on the porch! The very second that she felt that huge dick deep inside her she knew that she could cum any second.
Duration: 32m:56s
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Models: Garik, Irenka
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http://www.jiskha.com/display.cgi?id=1297960703 | 1,498,326,589,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320270.12/warc/CC-MAIN-20170624170517-20170624190517-00638.warc.gz | 566,746,099 | 4,518 | # Maths
posted by .
Hi I am really sorry I now people have already attempted to help me with this one already, however I am still not able to understand, I really have not done a lot of these equations and am struggling. I have put a little more information down this time, any help is much appreciated.
1- Find, in implicit form, the general solution of the differential equation:
Dy/dx = 2(e^x – e^-x) / y^2 (e^x + e^-x)^4 (y > 0).
2- Find the corresponding particular solution (in implicit form) that satisfies the initial; condition y = ½ when x = 0.
3- Find the explicit form of this particular solution.
4- What is the value of y given by this particular solution when x =1?
regards Claire
• Maths-differential equation -
This differential equation is separable, meaning that it can be transformed into two integrals, one involving only x, and the other, only y.
It can then be integrated to get the solution in terms of an integration constant. The initial conditions can be used to determine the constant.
Given:
Dy/dx = 2(e^x – e^-x) / y^2 (e^x + e^-x)^4 (y > 0)
transpose x and y terms to give:
y² dy = 2(e^x – e^-x) dx / (e^x + e^-x)^4
Note how the numerator is related to the derivative of the denominator, so we can use a substitution
u=e^x+e^(-x),
du=(e^x-e^(-x))dx
so the equation becomes:
y² dy = 2du /u^4
Integrate both sides:
y³/3 = -(2/3)u^(-3)+C'
where C is an integration constant.
The general solution is therefore:
y³=-2/(e^x+e^(-x))³+C
2. given y(0)=1/2 =>
(1/2)³=-2/(e^0+e^(-0))^3+C
1/8=-2/2³ + C
=>
C=1/8+1/4 = 3/8
The particular solution is therefore
y³=-2/(e^x+e^(-x))³+ 3/8
3. the explicit form is in the format
y=.....
=>
y=∛(-2/(e^x+e^(-x))³+ 3/8)
(y>0)
4. y(1)=? .... I'll leave that to you.
Do check my work. | 574 | 1,757 | {"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} | 4.375 | 4 | CC-MAIN-2017-26 | latest | en | 0.872768 |
https://alloutlotto.com/lottery-online-ga-buy-lottery-tickets-online-australia.html | 1,558,436,830,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256314.52/warc/CC-MAIN-20190521102417-20190521124417-00232.warc.gz | 375,781,196 | 4,603 | Powerball® tickets print the white ball numbers in numerical order of a given play. You can match the white ball numbers in any order of a given play to win a prize. The red Powerball number of a given play on your ticket must match the red Powerball drawn. Each play on a ticket is separately determined; players cannot crisscross play lines on a ticket or combine numbers from other tickets.
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```Powerball numbers are drawn from two sets of numbers, so the odds of winning a prize are calculated by combining the odds for both sets of numbers for all prize levels. The odds for matching just the Powerball are calculated by combining the odds of selecting the Powerball and the odds of not selecting any of the five numbers from the first set of numbers drawn.
```
USA Green Card Organization is not affiliated with the U.S. Government or any government agency. You can enter the U.S. Diversity Lottery for Free at www.state.gov during their open registration dates which typically start in early October. We are not a law firm, we do not provide legal advice, and are not a substitute for an attorney. This site provides a review and submission service that requires a fee. Access to and use of USA Green Card Organization website is subject to Terms of Use and Privacy Policy. Reproduction in whole or in part of this website is strictly prohibited.
^ Jump up to: a b Prizes are uniform in all Mega Millions jurisdictions, with the exception of California (where all prizes, including the jackpot, are pari-mutuel; payouts are based on sales and the number of winners of each prize tier.) All other Mega Millions members' second through ninth prizes are set amounts, although in rare cases they can be reduced.
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Powerball lottery draws are held every Wednesday and Saturday at 22h59 Eastern Time and are broadcast live. The results of each lotto draw are not official until they are audited by an accounting firm. There are nine prize tiers on the Powerball Lottery, with approximate prizes ranging from \$4 - \$1,000,000. Players' tickets for the US Powerball are purchased by our ticket purchasing offices based in Connecticut, USA.
The structure of the draw is one which regular lottery players will be very familiar with; players must pick 5 regular numbers from a pool with a total of 69 numbers and in addition to these regular number picks, you also choose one bonus ball (known as the Powerball) from a pool of 26. These two pools of numbers are mutually exclusive and remain completely separate throughout the drawing procedure. In order to jackpot the US Powerball, you need to match all 5 regular numbers and the Powerball. Do this and you are instant Powerball millionaire – it’s as simple as that! | 1,475 | 6,576 | {"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-2019-22 | latest | en | 0.92311 |
https://calculator.academy/bonus-points-value-calculator/ | 1,716,445,313,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058611.55/warc/CC-MAIN-20240523050122-20240523080122-00564.warc.gz | 125,086,978 | 55,061 | Enter the total bonus points and the conversion factor ($/point) into the Calculator. The calculator will evaluate the Bonus Points. ## Bonus Point Formula BPV = BP * CF Variables: • BPV is the Bonus Point ($)
• BP is the total number of bonus points
• CF is the conversion factor ($/point) To calculate the Bonus Points value, multiply the number of bonus points by the conversion factor. ## How to Calculate Bonus Points? The following steps outline how to calculate the Bonus Point. 1. First, determine the total number of bonus points. 2. Next, determine the conversion factor ($/point).
3. Next, gather the formula from above = BPV = BP * CF.
4. Finally, calculate the Bonus Point.
5. After inserting the variables and calculating the result, check your answer with the calculator above.
Example Problem :
Use the following variables as an example problem to test your knowledge.
total number of bonus points = 50000
conversion factor ($/point) = 0.1 ## Frequently Asked Questions (FAQ) What are bonus points and how are they used? Bonus points are reward points that businesses give to their customers as part of a loyalty program or promotion. These points can typically be converted into discounts, products, or other rewards based on a conversion factor that determines their dollar value. Can the conversion factor for bonus points vary between different programs? Yes, the conversion factor ($/point) can vary significantly between different programs and businesses. It is determined by the specific terms and conditions of each program, which dictate how much each point is worth in terms of dollar value or other rewards.
Is there a standard value for bonus points across all platforms?
No, there is no standard value for bonus points as each business or loyalty program can set their own conversion rates and rules for how points are earned and spent. This flexibility allows programs to tailor their rewards to fit their specific customer base and business model.
How can consumers maximize the value of their bonus points?
Consumers can maximize the value of their bonus points by understanding the specific conversion factors and redemption options of each program they participate in. Additionally, taking advantage of special promotions or offers that increase the value of points or provide additional points can also enhance the overall value of the rewards earned. | 462 | 2,394 | {"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} | 3.9375 | 4 | CC-MAIN-2024-22 | latest | en | 0.90934 |
http://douglaswoolley.com/puzzles/rubiks-cube-users-guide.html | 1,585,415,170,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370492125.18/warc/CC-MAIN-20200328164156-20200328194156-00250.warc.gz | 59,818,514 | 26,724 | COMPUTER ORIENTED SOLUTIONS
TO
THE RUBIK'S CUBE
IBM Version 1.0
USER'S GUIDE
Douglas E. Woolley
University of South Florida
September 1988
## Introduction
The Rubik's Cube, invented by Erno Rubik, started the historical puzzle craze in the early 1980's. Without help, solving the Rubik's Cube could take several months. For skilled mathematicians it could take several weeks to restore the cube to its original state. The most advanced computer would take 1.4 million years to examine the cube's 43,252,003,274,489,856,000 possible combinations. The solution to the Rubik's Cube, along with many other puzzles, is found by the use of abstract algebra, or group theory.
Being fascinated by computers and having the math skills necessary to take on the task, I embarked upon proving my philosophy: Anything that a person can logically solve or develop an algorithm for, a programmer can instruct a computer to do the same. With this thought in mind, I worked on manually solving the Rubik's Cube with the aid of some elementary group theory and a solution guide. Once an algorithm was developed for solving the puzzle from any given combination, I then coded the solution into a program written in Turbo Pascal.
This program entitled, Computer Oriented Solutions to the Rubik's Cube, allows the user to enter the appearance of the puzzle into the computer. The program will then display a step-by-step solution for the user to restore the puzzle to its original state. After each move of the solution is put on the screen, either a graphical or a textual sketch of the cube is displayed along with the colors on the recently moved cube.
The program named, RUBECUBE.COM, can be run on any IBM PC or compatible. It is recommended that a color monitor be used in order to appreciate the color displays. When the computer is in DOS (Disk Operation System), the program can be started by typing RUBECUBE and then pressing the RETURN key (ENTER key). At this point a title page will appear. To continue, press any key, and then the program menu will appear:
1. Graphical Solution
2. Text Solution
3. Instructions
4. Set Valid Colors
5. Test Program
6. Exit
Choose:_
Simply press the number of the option you desire to choose.
Introduction
Input for Graphical and Text Solutions
Graphical Solution
Text Solution
Instructions
Set Valid Colors
Test Program
Research - History
Research - Competitions
Research - Solution Manuals
Research - Group Theory
Research - Group Theory Solutions
Description of Program Solution Strategy
Partial Pascal Solution Listing
Name and Description of Procedures in Program
Appendix - Variations of the Rubik's Cube
Bibliography
Author's Biography
## Input for Graphical and Text Solutions
In order to obtain a solution (Graphical or Text), the user must describe the appearance of the puzzle to the computer. This is done by entering the color of the squares on each of the six sides of the cube into the computer. Each side of the Rubik's Cube contains 9 squares--a total of 54 squares. The original Rubik's Cube, produced by Ideal Toy Corporation, along with other similar brands, will have the following six unique colors on its squares: (B)lue, (G)reen, (O)range, (R)ed, (W)hite, and (Y)ellow. If your cube does not have all of these colors, then you first must choose option 4 from the main menu to "set valid colors" for the cube before you can input the cube's colors.
The input routine is the same for the graphic solution option and the text solution option. To begin, place the cube in front of you and designate one side to be the front--it will face you. To input colors on the Rubik's Cube, you will enter the color symbol pertaining to each of the squares, beginning with those on the top, then front, right, back, left, and bottom. For a graphic illustration of these sides, you may want to view the instructions in the program, option 3. If you have set your own valid color symbols through option 4, then the first line of instructions below will differ slightly after displaying "Press". The colors on each side are to be input from top to bottom, left to right by pressing the valid color symbol. If you make a mistake, press the left arrow and then press the correct color symbol. The input screen will look like the following after you have entered the colors on the top and part of the front:
+==== Input Colors on Rubik's Cube ====+
| |
| +-------+ |
| | W G G | Press: B, G, O, R, W, Y |
|T| | or Space to remove entry |
|o| G W R | or -> for next square |
|p| | or <- for previous square |
| | R B B | or ESC when finished. |
| +-------+ |
| +-------++-------++-------++-------+ |
|F| Y R _ || _ _ _ || _ _ _ || _ _ _ | |
|r| || || || | |
|o| _ _ _ || _ _ _ || _ _ _ || _ _ _ | |
|n| || || || | |
|t| _ _ _ || _ _ _ || _ _ _ || _ _ _ | |
| +-------++-------++-------++-------+ |
| +-------+ Right Back Left |
|B| _ _ _ | |
|o| | |
|t| _ _ _ | |
|t| | |
|o| _ _ _ | |
|m+-------+ |
+======================================+
If you press a valid color symbol and it does not display on the screen, then there are already 9 squares entered with that color. You probably mistyped one of the other squares. To make a correction, press the left or right arrow keys until the cursor appears at the square that needs to be corrected; then press the new color.
After entering the colors for the top, front, right, back, and left sides, then place the cube so that the front is facing you and the top side is facing up. To enter the color symbols on the bottom, tilt the cube so that the front becomes the top and the bottom becomes the front, temporarily. Now, enter the colors from top to bottom, left to right as you see them on the front. After doing so, tilt the cube back so that it has its original front and top. You may want to check to make sure that everything was entered correctly. If you encounter an error, then you will need to move the cursor to the positions that need to be changed. First clear them by pressing the space bar; then enter the correct color symbols. When you have finished entering, the screen will look similar to the following:
+==== Input Colors on Rubik's Cube ====+
| |
| +-------+ |
| | W G G | Press: B, G, O, R, W, Y |
|T| | or Space to remove entry |
|o| G W R | or -> for next square |
|p| | or <- for previous square |
| | R B B | or ESC when finished. |
| +-------+ |
| +-------++-------++-------++-------+ |
|F| Y R O || Y W W || R O O || B Y B | |
|r| || || || | |
|o| W O O || B B O || Y R Y || B G O | |
|n| || || || | |
|t| G B B || W G O || G Y G || Y R Y | |
| +-------++-------++-------++-------+ |
| +-------+ Right Back Left |
|B| O W R | |
|o| | |
|t| G Y W | |
|t| | |
|o| R R W | |
|m+-------+ |
+======================================+
At this point, you may press the ESC key. If you press the ESC key before all 54 squares have been filled, then the following message is displayed, "Input is incomplete. Try again? (Y/N)". If you enter 'N', then the program returns to the menu. You may continue inputting by pressing 'Y'. If you entered a side in an improper manner, especially the bottom, then this message will display, "A side is incorrect. Try again? (Y/N)". To correct your mistakes, press 'Y' and then clear the squares that are incorrect and then re-enter them.
## Graphical Solution
After ESC is pressed during the input routine, the computer will then generate the entire solution within the next 1 to 2 seconds, if the puzzle can be solved. It is possible to have a Rubik's Cube that is impossible to solve. This can happen if a person takes apart the pieces of the puzzle and puts them back in an improper order. If a person randomly puts back the pieces, then there is only a one in twelve chance that the puzzle has a solution. If no solution is possible the following message is displayed:
An error has been detected.
Possibly, you have an im-
possible cube.
Press any key.
If a solution is possible, then the screen will clear and a graphical sketch of the Rubik's Cube will appear along with a new set of instructions (shown below). To see the first move, you must press either the space bar, the right arrow, or the Enter key. One move of the solution will appear in large symbols, underneath the phrase, "Do the move:". The number of moves made will appear followed by the total number of moves required in the complete solution. The screen is formatted similarly as show below:
Graphical Solution
Do the move:
T+
------------ A graphical Rubik's
Cube will appear here
1 move with the colors currently
out of 134 on the front side after
----------------------------------------
Press: Space, ->, or Enter and perform
move shown above, then compare side(s);
or (T, F, R, P, L, B) to view sides;
or <- for previous move; or ESC to quit.
Each move will either: 1) turn one side independently of the others, or 2) rotate the entire cube so that a new side becomes the front.
First, most moves will consist of a sides' symbol (T, F, R, L, or B) and a rotation (+, 2, or -). This move means to turn the corresponding side clockwise (if a + follows), or counter- clockwise (if a - follows), or 2 times either way (if a 2 follows). Each move is made as if you are viewing it face to face. For Example, T+ means "turn the Top side clockwise." Furthermore, B- means "turn the Bottom side counter-clockwise" as if you are looking at it from the bottom. (See illustrations on next page)
ILLUSTRATIONS: The six faces are indicated by their initials: top (T), right (R), front (F), left (L), posterior (P), and bottom (B).
Keeping the front face oriented toward you—all clockwise rotations are shown. To make counter-clockwise turns, simply rotate the faces in the opposite directions. All rotations should be made as though you were viewing each face from the front. A clockwise turn would always be a twist to the right as if viewed from the front. See illustrations above.
(Text/Graphic illustrations taken from The Simple Solution to Cubic Puzzles, p. 52)
Another move might be to Rotate the Cube (RC). RC+ means to rotate the cube clockwise as if you were looking at the top. Therefore, the right side becomes the new front, the front becomes the left side, and the top remains the same. RC- and RC2 mean to rotate the cube counter-clockwise and twice, respectively, as if you are viewing the cube from the top.
After completing the move, compare the colors on the side displayed with the colors on the corresponding side of your cube. To view the colors on the other sides, press T, F, R, P, L, or B.
To see the next move, press either the space bar, right arrow, or Enter key. The colors on the new front side will appear, as a result of making the next move. Compare this side with yours and continue to make moves until your cube is solved. If you make a mistake, then undo the move you made. You may then need to either make the proper move or have to press the left arrow key to view the previous move.
After you have completed all the moves, the computer will "congratulate" you. If you want to return to the main menu without completing all the moves, then press the ESC key.
## Text Solution
After ESC is pressed during the input routine, the computer will then generate the entire solution within the next 1 to 2 seconds, if the puzzle can be solved. It is possible to have a Rubik's Cube that is impossible to solve. This can happen if a person takes apart the pieces of the puzzle and puts them back in an improper order. If a person randomly puts back the pieces, then there is only a one in twelve chance that the puzzle has a solution. If no solution is possible the following message is displayed:
An error has been detected.
Possibly, you have an im-
possible cube.
Press any key.
If a solution is possible, then a new set of instructions will be displayed. To see the first move, you must press either the space bar, the right arrow, or the Enter key. One move of the solution will appear between two arrows as displayed below, underneath the Right and Back sides. The number of moves made will appear followed by "move out of" and then the total number of moves required in the complete solution, as show below:
+============ Text Solution ===========+
| |
| +-------+ |
| | R G W | Press: Space, ->,or Enter |
|T| | and peform move shown |
|o| B W G | below then compare sides; |
|p| | or <- for previous move; |
| | B R G | or ESC to quit. |
| +-------+ |
| +-------++-------++-------++-------+ |
|F| Y W W || R O O || B Y B || Y R O | |
|r| || || || | |
|o| W O O || B B O || Y R Y || B G O | |
|n| || || || | |
|t| G B B || W G O || G Y G || Y R Y | |
| +-------++-------++-------++-------+ |
| +-------+ Right Back Left |
|B| O W R | |
|o| | |
|t| G Y W | --> T+ <-- |
|t| | |
|o| R R W | 1 move out of 134|
|m+-------+ |
+======================================+
Each move will either: 1) turn one side independently of the others, or 2) rotate the entire cube so that a new side becomes the front.
First, most moves will consist of a sides' symbol (T, F, R, L, or B) and a rotation (+, 2, or -). This move means to turn the corresponding side clockwise (if a + follows), or counter-clockwise (if a - follows), or 2 times either way (if a 2 follows). Each move is made as if you are viewing it face to face. For Example, T+ means "turn the Top side clockwise." Furthermore, B- means "turn the Bottom side counter-clockwise" as if you are looking at it from the bottom. (See illustrations of moves under the topic Graphical Solution)
Another move might be to Rotate the Cube (RC). RC+ means to rotate the cube clockwise as if you were looking at the top. Therefore, the right side becomes the new front, the front becomes the left side, and the top remains the same. RC- and RC2 mean to rotate the cube counter-clockwise and twice, respectively, as if you are viewing the cube from the top.
After completing the move, compare the colors on the sides of your cube with the colors on the corresponding sides displayed.
To see the next move, press either the space bar, right arrow, or Enter key. The colors on the new sides will appear, as a result of making the next move. Compare these sides with yours and continue to make moves until your cube is solved. If you make a mistake, then undo the move you made. You may then need to either make the proper move or have to press the left arrow key to view the previous move.
After you have completed all the moves, the computer will "congratulate" you. If you want to return to the main menu without completing all the moves, then press the ESC key.
## Instructions
Choosing option 3 from the main menu allows the user to view a brief description of each of the menu options. It will also emphasize that the first option that should be chosen (after reading the instructions) is number 4, Set Valid Colors. The program must know the exact colors used on your cube in order to provide a solution.
After option 4 is completed, you may then enter the color symbols on the cube into the computer by either choosing option 1 or 2. Both the Graphical Solution and the Text Solution use the same input routine. After describing the cube to the computer, the program will either remain in text mode (if a text solution was desired) or will enter graphics mode (if a graphical solution was desired).
## Set Valid Colors
The cube has six sides, each with a unique color in its original state. However, not all versions of the cube puzzle have the same six unique colors. The original Rubik's Cube, produced by Ideal Toy Corporation, has as its colors, blue, green, orange, red, white, and yellow. Other cubes produced by different corporations have colors such as purple, green, orange, pink, white, and yellow. There many other variations.
The Computer Oriented Solutions to the Rubik's Cube program solves the puzzle by placing each of the colors on its own side. Thus, the computer needs to know the exact colors used on your cube. Instead of entering the actual name of the color, the computer will accept a one letter color symbol, such as the first letter in the name of the color. Six unique color symbols must be entered in order to properly describe the appearance of the cube.
If your cube has the same six colors that are on the original Rubik's Cube, then you can press the Enter key six times so that the program defaults to the first letter of the following colors: (B)lue, (G)reen, (O)range, (R)ed, (W)hite, (Y)ellow. If your cube has colors other than these, then you must enter the six color symbols that are on your cube before you can obtain a Graphical or Text Solution. To do so, press a one letter color symbol to replace the default values that are shown that differ. If some of the color symbols match, then press the Enter key to select the default value shown.
## Test Program
Choosing option 5 brings you to the routine that tests the program Computer Oriented Solutions to the Rubik's Cube. Furthermore, this option provides statistical facts pertaining to this computer oriented solution. The first command requiring a response is:
Enter # of cubes to solve (1-14): --
You must enter a number between 1 to 14 indicating the number of imaginary cubes that you want the computer to solve. Suppose you enter 10. The computer will simulate 10 solved cubes within its memory. The next command displayed is:
Enter # of random turns (1-30): --
You must enter a number between 1 to 30 indicating the number of random turns (or moves) to make to each of the 10 solved puzzles. If you choose a relatively small number, less than 15, then the puzzle will not be very mixed up. However, you may want to gather statistics for solutions to cubes that are not mixed up well. For our sake, let's choose 30 random turns. Each turn is randomly chosen from the following moves:
T+, T2, T-, F+, F2, F-, R+, R2, R-,
L+, L2, L-, B+, B2, B-, RC+, RC2, RC-
The notation of moves are explained in the Graphical Solution section. With 10 cubes each randomly scrambled, the computer then attempts to solve each one. The program stores the number of moves it took to solve each cube, and then displays the cube number followed by the number of moves it took to solve the puzzle. The program then displays the "Average # of moves", "Most moves", and "Least moves".
Statistically speaking, the program will solve a randomly mixed up cube in an average of 140 moves (including Rotating Cube moves). Without counting Rotation moves, the program will solve a cube in an average of 125 moves (assuming 15 rotations). Other authors of algorithm solutions do not include Rotating Cube moves in their statistics, but only include those moves that permute the colors on the cube. The program solution seems to have an upper limit of approximately 170 moves (including RC's). The least number of moves needed to solve the cube is trivially 0, if the random moves leave the cube unchanged.
At this point in time, the program will always solve a valid Rubik's Cube. However, in the developmental stages of this program, the computer did not always solve the puzzle correctly. In such cases, this Test Program routine displays the set of random moves the computer made to its puzzle in order for the user to recreate the error and detect the part of the program needing correction.
## Research - History
The Rubik's Cube, which is also called the Magic Cube, is a three by three by three inch cube. A cube has six sides or faces which contain a different color. Each face is divided up into nine squares. A brilliant inner mechanism of a spring loaded spindle allows sides of the cube to be rotated either vertically or horizontally and independent of the other sides. In the cube's pristine condition it has a solid color on each face, but is mixed up by just a few random moves. The cube comes in six different colors which can then be mixed up. The original Rubik's Cube has these six colors: white, yellow, orange, red, green, and blue. The object is to get the cube back to its original positions so that all six sides have nine identical squares with respect to their colors.
This great puzzle, the Rubik's Cube, was invented by Erno Rubik, who has held degrees in architectural engineering and interior design and teaches at the Academy of Applied Arts in Budapest, Hungary. From this job of teaching he only earned an estimated \$1,820 a year. Although he was licensed the right to the cube at a modest fee, he made \$800,000 for 1980-81.
The summer of 1974 was when Rubik had thought about the laws of geometry and had started playing with the ideas of a three dimensional object that could rotate around any of the three axes. He kept turning the object in his mind till he could turn the theoretical construction gloating in his head into something real. He had a problem and wanted to make a mechanical devise capable of converting all of the complicated movements into simple twists and turns that would change the arrangement of squares on the faces of an actual cube. After working hard on his model he gave a few turns and then found out that he made a puzzle to solve.
Weeks went by and he finally got some of the different sections of the cube back to their starting point. What had started to be an aid for his students to give greater experience in dealing with three dimensional objects turned out to be much more.
In 1975, he wrote up the details of the cube's construction and obtained a patent. Rubik then interested a small toy-manufacturing co-operative, Polutechnika, in his invention and soon it was in mass production. The original production run of 5000 was shipped to Hungarian stores just before Christmas 1977. Within a few days they had sold out, and 7000 more cubes were rushed off the production line.
In 1980, cubes were bought at prices from \$5 to \$15. A total of 30 million of the cubes were purchased worldwide by 1983. These cubes were sold in all kinds of shops. In Rubik's homeland, Hungary, a controlled economy had not kept up with demand for manufacture of the cube, and they tended to buy three or four at a time when they were available. In the United States, Hong Kong, and the Caribbean, by arrangement with Komsumex, one of Hungary's import/export agencies, the Ideal Toy Corp. manufactured 1.5 million cubes every month. Logical Games Inc., of Haymarket, Va., which introduced the puzzle to the United States under the name Magic Cube in 1979, has manufactured the Cube.
One reason why Ideal Toy Corporation was so successful with the selling of the Rubik's Cube was because they advertised. Helfgart, Towne & Silverstein, were the company's agency for 16 years. The Rubik's Cube showed up at a Toy Fair for the first time in February 1980, but its reception was less than spectacular. Beginning the 4th of July weekend, advertising started with two weeks of television exposure using the 30-second commercial made up of three vignettes, and then four weeks using three 10-second spots, each of which was one of the vignettes. The television advertising was used in four major markets along with small-space newspaper ads. After a respite during the summer, advertising started again in September 1981, and was in 100 spot markets by December 1981.
Ideal Toy Corporation agreed to be acquired by the toy division of CBS Inc. in a transaction valued at about \$58 million. Under terms of the proper acquisition, CBS would pay \$14.85 a share for the 3.9 million outstanding common shares of Ideal, which is based in Hollis, Queens. For the fiscal year 1981, the first full year to include results from sales of the cube, Ideal reported a net income of \$19.1 million or \$4.94 a share versus a loss of \$15.45 million in 1980 and income of \$3.7 million in 1979.
Magic Puzzler, Wonderful Puzzler, and Le Cube, are all variations of Rubik's Cubes except that they do not bear the trademark of the Rubik's Cube and are made by companies other than Ideal. These puzzles retail one-third the price of Rubik's Cube and were selling very well. Ideal had filed dozens of civil suits against cube distributors and retailers across the country. In Federal District Court of Manhattan alone there were seven suits pending, one against the largest distributors, the John Hansen Company of Milbrae California. In May of 1981, Ideal charged Hansen with packaging its cube so that customers would mistake the puzzle for Rubik's Cube. Hansen then sold his cube under the name of Le Cube. Ideal Toy Corporation also filed a suit against United Supply Corporation, claiming unfair competition and trade dress copying of Ideals popular Rubik's Cube. Ideal brought Macy's to court in April of 1981, for displaying hundreds of nameless cubes piled right next to a video screen continuously showing an advertisement for Rubik's Cube. Macy's agreed not to sell nameless cubes in that way anymore, and in return Ideal agreed not to discuss the case publicly.
The privately held Moleculon Research Corporation of Cambridge filed a \$60 million patent infringement suit against Ideal in Federal District Court in Delaware. Moleculon contends that Mr. Nichol's 1972 American patent for a cubic "pattern-forming puzzle" which closely resembles the 30 million cubes that Ideal sold worldwide since 1980, beat Rubik's invention by 2 years. Rubik holds a 1974 patent for his devise in Hungary but never applied for one in the United States or elsewhere. Mr. Nichol's said he has been trying to sell his puzzle since 1969 but was rejected because Moleculon did not want to show a prototype until it received a patent protection. Morris & Safford, of Ideal say Moleculon's 1972 patent was invalid because the ideas contained in it were not original and that Mr. Nichol's patent, which primarily describes a cube held together by magnets, did not cover the mechanical devise that enables puzzle mechanical solvers to move each piece of Rubik's Cube independently.
## Research - Competitions
The first organized cube competition began on January 4, 1980. Kate Fried organized the competition in Budapest, Hungary at the Youth Mathematical Circle and at a Magic Cube Fans Club. Each competitor had to supply his own cube and put it in a cardboard box with his name on it. The judge(s) then scrambled the cubes identically and put them back in their boxes. Viktor Toth, a student, won this first competition in 55 seconds. After this competition, the winner and the runner-up had a play-off in which each scrambled the other's cube. Toth won this competition also.
The second competition was conducted in March, and over a thousand people came, packing into several rooms. This competition consisted of those people who had previously shown that they can solve the cube in less than 90 seconds. Nine people competed in the competition. The winner solved his cube in 40 seconds.
The first to try her skill publicly with the cube in America was Zsa Zsa Gabor, who was hired by the Ideal Toy Corporation, to promote the creation of her fellow Hungarian. The first regional competition of the title of the United States Rubik's Cube championship was held on July 25, 1981.
## Research - Solution Manuals
Without help, solving the Rubik's Cube could have taken several months. For skilled mathematicians it would have taken several weeks to complete a cube. The most advanced computer would have taken 1.4 million years to examine all its 43,252,003,274,489,856,000 possible combinations. The solution to the Rubik's Cube, along with many other puzzles, was found by the use of abstract algebra, or group theory. A person did not have to know group theory to solve these puzzles. All a person had to do was learn or memorize a solution given by authors that have written books about the solutions of the puzzles. Once a person had read and learned the book well, that person could usually solve a puzzle with great ease.
James G. Nourse, a Stanford University Chemist who worked out his formula during his Christmas vacation, gets the Rubik's Cube back to its proper place in one minute. Nourse has sold over one million copies of his solution book, The Simple Solution to Rubik's Cube. He is also the author of The Simple Solution to Cubic Puzzles, which teaches the reader how to solve the Pyraminx, Missing Link, the Barrel, Picture Cube, Rubik's Magic Snake, and the Octagon.
The Hawaiian high school student, Minh Thai, wrote The Winning Solution, which reveals his strategy and racing secrets for the Rubik's Cube. He also wrote a solution book for Rubik's Revenge, a four by four by four cube puzzle. Minh's book, The Winning Solution to Rubik's Revenge, also reveals his strategy and racing secrets for a puzzle. Minh became the U.S. National Champion at the Rubik's Cube-a-Thon aired on ABC's That's Incredible on December 7, 1981, by unscrambling Rubik's Cube in 26.04 seconds. He also restored seven scrambled cubes in under 30 seconds each in a public exhibition in Los Angeles.
Another author of a solution book was Patrick Bassert. Bassert was only 13 years old when he sold 750,000 copies of his solution book, You Can Do The Cube, in London and earned more than \$60,000. The appeal of his book is its relative clarity, and he accompanied his instructions with a system of diagrams and symbols showing how to rotate and move the squares. He and his cousin began charting basic moves on a tattered piece of graph paper that grew quickly into his book. He claims that "doing the cube depends not so much on math as on logic. You need to think logically and see what's happening to see the logic of the relationships between the various pieces, then it becomes clear."
## Research - Group Theory
Most of the people who wrote solutions to various puzzles used group theory. The following is a brief introduction to the group theory used to solve puzzles. A group, G, is a set of elements with a well defined binary operation, (+), such that:
1) If a, b are elements of Group G, then so is a(+)b (closure under operation);
2) There exists an element e of Group G such that a(+)e= a =e(+)a for all a an element of G (identity);
3) If a is an element of G, there exists an element a' of G such that a(+)a'= e =a’(+)a (inverse);
4) If a, b, and c are elements of G, then (a(+)b)(+)c= a(+)(b(+)c) (associative law).
A permutation of the cube is a sequence of moves that rearrange the pieces and color squares of the puzzle. There are over 4.3 quintillion unique permutations for a standard Rubik’s Cube. Each permutation is considered an element of Group G. The group consists of the collection of all unique permutations of the cube. “There are six basic moves or permutations of the cube:”
T+, F=, R+, P+, L+, B+,
Where each letter represents a side of the cube,
And + indicates a 90 degree turn clockwise.
By combining the above moves, all the other elements of the group can be generated. Some other elements include: R+F+, L+B+B+T+.
The binary operation, (+), is defined as combining two elements in the group, which means to perform them in sequence. If this is a group, then the combination of any two members will result in a permutation that is a member of the collection.
The permutation that results in no change is called the identity permutation. The order of the permutation is the least number of times that the permutation needs to be performed to get the identity permutation. For example, the permutation element R2F2 has an order of 6 since the identity element is obtained by combining the moves R2F2 six times. No element has an order larger than 1260. The permutation which undoes a given permutation and gives the identity is called its inverse. A group does not have to be commutative, but when the order in which two permutations are combined is not important, it is said to commute. The Group G, of permutations, is not commutative.
All permutations have cycles which switch certain pieces around; and if this permutation is done enough times, the original permutation shows up again. The length of a cycle is the number of elements in the cycle, and the order of a cycle is its length. A transposition is a cycle of length two. Two cycles are disjoint if there are no common positions which are formed in both. A permutation which is the product of an even number of transpositions is called even, and one which is the product of an odd number of transpositions is called odd. If every completed move of a puzzle is an even permutation and its eventual desired permutation is odd, then the puzzle is impossible to solve.
## Research - Group Theory Solutions
The basic mathematical objective is to restore the cube from any scrambled pattern back to its original position, where each side has a solid color. In general, it is desirable to obtain a set of moves to obtain a particular pattern from any other valid pattern.
“The group of all possible permutations is as follows. The 8 corner pieces can be permuted among themselves in any way, giving 8! ways (! means factorial), and the 12 edge pieces can be permuted among themselves in any of 12! ways, except that the total permutation of corners and edges must be even. Further, independently of the movement of pieces, it is possible to flip the orientations of any two edge pieces and twist any two corner cubes in opposite directions. This means that it is possible to orientate all but one of the edges or corners--the orientation of the last one being forced. This means there are a total of
8 12
8! 12! 3 2 19
N = ------ x --- x --- = 43252003274489856000 ~ 4.3 x 10
2 3 2
27 14 3 2
= 2 x 3 x 5 x 7 x 11
different patterns.” The total number of ways a person can reassemble the cube after taking it apart is indicated by the numerator N. The denominator of 12 indicates that there are “12 distinct orbits of constructible patterns. (An orbit is the set of all patterns reachable from a given pattern by application of our group. It is impossible to get out of one orbit into another by use of the group of motions.)” If the cube is taken apart and randomly reassembled, then there is only a 1/12 chance of being able to solve the cube.
The following people are known for their developments of algorithms to solve the Rubik's Cube by using group theory. Morwen B. Thistlethwaite has done much work in the area of computing to find efficient algorithms. He had his algorithm down to 52 moves to solve the cube, but he was hoping to lower this to 50 by doing more computing. Thistlethwaite believes that it may be reducible to 45 with a lot of effort and researching.
In an article in the Journal of Recreational Mathematics, John Conway and Dave Benson showed how to always restore the cube in at most 100 moves. They did bottom edges, then bottom corners and middle edges together and then use their tables to first position and then orient the top layer, in at most 13 to 19 moves respectively. Roy Nelson solves the cube by doing the bottom and middle, then orienting top corners, then positioning them, then orienting top edges, then positioning them. Without having a cube, Hanke Bremer produced a solution. He gets corners in place, then oriented, then edges in place then oriented. Michel Daughpin, a sixth year student at a lycee in Luxembourg, has a method similar to Bremer's.
3-D Jackson has compiled "The Cube Dictionary" giving about 177 one layer processes. He claims that these allow the following algorithm to produce at most 92 moves: solve bottom edges (10 moves); bottom corners (22); middle edges (28); upper corners in place (8); then orient (14); upper edges (10).
Gerzson Keri says he can do two layers in at most 57 moves and that he believes any one layer process can be done in at most 18 moves. He claims to have over 100 one layer processes of at most 15 moves. He has published a lengthy two-part article in Hungary about his solution. One of his methods is entirely tabular--proceeding one piece at a time, seeing where it is and then looking up a process to get it correctly in place. The table contains 24 + 21 + 18 + ... + 6 + 3 + 24 + 22 + 20 + ... + 4 + 2 = 264 entries of which 5 are impossible and 20 are trivial. There are at most 182 moves that are required.
Due to Adam Kertesz, Bill McKeeman has an algorithm which does two layers, then upper corners, then orients upper edges and then upper corners. Kersten Meier has a layer by layer algorithm but does not detail the working for the last layer. Dame Kathleen Ollerenshaw's method does the bottom face, then the top corners, then the middle slice edges, then top edges. She claims that the algorithm produces an average of 80 moves.
Based on Penrose’s algorithm, solving bottom edges, middle slice edges, top edges, then corners in place then oriented, Zoltan Perjes wrote out his own algorithm. He also outlined Rubik's original fast method: down corners, upper corners in place, then oriented, three down edges, three upper edges, the other down and upper edges, then the middle slice edges. Don Taylor does down edges, down corners, middle edges, upper corners in place, upper edges in place, upper corners oriented, then upper edges oriented.
Michael Vaughan-Lee has made a careful study of an edge first process: down edges (13), three middle edges (14), upper edges and last middle edge (17), corners in place (30) and then in orientation (38), but the last two steps can be reduced to 26 + 24, 32 + 22 or 28 + 22, giving a maximum of 98 moves.
David Sigmaster's algorithm looks like this:
1) Put all bottom edges correctly in place.
2) Put all bottom corners correctly in place.
3) Put middle slice edges correctly in place.
4) Flip top edges so all upper faces are up.
5) Make top orientation correct.
6) Put top edges correctly in place.
7) Put top corners in their right positions.
8) Twist top corners into their correct orientations.
Will there ever be an algorithm that is the shortest? Trivially, some positions of the cube can be unscrambled in 0, 1, 2, or 3 moves. Given any scrambled cube, what is the least number of moves required to restore the cube to its six sides of solid colors? There are several ways to show that at least 17 moves are required for the most scrambled cube. Can every cube be solved in at most 17 moves? The algorithm that obtains the shortest set of moves is know as God’s algorithm.
## Description of Program Solution Strategy
If a person is able to solve a puzzle himself in a logical fashion, then a person probably can write a program to solve the puzzle. With this thought as a basis, how does a person solve the Rubik's Cube? There are 12 corner cubes, 8 edge cubes, and 6 center cubes that need to be positioned and oriented correctly so that each of the 6 sides contain a solid color. First, you need an overall strategy or algorithm before considering the smaller details. The following is a brief sketch of the algorithm that I use to solve the Rubik's Cube:
1) Position and orient Top edges
2) Position and orient Top corners
3) Position and orient Vertical edges
4) Position Bottom corners
5) Orient Bottom corners
6) Position Bottom edges
7) Orient Bottom edges
The center cube on each side does not "move", and they determine the color to be restored on their sides'. For each of the above steps, particular cube edges or cube corners need to be located and then positioned and oriented in a certain spot. There are a finite number of places that this particular piece could be in. For each spot, group theory can aid in determining the set of moves to make to the cube to position and/or orient a particular piece without "messing up" too much of the rest of the puzzle. Once one piece is put in place, another piece needs to be located on this puzzle and then positioned and oriented.
With this manual algorithm, I could program the computer to solve the Rubik's Cube. The two biggest questions that I needed answers to were, "how do I let the computer know what the puzzle looks like?" and "how do I tell the computer to move the cube?"
The computer needs to know the colors on each of the 54 squares of the puzzle. The program accepts a letter as a color symbol for each of the squares and stores it in the designated array position. The color symbols on the top side are stored in A[1..9], the front side colors are stored in A[10..18], the right in A[19..27], the posterior (back) in A[28..36], the left in A[37..45], and the bottom in A[46..54].
With the appearance of the puzzle, the computer does a step of the program's overall algorithm and finds a spot that does not have the correct piece. The program then searches for the misplaced cube by asking itself logical "if-then" statements regarding the positions and colors of the squares. When it finds the correct piece, it reads a set of moves to position and/or orient this piece. The program then branches to a subroutine that "moves" the puzzle. To make a move, the program will actually exchange some of the variables in the array.
After all the moves have been made to place the desired piece, the computer will then know the exact appearance of the new puzzle as if a person had actually performed the moves. The program then does the next part of its algorithm and repeats the above process until it has completed each of the seven steps.
## Partial Pascal Solution Listing
{ -- SOLUTION.INC}
procedure SolveTopEdges ({giving} var MoveError: Boolean);
{ -- This procedure will store the moves in Solution to restore top edges. }
{ -- Strategy: Work on front-top edge then rotate cube to next face. }
var
Face: Integer;
Tmid, Fmid: Char;
begin
for Face := 1 to 4 do begin
Tmid := A[5]; { -- Has Color symbol of Top middle square. }
Fmid := A[14]; { -- Has color symbol of Front middle square. }
if (A[8] = Fmid) and (A[11] = Tmid) then { -- Orient FT }
AddToSolution ('F+ T- R+ T+ ')
else if (A[6] = Fmid) and (A[20] = Tmid) then { -- Move RT to FT }
else if (A[6] = Tmid) and (A[20] = Fmid) then { -- Move RT to FT }
AddToSolution ('R- T- R+ T+ ')
else if (A[2] = Fmid) and (A[29] = Tmid) then { -- Move PT to FT }
AddToSolution ('T+ R- T- F- ')
else if (A[2] = Tmid) and (A[29] = Fmid) then { -- Move PT to FT }
AddToSolution ('T+ R2 T- B- F2 ')
else if (A[4] = Fmid) and (A[38] = Tmid) then { -- Move LT to FT }
else if (A[4] = Tmid) and (A[38] = Fmid) then { -- Move LT to FT }
else if (A[15] = Fmid) and (A[22] = Tmid) then { -- Move FR to FT }
else if (A[15] = Tmid) and (A[22] = Fmid) then { -- Move FR to FT }
AddToSolution ('R- B- R+ F2 ')
else if (A[24] = Fmid) and (A[31] = Tmid) then { -- Move PR to FT }
AddToSolution ('R+ B- R- F2 ')
else if (A[24] = Tmid) and (A[31] = Fmid) then { -- Move PR to FT }
else if (A[33] = Fmid) and (A[40] = Tmid) then { -- Move LP to FT }
else if (A[33] = Tmid) and (A[40] = Fmid) then { -- Move LP to FT }
AddToSolution ('L- B+ L+ F2 ')
else if (A[13] = Fmid) and (A[42] = Tmid) then { -- Move FL to FT }
else if (A[13] = Tmid) and (A[42] = Fmid) then { -- Move FL to FT }
else if (A[17] = Fmid) and (A[47] = Tmid) then { -- Move BF to FT }
else if (A[17] = Tmid) and (A[47] = Fmid) then { -- Move BF to FT }
AddToSolution ('F+ T+ L- T- ')
else if (A[26] = Fmid) and (A[51] = Tmid) then { -- Move BR to FT }
else if (A[26] = Tmid) and (A[51] = Fmid) then { -- Move BR to FT }
else if (A[35] = Fmid) and (A[53] = Tmid) then { -- Move BP to FT }
else if (A[35] = Tmid) and (A[53] = Fmid) then { -- Move BP to FT }
AddToSolution ('B+ L- F+ L+ ')
else if (A[44] = Fmid) and (A[49] = Tmid) then { -- Move BL to FT }
else if (A[44] = Tmid) and (A[49] = Fmid) then { -- Move BL to FT }
if (A[8] = Tmid) and (A[11] = Fmid) and (Face < 4) then { --Rotate Cube }
else If (A[8]<>Tmid) or (A[11]<>Fmid) then begin { -- Fatal error }
ErrorInMoves; MoveError := True; Exit; end;
end; { -- for Face }
end;
procedure SolveTopCorners ({giving} var MoveError: Boolean);
{ -- This procedure will store the moves in Solution to restore top corners. }
{ -- Strategy: Work on front-right-top corner then rotate cube to next face. }
var
Face: Integer;
Tmid, Fmid: Char;
begin
for Face := 1 to 4 do begin
Tmid := A[5]; { -- Has Color symbol of Top middle square. }
Fmid := A[14]; { -- Has color symbol of Front middle square. }
if (A[9] = Fmid) and (A[19] = Tmid) then { -- Orient FRT }
AddToSolution ('R- B2 R+ F+ B2 F- ')
else if (A[12] = Tmid) and (A[19] = Fmid) then { -- Orient FRT }
AddToSolution ('F+ B2 F- R- B2 R+ ')
else if (A[3] = Tmid) and (A[21] = Fmid) then { -- Move PRT to FRT }
AddToSolution ('R+ B+ R- F+ B2 F- ')
else if (A[3] = Fmid) and (A[28] = Tmid) then { -- Move PRT to FRT }
AddToSolution ('R+ B- R- F+ B- F- ')
else if (A[21] = Tmid) and (A[28] = Fmid) then { -- Move PRT to FRT }
AddToSolution ('R+ B2 R2 B+ R+ ')
else if (A[1] = Tmid) and (A[30] = Fmid) then { -- Move PLT to FRT }
AddToSolution ('L- B2 L+ B- R- B+ R+ ')
else if (A[30] = Tmid) and (A[37] = Fmid) then { -- Move PLT to FRT }
AddToSolution ('L- B+ L+ R- B2 R+ ')
else if (A[1] = Fmid) and (A[37] = Tmid) then { -- Move PLT to FRT }
AddToSolution ('L- B- L+ F+ B- F- ')
else if (A[7] = Tmid) and (A[39] = Fmid) then { -- Move FLT to FRT }
AddToSolution ('L+ B2 L- F+ B- F- ')
else if (A[10] = Fmid) and (A[39] = Tmid) then { -- Move FLT to FRT }
AddToSolution ('L+ R- B+ L- R+ ')
else if (A[7] = Fmid) and (A[10] = Tmid) then { -- Move FLT to FRT }
AddToSolution ('F- B2 F2 B- F- ')
else if (A[18] = Tmid) and (A[48] = Fmid) then { -- Move FRB to FRT }
else if (A[25] = Fmid) and (A[48] = Tmid) then { -- Move FRB to FRT }
AddToSolution ('R- B+ R+ F+ B2 F- ')
else if (A[18] = Fmid) and (A[25] = Tmid) then { -- Move FRB to FRT }
else if (A[27] = Fmid) and (A[34] = Tmid) then { -- Move PRB to FRT }
else if (A[27] = Tmid) and (A[54] = Fmid) then { -- Move PRB to FRT }
AddToSolution ('B2 R- B+ R+ ')
else if (A[34] = Fmid) and (A[54] = Tmid) then { -- Move PRB to FRT }
AddToSolution ('B- R- B+ R+ F+ B2 F- ')
else if (A[36] = Tmid) and (A[52] = Fmid) then { -- Move PLB to FRT }
else if (A[36] = Fmid) and (A[43] = Tmid) then { -- Move PLB to FRT }
else if (A[43] = Fmid) and (A[52] = Tmid) then { -- Move PLB to FRT }
AddToSolution ('B2 R- B+ R+ F+ B2 F- ')
else if (A[45] = Tmid) and (A[46] = Fmid) then { -- Move FLB to FRT }
else if (A[16] = Tmid) and (A[45] = Fmid) then { -- Move FLB to FRT }
AddToSolution ('B2 F+ B- F- ')
else if (A[16] = Fmid) and (A[46] = Tmid) then { -- Move FLB to FRT }
AddToSolution ('B+ R- B+ R+ F+ B2 F- ');
if (A[9] = Tmid) and (A[12] = Fmid) and (Face < 4) then { --Rotate Cube }
else if (A[9]<>Tmid) or (A[12]<>Fmid) then begin { -- Fatal error }
ErrorInMoves; MoveError := True; Exit; end;
end; { -- for Face }
end;
procedure SolveVerticalEdges ({giving} var MoveError: Boolean);
{ -- This procedure will store the moves in Solution to restore vert. edges. }
{ -- Strategy: Work on front-right edge then rotate cube to next face. }
var
Face: Integer;
Fmid, Rmid: Char;
begin
for Face := 1 to 4 do begin
Fmid := A[14]; { -- Has color symbol of Front middle square. }
Rmid := A[23]; { -- Has color symbol of Right middle square. }
if (A[15] = Rmid) and (A[22] = Fmid) then { -- Orient FR }
AddToSolution ('R- B+ R+ B+ F+ B- F- B+ R- B+ R+ B+ F+ B- F- ')
else if (A[24] = Fmid) and (A[31] = Rmid) then { -- Move PR to FR }
AddToSolution ('RC+R- B+ R+ B+ F+ B- F- RC-B- F+ B- F- B- R- B+ R+ ')
else if (A[24] = Rmid) and (A[31] = Fmid) then { -- Move PR to FR }
AddToSolution ('RC+R- B+ R+ B+ F+ B- F- RC-R- B+ R+ B+ F+ B- F- ')
else if (A[33] = Fmid) and (A[40] = Rmid) then { -- Move PL to FR }
AddToSolution ('RC2R- B+ R+ B+ F+ B- F- RC2B2 F+ B- F- B- R- B+ R+ ')
else if (A[33] = Rmid) and (A[40] = Fmid) then { -- Move PL to FR }
AddToSolution ('RC2R- B+ R+ B+ F+ B- F- RC2B- R- B+ R+ B+ F+ B- F- ')
else if (A[13] = Fmid) and (A[42] = Rmid) then { -- Move FL to FR }
AddToSolution ('RC-R- B+ R+ B+ F+ B- F- RC+B2 R- B+ R+ B+ F+ B- F- ')
else if (A[13] = Rmid) and (A[42] = Fmid) then { -- Move FL to FR }
AddToSolution ('RC-R- B+ R+ B+ F+ B- F- RC+B+ F+ B- F- B- R- B+ R+ ')
else if (A[17] = Fmid) and (A[47] = Rmid) then { -- Move FB to FR }
AddToSolution ('B- R- B+ R+ B+ F+ B- F- ')
else if (A[17] = Rmid) and (A[47] = Fmid) then { -- Move FB to FR }
AddToSolution ('B2 F+ B- F- B- R- B+ R+ ')
else if (A[26] = Rmid) and (A[51] = Fmid) then { -- Move RB to FR }
AddToSolution ('B+ F+ B- F- B- R- B+ R+ ')
else if (A[26] = Fmid) and (A[51] = Rmid) then { -- Move RB to FR }
AddToSolution ('B2 R- B+ R+ B+ F+ B- F- ')
else if (A[35] = Fmid) and (A[53] = Rmid) then { -- Move PB to FR }
AddToSolution ('B+ R- B+ R+ B+ F+ B- F- ')
else if (A[35] = Rmid) and (A[53] = Fmid) then { -- Move PB to FR }
AddToSolution ('F+ B- F- B- R- B+ R+ ')
else if (A[44] = Fmid) and (A[49] = Rmid) then { -- Move LB to FR }
AddToSolution ('R- B+ R+ B+ F+ B- F- ')
else if (A[44] = Rmid) and (A[49] = Fmid) then { -- Move LB to FR }
AddToSolution ('B- F+ B- F- B- R- B+ R+ ');
if (A[15] = Fmid) and (A[22] = Rmid) and (Face < 4) then { --Rotate Cube }
else if (A[15]<>Fmid) or (A[22]<>Rmid) then begin { -- Fatal error }
ErrorInMoves; MoveError := True; Exit; end;
end;
end;
procedure SolveBottomCorners ({giving} var MoveError: Boolean);
{ -- This procedure will store the moves in Solution to restore bot corners. }
{ -- Strategy: First "position" all 4 corners, then "orient" 4 corners. }
var
Face: Integer;
Fmid, Rmid, Pmid, Lmid, Bmid: Char;
BFL, BFR, BPR, BLP: Boolean; { -- True if this corner exists. }
All4Positioned, All4Oriented: Boolean;
begin
{ -- ****************** Position all 4 corners *************** }
Face := 1;
repeat
If Face > 4 then begin { -- Fatal error }
ErrorInMoves; MoveError := True; Exit; end;
Fmid := A[14]; Rmid := A[23]; Pmid := A[32]; Lmid := A[41];
{ -- Determine which of the 4 Bottom corners exist (= True). }
BFL := ((A[16] = Fmid) and (A[45] = Lmid)) or
((A[16] = Lmid) and (A[46] = Fmid)) or
((A[45] = Fmid) and (A[46] = Lmid));
BFR := ((A[18] = Fmid) and (A[25] = Rmid)) or
((A[18] = Rmid) and (A[48] = Fmid)) or
((A[25] = Fmid) and (A[48] = Rmid));
BPR := ((A[27] = Rmid) and (A[34] = Pmid)) or
((A[34] = Rmid) and (A[54] = Pmid)) or
((A[27] = Pmid) and (A[54] = Rmid));
BLP := ((A[36] = Pmid) and (A[43] = Lmid)) or
((A[36] = Lmid) and (A[52] = Pmid)) or
((A[43] = Pmid) and (A[52] = Lmid));
All4Positioned := True; { -- Will be after this set, unless no matches. }
if (BFL and BFR) or (BFR and BPR) or (BPR and BLP) or (BLP and BFL) then
{ -- Either all 4 match, or only 2 corners match. }
if not (BFL and BFR and BPR and BLP) then { -- only 2 corners match }
begin
if (BFL and BFR) then
else if (BFR and BPR) then
else if (BLP and BFL) then
{ -- Exchange adjacent sides BFL and BFR, which are out of place. }
AddToSolution ('R- B- R+ F+ B+ F- R- B+ R+ B2 ');
end
else { -- null else- since all 4 match, skip to next section. }
else if (BFL and BPR) or (BFR and BLP) then {--Pair of diagonals match. }
begin
{ -- Exchange diagonals BFL and BPL}
AddToSolution ('F- B- R- B+ R+ F+ ');
{ -- Turn Bottom one rotation left or right to match all 4 corners. }
if (BFL and BPR) then
else
end
else { -- No matches found with current Bottom rotation, try another. }
begin
AddToSolution ('B+ '); All4Positioned := False;
end;
Face := Face + 1;
until All4Positioned;
{ -- ****************** Orient 4 corners ****************** }
{ -- Rotate Cube until 1 of 7 patterns appear, or all 4 are oriented. }
{ -- Perform set of moves. If pattern is BC1 or BC2, cube is oriented. }
All4Oriented := False; Face := 1;
repeat
If Face > 4 then begin { -- Fatal error, Pattern not found after 4 turns.}
ErrorInMoves; MoveError := True; Exit; end;
Bmid := A[50];
if (A[46] = Bmid) and (A[48] = Bmid) and (A[52] = Bmid) and (A[54] = Bmid)
then All4Oriented := True
else
if (A[46] = Bmid) and (A[43] = Bmid) and (A[34] = Bmid) and (A[25] = Bmid)
then begin { -- BC2 pattern }
AddToSolution ('B2 R- B2 R+ B+ R- B+ R+ '); Face := 1;
end
else { -- 6 other patterns possible if the cube is at proper rotation. }
if (A[46] = Bmid) and (A[36] = Bmid) and (A[27] = Bmid) and (A[18] = Bmid)
or (A[45] = Bmid) and (A[36] = Bmid) and (A[34] = Bmid) and (A[25] = Bmid)
or (A[16] = Bmid) and (A[36] = Bmid) and (A[54] = Bmid) and (A[48] = Bmid)
or (A[16] = Bmid) and (A[52] = Bmid) and (A[54] = Bmid) and (A[18] = Bmid)
or (A[16] = Bmid) and (A[52] = Bmid) and (A[27] = Bmid) and (A[48] = Bmid)
or (A[45] = Bmid) and (A[43] = Bmid) and (A[27] = Bmid) and (A[25] = Bmid)
then begin { -- BC1 or BC3 or BC4 or BC5 or BC6 or BC7 pattern. }
AddToSolution ('R- B- R+ B- R- B2 R+ B2 '); Face := 1;
end
else begin { -- Cube is not yet at proper rotation to match a pattern. }
AddToSolution ('RC+'); Face := Face + 1;
end;
until All4Oriented;
end;
procedure SolveBottomEdges ({giving} var MoveError: Boolean);
{ -- This procedure will store the moves in Solution to restore bot edges. }
{ -- Strategy: First "position" all 4 edges, then "orient" the 4 edges. }
var
Face: Integer;
Fmid, Rmid, Pmid, Lmid, Bmid: Char;
BF, BR, BP, BL: Boolean; { -- True if this corner exists. }
All4Positioned, All4Oriented: Boolean;
Pattern1, Pattern2, Pattern3: Boolean;
begin
{ -- ****************** Position all 4 edges *************** }
All4Positioned := False; Face := 1;
repeat
If Face > 4 then begin { -- Fatal error }
ErrorInMoves; MoveError := True; Exit; end;
Fmid := A[14]; Rmid := A[23]; Pmid := A[32]; Lmid := A[41];
{ -- Determine which of the 4 Bottom edges are positioned (= True). }
BF := (A[17] = Fmid) or (A[47] = Fmid);
BR := (A[26] = Rmid) or (A[51] = Rmid);
BP := (A[35] = Pmid) or (A[53] = Pmid);
BL := (A[44] = Lmid) or (A[49] = Lmid);
if BF and BR and BP and BL then { -- All 4 edges are positioned. }
All4Positioned := True
else { -- 0 or 1 egde is positioned. There are 2 ways to position them.}
begin
else if BP then AddToSolution ('RC2')
else if BL then AddToSolution ('RC-');
{ -- if 1 edge correctly positioned then it is in the BF position. }
Lmid := A[41]; { -- Cube may have rotated and changed colors. }
if (BF or BR or BP or BL) and ((A[51] = Lmid) or (A[26] = Lmid)) then
{ -- Permute 3 bottom edges: BR -> BL -> BP -> BR }
AddToSolution ('L- R+ F- L+ R- B2 L- R+ F- L+ R- ')
else { -- Either 0 edges positioned, or 3 non-positioned need this. }
{ -- Permute 3 bottom edges: BR -> BP -> BL -> BR }
AddToSolution ('L- R+ F+ L+ R- B2 L- R+ F+ L+ R- ');
end;
Face := Face + 1;
until All4Positioned;
{ -- ****************** Orient 4 edges ****************** }
{ -- Rotate Cube until 1 of 7 patterns appear, or all 4 are oriented. }
{ -- Perform set of moves. If pattern is BC1 or BC2, cube is oriented. }
Bmid := A[50];
{ -- Check if all edges are oriented. }
All4Oriented := (A[47] + A[49] + A[51] + A[53]) = (Bmid+ Bmid+ Bmid+ Bmid);
if not All4Oriented then begin { -- Edges will be in 1 of 3 patterns. }
Pattern1 := (A[47] <> Bmid) and (A[49] <> Bmid) and
(A[51] <> Bmid) and (A[53] <> Bmid);
Pattern2 := (A[47] = Bmid) and (A[53] = Bmid) or
(A[49] = Bmid) and (A[51] = Bmid);
Pattern3 := (A[47] = Bmid) and (A[49] = Bmid) or
(A[49] = Bmid) and (A[53] = Bmid) or
(A[51] = Bmid) and (A[53] = Bmid) or
(A[47] = Bmid) and (A[51] = Bmid);
if Pattern1 then { -- Permute each edge (swap the colors on each) }
AddToSolution('L- R+ F2 L+ R- B2 L- R+ F+ L+ R- B2 L- R+ F2 L+ R- B- ')
else if Pattern2 then begin { -- Permute edges across from each other. }
if (A[47] = Bmid) and (A[53] = Bmid) then { -- Rotate Cube }
AddToSolution('L- R+ F+ L+ R- B+ L- R+ F+ L+ R- B+ L- R+ F2 L+ R- B+ ');
AddToSolution('L- R+ F+ L+ R- B+ L- R+ F+ L+ R- B2 ');
end
else if Pattern3 then begin { -- 4 possible orientations, Rotate cube. }
if (A[47] = Bmid) and (A[49] = Bmid) then
else if (A[49] = Bmid) and (A[53] = Bmid) then
else if (A[51] = Bmid) and (A[53] = Bmid) then
{ -- Rotate BP -> BF -> BL -> BP, then orient 3 edges. }
AddToSolution ('L- R+ F+ L+ R- B- L- R+ F- L+ R- B- L- R+ F2 L+ R- ');
AddToSolution ('RC+L- R+ F+ L+ R- B2 L- R+ F+ L+ R- ');
end
else begin { -- Fatal error, none of the patterns were detected. }
ErrorInMoves; MoveError := True; Exit; end;
end; { -- If not All4Oriented }
end;
## Name and Description of Procedures in Program
program RubeCube;
{ -- Project Name: RubeCube.Pas
-- Author: Douglas E. Woolley
-- Date Started: 8/15/88
-- Last Update: 9/27/88
-- This program will accept the appearance of the Rubik's Cube as input,
-- and output a step-by-step text/graphical solution to restore the puzzle.
}
procedure DisplayBorder ({using} Title: String30; TM: Integer);
{ -- This procedure will display special characters around the perimeter
-- and centers the Title on the top line; TM=0 is width 80, = 1 width 40. }
procedure DisplayTitlePage;
{ -- This procedure will display program name, author, date, etc. }
procedure DisplayMenu ({giving} var Option: Integer);
{ -- This procedure will display Menu and accept valid option. }
procedure Beep;
{ -- This procedure makes a beep for errors. }
procedure InitValidColors;
{ -- This procedure will assign 6 common colors on the Rubik's Cube
-- to the captial global variable ValidColor. }
procedure GetValidColors;
{ -- This procedure will accept 6 valid colors to use for the cube. }
procedure InitArrayOfColors; { -- output is global array of colors. }
{ -- This procedure will set all array items to spaces for color. }
procedure DisplayBox ({using} Side: Integer);
{ -- This procedure displays a box for corresponding side at row,col. }
procedure DisplayColors ({using} Side: Integer);
{ -- This procedure displays the color contents of squares on the Side. }
procedure DisplayTextInstr;
{ -- This procedure will display commands needed for displaying the solution.}
procedure DisplayInputInstr;
{ -- This procedure will display commands needed for inputting colors. }
procedure DisplayBoxes;
{ -- This procedure will display the 6 sides of the cube. }
function Match({using} Cor: String3; A1, A2, A3: Char): {giving} Boolean;
{ -- This function returns True if Cor matches a permutation of A1,A2,A3. }
function InputCorrect: {giving} Boolean;
{ -- This function will output True if 12 unique edges and 8 unique corners
-- are entered with 6 unique middle squares on the 6 sides. }
procedure GetColors;
{ -- This procedure gets the color of squares on each Side.
-- Output: AInit[1..54] are assigned; InputCorrect is True or False. }
procedure GetRCInput;
{ -- This procedure will assign colors to array variables for Rubik's Cube. }
{ -- ************ Computer Coded Solution **************** }
procedure MoveSide ({given} SideToMove: String1; NumOfRot: Integer);
{ -- This procedure will turn Side a total of NumOfRot rotations clockwise. }
procedure MakeMove ({using} NextMove: String3);
{ -- This procedure will process one move by swapping the A array
-- corresponding to the Side being moved. }
procedure Combine ({using} FirstMove: String3);
{ -- This procedure will add the first move to the Solution array, compacting}
{ -- This procedure will first internally move the cube as directed, then
-- Add the set of moves to the entire Solution array, with compacting. }
procedure ErrorInMoves;
{ -- This procedure will display an error message about the solution. }
procedure SolveTopEdges ({giving} var MoveError: Boolean);
{ -- This procedure will store the moves in Solution to restore top edges. }
{ -- Strategy: Work on front-top edge then rotate cube to next face. }
procedure SolveTopCorners ({giving} var MoveError: Boolean);
{ -- This procedure will store the moves in Solution to restore top corners. }
{ -- Strategy: Work on front-right-top corner then rotate cube to next face. }
procedure SolveVerticalEdges ({giving} var MoveError: Boolean);
{ -- This procedure will store the moves in Solution to restore vert. edges. }
{ -- Strategy: Work on front-right edge then rotate cube to next face. }
procedure SolveBottomCorners ({giving} var MoveError: Boolean);
{ -- This procedure will store the moves in Solution to restore bot corners. }
{ -- Strategy: First "position" all 4 corners, then "orient" 4 corners. }
procedure SolveBottomEdges ({giving} var MoveError: Boolean);
{ -- This procedure will store the moves in Solution to restore bot edges. }
{ -- Strategy: First "position" all 4 edges, then "orient" the 4 edges. }
function CubeIsSolved: Boolean;
{ -- This function returns True if each side has 9 squares of the same color.}
procedure GetRCSolution ({using/giving} var MoveError: Boolean);
{ -- This procedure will store all the moves to restore the puzzle in the
-- array Solution[1..200] of string[3], with Solution[LastSolIndex]
-- having the last move of the solution. }
procedure DisplayTextSolution ({using} Option1or2: Integer);
{ -- This procedure will display step-by-step the moves to restore the puzzle
-- in the array Solution[1..200] of string[3], with Solution[LastSolIndex]
-- having the last move of the solution.
-- If Option1or2 = 2 then display is text. }
procedure MagnifyLetter (L: Char; {by} M: Integer; {at} X, Y: Integer;
{using} Co: Integer);
{ -- This procedure will magnify the letter L by M times at position X, Y.
-- If the character L is not available (such as a space), then no display. }
procedure DisplayInstr;
{ -- This procedure will display a brief description of each menu option. }
procedure DisplayGraphicsInstr;
{ -- This procedure will display commands needed for displaying the solution.}
{ -- 4 colors available (Graphics and Text):
0 = Background, 1 = LightCyan, 2 = LightMagenta, 3 = White. }
procedure DisplayGraphics;
{ -- This procedure will display the Graphical cube. }
procedure DisplayGraphicsColors ({using} Side, Co: Integer);
{ -- This procedure displays the color contents of squares on the Side
if Co is 1, 2, or 3. If Co = 0 then the previous colors are erased. }
procedure DisplayGraphicalSolution;
{ -- This procedure will display step-by-step the moves to restore the puzzle
-- in the array Solution[1..200] of string[3], with Solution[LastSolIndex]
-- having the last move of the solution. One side of a graphical cube is
-- displayed. }
{ -- **************** Test Program routine **************** }
procedure PutColorsOnCube; { -- output is global array of colors. }
{ -- This procedure will set all array items to Valid colors for a cube. }
procedure GetNumber ({using} Col, Row: Integer; {giving} var Number: Integer);
{ -- This procedure will accept as input a 2 digit number. }
procedure TestProgram ({using} TestOption: Integer);
{ -- This procedure will display statistics for solving random cubes. }
{ -- *********** Main Flow of Program ************* }
begin
DisplayTitlePage;
InitValidColors;
repeat
case Option of
1, 2: begin
GetRCInput;
if CorrectInput then begin
GetRCSolution (MoveError);
if not MoveError then
if Option = 1 then { -- Graphical Solution }
DisplayGraphicalSolution
else { -- Option = 2 Text Solution }
DisplayTextSolution (Option)
end;
end;
3: DisplayInstr;
4: GetValidColors;
5: TestProgram (Option);
end;
until Option = LastOption;
ClrScr;
end.
## Appendix - Variations of the Rubik's Cube
The Picture Cube, although mechanically similar to Rubik's Cube, has pictures on the six sides instead of solid colors. One such cube is called the Pac Man Picture Cube which has Pac Men, ghosts, and keys on the square patches. The difference between the Rubik's Cube and this Picture Cube is that the latter's center could be incorrectly oriented. Because of this variation, this Cube is harder to solve than the Rubik's Cube and has 88,580,102,706,155,225,088,000 possible combinations.
Rubik's Pocket Cube is a two by two by two inch cube that has six sides which contain a different color. The sides of this puzzle move independently of each other. Rubik's Pocket Cube has no center piece on any of its sides and has only 3,674,160 different arrangements.
Rubik's Revenge is a four by four by four cube. This is the hardest puzzle on the market because of its incredible number of arrangements: 3.7 x (10 to the 45th power). The Rubik's Cube has approximately 4.3 x (10 to the 19th power). Rubik's Revenge has 8 corner pieces, 24 middle edge pieces, and 24 center pieces.
## Bibliography
Alexander, Ron, "A Cube Popular in all Circles", New York Times, July 21, 1981, III, 6:2.
Borders, William, "Best-Selling Author, 13, Thanks Rubik's Cube", New York Times, October 17, 1981, I, 3:5.
Carmichael, Jane, "Figure This One Out", Forbes, p. 34, May 1981.
Dougherty, Phillip H., "Ideal Toy's Son of Rubik Cube", New York Times, July 30, 1981, IV, 19:1.
Dubisch, Roy, Introduction to Abstract Algebra, p. 70, Seattle, Washington, John Wiley & Sons, Inc., 1965.
Editors of Discover, "Robot Cubist", Discover, June 1982.
Editors of Time, "Hot-Selling Hungarian Horror", Time, p. 83, March 23, 1981.
Ewing, John, and Kosniowski, Czes, Puzzle It Out, New York, NY, Cambridge University Press, 1982.
Faludi, Susan C., "CBS Unit Offers \$58 Million for Makers of Rubik's Cube", New York Times, April 23, 1982, IV, 3:1.
Hauptfuhrer, Fred, "An Obscure Hungarian Professor Transforms America into a Nation of Cubic Rubes", People, p. 31, May 1981.
Nourse, James G., The Simple Solution to Rubik's Cube, New York, NY, Bantom Books 1981.
Nourse, James G., The Simple Solution to Cubic Puzzles, New York, NY, Bantom Books 1981.
Roman, Mark, "Rubik's Cube: Ideal Toy Takes on the Knock-Offs", New York Times, October 4, 1981, III, 21:1.
Sanger, David E., "Rubik's Rival", NY Times, June 6, 1982, III, 19:4.
Sigmaster, David, Notes on Rubik's Magic Cube, Hillside, NJ, Enslow Publishers, 1981.
Thai, Minh, The Winning Solution to Rubik's Revenge, pp. 6,7,9, Wayne, Pennsylvania, Banbury Books, Inc., 1982.
Warshofsky, Fred, "Rubik's Cube: Madness for Millions", Readers Digest, pp. 138-139, May 1981.
## Author's Biography
Douglas E. Woolley is currently a senior at the University of South Florida in Tampa, majoring in Computer Science and minoring in Mathematics. Doug is a member of three national honor societies and is active in intramural sports. He is a teacher and leader in a Christian church and has been a guest speaker at high schools, banquets, churches, and conventions.
PROGRAMMING EXPERIENCE
1984-present
Florida Center for Instructional Computing, Tampa, Fl
Senior Computer Programmer
· Developed and maintained code for Instructional Management System which is designed for prescription learning and providing student and administrative reports. It is presently being used in Federally funded instructural environments.
· Author and official judge of the annual Florida High School Computer Contest items.
· Developed and maintained code for The Micro Goes to School (IBM and Apple versions). The four diskettes that accompany these books demonstrate the integration of instructional software in existing school curricula.
· Developed and maintained code for the "Computer Contest Database" which stores attributes of contest items and creates contests.
· Guest speaker at annual statewide computer conventions to demonstrate computer programs developed and maintained.
AWARDS AND HONORS
Key note speaker at State Mu Alpha Theta (MAT) convention (1985)
** Ranked number one in Florida in computer science (1984)
** Ranked number one in Florida in mathematics (1984)
Third at national (MAT) convention in mathematics (1984)
Fifth at national (MAT) convention in computer programming (1984)
Captain of first place Florida computer team (1984)
Recognized as one of the top 300 high school scientists in America by Westinghouse Science Talent Search (1984)
PUBLICATIONS
The Micro Goes to School: Instructional Applications of Microcomputer Technology, IBM and Apple Versions (Brooks/Cole Publishing Company) | 18,747 | 69,825 | {"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} | 3.6875 | 4 | CC-MAIN-2020-16 | latest | en | 0.902843 |
https://cstheory.stackexchange.com/questions/42018/is-this-a-knapsack-problem | 1,718,653,614,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861737.17/warc/CC-MAIN-20240617184943-20240617214943-00688.warc.gz | 158,104,004 | 39,133 | # Is this a knapsack problem?
I have a set of $$K$$ keywords. Each of this keywords can have a set of bids from $$1\,\ldots, N\$$. For each bid for a keyword, it will get a specific amount of clicks and a specific cost. Clicks and Cost are monotonic in bids. Now, the problem is:
Select a bid for each of the keyword such that overall clicks across the keywords are maximized such that the cost is still within a given budget $$B$$. Let $$\text{Clicks}(W_i)$$ and $$\text{Cost}(W_i)$$ denote the clicks and cost on the keyword $$i$$. Then (informally) the problem is to select a value of bid for each keyword such that \begin{align}\max\sum_{i=1}^{K}&\text{Clicks}(W_i) \\s.t.&~~\sum_{i=1}^{K}\text{Cost}(W_i)\leq B\end{align} Is this a type of knapsack problem? Are there any known heuristic in the literature.
Following is not important for the question (and thus can be skipped), but is for the curious ones to know how I am solving this currently.
Let $$S_{ij}$$ denote the cost for using bid $$j$$ for keyword $$i$$ and let $$C_{ij}$$ be the clicks for the same. Let $$x_{ij}$$ be the indicator variable for using bid $$j$$ for keyword $$i$$. Then among variables $$x_{i1},\dots,x_{iN}$$ only one can be non-zero (i.e. $$x_{ij}=1$$) whereas the rest is zero (since only one bid can chosen for keyword).
Thus, the whole problem can be formulated mathematically as
\begin{align}\max_{x_{ij}}~&\sum_{i=1}^{K}\sum_{j=1}^{N}x_{ij}C_{ij} \\s.t.~~\sum_{i=1}^{K}\sum_{j=1}^{N}&x_{ij}S_{ij}\leq B\\\sum_{j=1}^{N}x_{ij}=1&~~\forall i\in \{1,\dots,K\}~~,~~x_{ij}\in\{0,1\}\end{align}
Last, two constraints ensure they are indicator variables and only one bid is chosen for a given keyword. Solving this should solve the original problem.
Yes, it is called a multiple-choice knapsack problem. It can be solved in time $$O(N B)$$ by dynamic programming. The state of the dynamic program is given by a pair $$(i, b)$$ and the associated value is the maximum value (clicks) you can get with keywords from $$1$$ to $$i$$ with a budget at most $$b$$. | 622 | 2,046 | {"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": 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": 22, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-26 | latest | en | 0.804816 |
http://myriverside.sd43.bc.ca/gracem2015/2018/02/ | 1,686,127,482,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653631.71/warc/CC-MAIN-20230607074914-20230607104914-00673.warc.gz | 33,415,816 | 10,746 | # Exploring Quadratic Functions
1. Definition: A quadratic function is the degree of 2. Can be written in this form ( y = a$x^2$ + bx + c)
graph of every quadratic function is called a parabola.
2. examples
a) function with:
b) function without:
3. desmos
a) see what happens:
graph forms a dip like thing on the graph. when you move the slider it changes in width, length and height.
b) Start with slider values a=1,b=0,c=0. Describe any symmetry you notice:
This turned the graph to be exactly center on the 0 line on the graph. Everything looked organized and perfectly lined up.
4. Compare two functions:
x² + 7x + 12 = 0
find two numbers which come to 7 and multiply to be 12. You immediately come up with (3, 4). which would be both your x values.
5. Describe what happens to the graph when:
a) a<0 Does the graph have a maximum point or a minimum point? - no? b) a>0
Does the graph have a maximum point or minimum point?
– no?
c) -11 or a<-1 - i still dont know 6. Make two statements that describes the relationship between the sign of a (positive and negative) and whether the vertex is a maximum or minimum: a) im still very confused by this b) same here 7. ? | 311 | 1,180 | {"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": 1, "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} | 3.859375 | 4 | CC-MAIN-2023-23 | latest | en | 0.877367 |
http://math.stackexchange.com/questions/9225/probability-of-the-propagation-of-a-rumor-problem | 1,469,592,059,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825365.1/warc/CC-MAIN-20160723071025-00244-ip-10-185-27-174.ec2.internal.warc.gz | 148,083,337 | 18,694 | # Probability of the propagation of a rumor problem
On a small island there are 25 inhabitants. One of these inhabitants, named Jack, starts a rumor which spreads around the isle. Any person who hears the rumor continues spreading it until he or she meets someone who has heard the story before. At that point, the person stops spreading it, since nobody likes to spread stale news.
a) Do you think all 25 inhabitants will eventually hear the rumor or will the rumor die out before that happens? Estimate the proportion of inhabitants who will hear the rumor.
How would one go about calculating the probability? I would try to block it out, but there are so many potential branches. This problem is in the simulations portion of our statistics book, but I'm curious if there is any other way that doesn't involve simulating the situation.
-
Take a look at the logistic function. – a.r. Nov 7 '10 at 3:00
Mike Spivey's remark that there are only three classes of people is quite relevant if you want to simulate this. At the start, there are 24 susceptible people and one infected. Call the number of each category S, I, and R. If two I meet, they both become R, so I goes down by 2 and R goes up by 2. If an S meets an I, the S becomes I, so S goes down by 1 and I goes up by 1. If R meets I, I goes down by 1 and R goes up by 1. Other meetings have no result. My guess is that somebody doesn't hear, because as S falls more and more meetings deplete the I's.
-
This description also gives a dynamic programming algorithm (q.v.) that can find the exact probability in $O(n^2)$, where $n = 25$ in our case. – Yuval Filmus Nov 7 '10 at 5:52
An SIR model is also appropriate because of this aspect of the problem: "Any person who hears the rumor continues spreading it until he or she meets someone who has heard the story before. At that point, the person stops spreading it, since nobody likes to spread stale news." Here, susceptible = hasn't heard rumor, infected = has heard rumor but is still willing to spread it, and recovered = has heard rumor but is no longer willing to spread it.
I see you're in 9th grade, and so the differential equations in the SIR model may not be something you've seen yet. Still, you may get something out of reading the explanation of the model on the Wikipedia page. Also, the standard SIR model doesn't quite fit this scenario; the equations for the infected and recovered populations for your scenario should be
$$\frac{dI}{dt} = \beta SI - \gamma RI,$$ $$\frac{dR}{dt} = \gamma RI.$$
I know you said you only wanted hints, but I'm giving you a fuller explanation of the SIR model because, according to this paper, "While the formal model is simple, in general it cannot be solved analytically." So this may just be one of those problems for which simulation is the best solution we have now.
O.K., I have to add this: For fun, you should also check out the SIRZ (susceptible-infected-recovered-zombie) model. We can all be thankful that the authors say, "Most of the analysis concluded that, under reasonable circumstances, a zombie apocalypse is unlikely."
- | 734 | 3,108 | {"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} | 3.671875 | 4 | CC-MAIN-2016-30 | latest | en | 0.970394 |
https://www.engineeringtoolbox.com/amp/equilibrium-d_943.html | 1,718,486,837,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00829.warc.gz | 687,592,414 | 7,322 | Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications!
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Support Reactions - Equilibrium
Equilibrium of a body requires both a balance of forces to prevent the body from translating or having accelerated motion along a straight or curved path - and a balance of moments to prevent the body from rotating.
Any static force system will be in equilibrium if the resultant force and resultant moment both are equal to zero.
Static equilibrium in a three dimensional system can be expressed as
Σ F = Σ F x = Σ F y = Σ F z = 0 (1)
Σ M = Σ M x = Σ M y = Σ M z = 0 (2)
where
F = force (N, lb)
M = moment (Nm, ft lb)
x, y, z = orthogonal axes
Often the loading of a body can be simplified to a two dimensional system with co-planar forces in the x-y plane. Eq. 1 and 2 can be reduced to
Σ F = Σ F x = Σ F y = 0 (3)
Σ M = Σ M z = 0 (4)
The best way to account for all forces acting on a body is to draw the body's free-body diagram . A free-body diagram shows the relative magnitude and direction of all forces acting upon an object in a given situation.
Free-body diagram example - gravity and friction forces acting on a body on an inclined plane
.
Example - Support Reactions on a Beam with Eccentric Load
A beam with length 6 m has an eccentric load of 9000 N 4 m from support 1 . Applying the equations of equilibrium we have
F x = R 1x = R 2x = 0 (5)
F y - (R 1y + R 2y ) = 0 (6)
M1 = F y a - R 2y (a + b) = 0 (7)
Rearranging (7) to express R 2y
R 2y = F a / (a + b) (7b)
Eq. (7b) with values
R2= (9000 N) (4 m) / ((4 m) + (2 m))
= 6000 N
= 6 kN
Rearranging (6) for R 1y
R 1y = F y - R 2y (6b)
Eq. (6b) with values
R 1y = (9000 N) - (6000 N)
= 3000 N
= 3 kN
.
Example - Reaction Forces on a Structural Frame
A weight F (1000 N) is hanging in a structural frame as shown in the figure above. A structural analyses can be done with the following equations:
∑M A = R By (230 cm) - (1000 N) (100 cm) = 0 (1)
∑M B = R Ay (230 cm) - (1000 N) (130 cm) = 0 (2)
∑M C-B = R By (130 cm) - R Bx h = 0 (3)
F = R Ay + R By (4)
R Ax = R Bx (5)
Eq. 1 can be rearranged to
R By = (1000 N) (100 cm) / (230 cm)
= 435 N
Eq. 2 can be rearranged to
R Ay = (1000 N) (130 cm) / (230 cm)
= 565 N
The height h in eq. 3 can be calculated as
h = (1532- 1302) 1/2
= 80.7 cm
Eq. 3 can then be rearranged to
R Bx = (435 N) (130 cm) / (80.7 cm)
= 700 N
From eq. 5
R Ax = R Bx
= 700 N
Related Topics
• Mechanics
The relationships between forces, acceleration, displacement, vectors, motion, momentum, energy of objects and more.
• Statics
Forces acting on bodies at rest under equilibrium conditions - loads, forces and torque, beams and columns.
Related Documents
Acceleration of Gravity and Newton's Second Law
Acceleration of gravity and Newton's Second Law - SI and Imperial units.
Bodies Moving on Inclined Planes - Acting Forces
Required forces to move bodies up inclined planes.
Force
Newton's third law - force vs. mass and acceleration.
Impulse and Impulse Force
Forces acting a very short time are called impulse forces.
Mass Moment of Inertia
The Mass Moment of Inertia vs. mass of object, it's shape and relative point of rotation - the Radius of Gyration.
Mass vs. Weight
Mass vs. weight - the Gravity Force.
Torque or Moment of Force - Online Converter
Torque or moment - the tendency of a force to rotate an object.
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Add standard and customized parametric components - like flange beams, lumbers, piping, stairs and more - to your Sketchup model with the Engineering ToolBox - SketchUp Extension - enabled for use with the amazing, fun and free SketchUp Make and SketchUp Pro . Add the Engineering ToolBox extension to your SketchUp from the Sketchup Extension Warehouse! | 1,191 | 4,278 | {"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-2024-26 | latest | en | 0.766411 |
http://forums.wolfram.com/mathgroup/archive/2001/Feb/msg00308.html | 1,527,343,342,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867417.75/warc/CC-MAIN-20180526131802-20180526151802-00355.warc.gz | 119,542,952 | 7,777 | Re: Problem: a matrix and a box...
• To: mathgroup at smc.vnet.net
• Subject: [mg27277] Re: [mg27268] Problem: a matrix and a box...
• From: Tomas Garza <tgarza01 at prodigy.net.mx>
• Date: Sun, 18 Feb 2001 02:52:12 -0500 (EST)
• References: <[email protected]>
• Sender: owner-wri-mathgroup at wolfram.com
```Let mat be your n x m matrix. Try
In[1]:=
Table[Plus @@ (Part[mat, {i, i + 1}, {j, j + 1}] // Flatten), {i, 1,
Dimensions[mat][[1]] - 1}, {j, 1, Dimensions[mat][[2]] - 1}]
Example:
In[2]:=
mat = {{a1, a2, a3, a4, a5}, {b1, b2, b3, b4, b5}, {c1, c2, c3, c4, c5},
{d1,
d2, d3, d4, d5}, {e1, e2, e3, e4, e5}, {f1, f2, f3, f4, f5}};
In[3]:=
Table[Plus @@ (Part[mat, {i, i + 1}, {j, j + 1}] // Flatten), {i, 1,
Dimensions[mat][[1]] - 1}, {j, 1, Dimensions[mat][[2]] - 1}]
Out[3]=
{{a1 + a2 + b1 + b2, a2 + a3 + b2 + b3, a3 + a4 + b3 + b4,
a4 + a5 + b4 + b5}, {b1 + b2 + c1 + c2, b2 + b3 + c2 + c3,
b3 + b4 + c3 + c4, b4 + b5 + c4 + c5}, {c1 + c2 + d1 + d2,
c2 + c3 + d2 + d3, c3 + c4 + d3 + d4,
c4 + c5 + d4 + d5}, {d1 + d2 + e1 + e2, d2 + d3 + e2 + e3,
d3 + d4 + e3 + e4, d4 + d5 + e4 + e5}, {e1 + e2 + f1 + f2,
e2 + e3 + f2 + f3, e3 + e4 + f3 + f4, e4 + e5 + f4 + f5}}
I don't claim this is the fastest but it works. For example, for a matrix
about the size you have (in a PC at 800 MHz),
In[4]:=
mat = Table[Random[Integer, {0, 9}], {3000}, {3005}];
In[4]:=
Table[Plus @@ (Part[mat, {i, i + 1}, {j, j + 1}] // Flatten), {i, 1,
Dimensions[mat][[1]] - 1}, {j, 1,
Dimensions[mat][[2]] - 1}]; // Timing
Out[59]=
{57.45 Second, Null}
Tomas Garza
Mexico City
----- Original Message -----
From: "Bedrosian Baol" <Bedrosian at MailAndNews.com>
To: mathgroup at smc.vnet.net
Subject: [mg27277] [mg27268] Problem: a matrix and a box...
> Hi
> this is the problem: I have a n x m matrix (with n and m ~3000 elements or
> more)
>
> a1 a2 a3..
> b1 b2 b3..
> c1 c2 c3..
> . . .
> . . .
>
> the algorythm is: compute the summ of the the a1, a2, b1 and b2 elements
(a
> ''gliding box'' 2x2), shifing the ''box'' for 1 column and repeat the sum
of
> the elements (now a2, a3, b2, b3), and so on...
> at the and of the first line, the box return on the second line and
> repeat...
>
> the final resuls would be another matrix like this:
>
> [a1+a2+b1+b2] [a2+a3+b2+b3]...
> [b1+b2+c1+c2] [b2+b3+c2+c3]...
> ...
>
> I'm not very experienced with Math4: there is a simple and easy way for
> reach
> the result?
>
> thanks
> Bed
>
>
```
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• Next by thread: Re: Problem: a matrix and a box... | 1,158 | 2,705 | {"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-2018-22 | latest | en | 0.380965 |
https://admin.clutchprep.com/physics/practice-problems/138193/an-underwater-diver-sees-the-sun-60-above-horizontal-how-high-is-the-sun-above-t | 1,596,910,245,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738015.38/warc/CC-MAIN-20200808165417-20200808195417-00077.warc.gz | 198,359,684 | 21,624 | # Problem: An underwater diver sees the sun 60° above horizontal.How high is the sun above the horizon to a fisherman in a boat above the diver? Express your answer using two significant figures.
###### FREE Expert Solution
The angle above horizontal = 60°
We'll define the angle of incidence as the angle at which the ray from the sun strikes the interface.
The boat is at the boundary. Therefore, the target is angle relating h1.
The angle of refraction = 90 - 60 = 30°
Use Snell's Law to relate the Angle of Incidence with the Angle of Refraction:
###### Problem Details
An underwater diver sees the sun 60° above horizontal.
How high is the sun above the horizon to a fisherman in a boat above the diver? Express your answer using two significant figures. | 176 | 768 | {"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} | 3.6875 | 4 | CC-MAIN-2020-34 | latest | en | 0.778695 |
https://goprep.co/ex-3.3-q3-number-of-children-in-six-different-classes-are-i-1njssy | 1,590,962,485,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413786.46/warc/CC-MAIN-20200531213917-20200601003917-00529.warc.gz | 386,666,285 | 27,901 | # Number of children in six different classes are given below. Represent the data on a bar graph.(a) How would you choose a scale?(b) Answer the following questions:(i) Which class has the maximum number of children? And the minimum?(ii) Find the ratio of students of class sixth to the students of class eighth.
The bar graph of the given data is represented below:
(a) Here,
We will choose a scale as follows:
1 unit = 10 children
We will take this scale because with this scale we can represent a better and clear difference among the number of students of class 7th and class 9th.
(b) .
(i) We can see that,
The tallest bar is of class fifth
Hence,
We can say that,
The maximum number of students are in class fifth.
Similarly,
We can observe that,
The smallest bar is of class tenth
Hence,
We can say that,
There are least number of students in class tenth.
(ii) Now,
Here,
We can clearly see that,
There are 120 students in class sixth
And,
There are 100 students in class eighth
Hence,
The ratio between the number of students of class sixth and the number of students of class eighth can be calculated as follows:
=
=
= 6 : 5
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better. | 294 | 1,289 | {"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 | longest | en | 0.9575 |
https://yuihuang.com/zj-d813/ | 1,713,464,224,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817222.1/warc/CC-MAIN-20240418160034-20240418190034-00544.warc.gz | 1,020,990,728 | 16,760 | # 【題解】ZeroJudge d813: 10583 – Ubiquitous Religions
【題目敘述】https://zerojudge.tw/ShowProblem?problemid=d813 (考生答對率: 2.46%)
【解題想法】並查集
```#include <iostream>
using namespace std;
const int maxn = 50000+5;
int pa[maxn];
int Find(int x){
return pa[x] == x ? x : pa[x] = Find(pa[x]);
}
bool Union(int x, int y){
int g1 = Find(x);
int g2 = Find(y);
if (g1 == g2) return false;
else{
pa[g2] = g1;
return true;
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int n, m, x, y, Case = 1;
while (cin >> n >> m && n+m){
for (int i = 0; i < maxn; i++){
pa[i] = i;
}
int ans = n;
while (m--){
cin >> x >> y;
if (Union(x, y)) {
ans--;
}
}
cout << "Case " << Case++ << ": ";
cout << ans << "\n";
}
return 0;
}
```
Python code (credit: Amy Chou)
• 注意：測資最後並非以 0 0 結束。
```import sys
def Find(x):
if (pa[x] == x):
return x
else:
pa[x] = Find(pa[x])
return pa[x]
def Union(x, y):
g1 = Find(x)
g2 = Find(y)
if (g1 == g2):
return False
else:
pa[g2] = g1
return True
idx = -1
Case = 0
while True:
try:
idx += 1
n, m = map(int, lines[idx].split())
if n == 0 and m == 0:
break
Case += 1
pa = []
for i in range(n+1):
pa.append(i)
ans = n
for i in range(m):
idx += 1
x, y = map(int, lines[idx].split())
if (Union(x, y)):
ans -= 1
print(f"Case {Case}: {ans}")
except:
break
``` | 497 | 1,267 | {"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-2024-18 | latest | en | 0.240417 |
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# 1115 ch 11 day 15
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### 1115 ch 11 day 15
1. 1. 11.6 The Binomial Theorem Day TwoHebrews 4:16 "Let us therefore come boldly unto the throneof grace that we may obtain mercy, and find grace to help intime of need."
2. 2. From the Binomial Theorem, a specificterm is represented by: ⎛ n ⎞ n−r r ⎜ r ⎟ x y ⎝ ⎠
3. 3. From the Binomial Theorem, a specificterm is represented by: ⎛ n ⎞ n−r r ⎜ r ⎟ x y ⎝ ⎠ (note: r is 1 less than the term number)
4. 4. From the Binomial Theorem, a specificterm is represented by: ⎛ n ⎞ n−r r ⎜ r ⎟ x y ⎝ ⎠ (note: r is 1 less than the term number)This formula is a variation of what is in your book.
5. 5. Example: Find the 5th term of ( a + b ) 6
6. 6. Example: Find the 5th term of ( a + b ) 6 n=6 r=4
7. 7. Example: Find the 5th term of ( a + b ) 6 n=6 r=4 ⎛ 6 ⎞ 2 4 ⎜ 4 ⎟ ab ⎝ ⎠ 2 4 15a b
8. 8. Example: Find the 5th term of ( a + b ) 6 n=6 r=4 ⎛ 6 ⎞ 2 4 ⎜ 4 ⎟ ab ⎝ ⎠ 2 4 15a b 6 5 1 4 2 3 3 2 4 1 5 6 a + 6a b + 15a b + 20a b + 15a b + 6a b + b term 5
9. 9. Groups: Find the 5th term of ( 3x − 5y ) 10
10. 10. Groups: Find the 5th term of ( 3x − 5y ) 10 n = 10 r=4 ⎛ 10 ⎞ 6 4 ⎜ 4 ⎟ ab ⎝ ⎠ ⎛ 10 ⎞ 6 4 ⎜ 4 ⎟ ( 3x ) ( −5y ) ⎝ ⎠ ⎛ 10 ⎞ 6 4 6 4 ⎜ 4 ⎟ ( 3) ( −5 ) x y ⎝ ⎠ 6 4 95,681,250x y
11. 11. Groups: Find the term that contains x in 4 the expansion of ( 3x + y ) 15
12. 12. Groups: Find the term that contains x in 4 the expansion of ( 3x + y ) 15 n = 15 r = 11 ⎛ 15 ⎞ 4 11 ⎜ 11 ⎟ ( 3x ) y ⎝ ⎠ 4 11 110,565x y
13. 13. HW #12“If we take people as we find them, we may make themworse, but if we treat them as though they are whatthey should be, we help them to become what they arecapable of becoming.” Johann Wolfgang von Goethe | 1,049 | 1,963 | {"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-2018-09 | longest | en | 0.789087 |
http://www.jiskha.com/display.cgi?id=1289954555 | 1,498,468,541,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320695.49/warc/CC-MAIN-20170626083037-20170626103037-00263.warc.gz | 572,488,761 | 3,731 | # calculus
posted by .
With waht speed must a baseball be thrown upward from the ground in order to reach the top of the Washington Monment? (The top of the monument is 55o ft)
• calculus -
Well, we would have to ignore air resistance which would have a major effect here but (using english units because you used feet):
v = Vo - 32 t
v = 0 at top
Vo = 32 t
h = 0 + Vo t - 16 t^2
550 = 32 t^2 - 16 t^2 = 16 t^2
t^2 = 34.4
t = 5.86 seconds
Vo = 32 t = 187 feet/second | 156 | 473 | {"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-2017-26 | latest | en | 0.920708 |
http://www.mathapachi.com/2016/03/consecutive-odd-number-cat-1997-question/ | 1,524,317,237,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945222.55/warc/CC-MAIN-20180421125711-20180421145711-00514.warc.gz | 444,538,424 | 5,872 | Home » CAT Questions » Consecutive Odd Number – CAT 1997 Question
# Consecutive Odd Number – CAT 1997 Question
Q.)P,Q and R are three consecutive odd numbers in ascending order. If the value of three times P is 3 less than two times R, find the value of R. [CAT 1997 Question]
(a)5
(b)7
(c)9
(d)11
Solution :
Let the three consecutive odd numbers P,Q and R be X , X+2 and X+4.
value of three times P is 3 less than two times R
=> 2(X+4) – 3X =3
=> 2X + 8 – 3X = 3
=> 8 – 3 – X = 0
=> 5 = X
Hence , the value of R = X+4 = 5+4 = 9 | 192 | 538 | {"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 | latest | en | 0.855416 |
http://cn.metamath.org/mpeuni/fiprc.html | 1,653,454,557,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662578939.73/warc/CC-MAIN-20220525023952-20220525053952-00418.warc.gz | 11,269,329 | 4,873 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > fiprc Structured version Visualization version GIF version
Theorem fiprc 8080
Description: The class of finite sets is a proper class. (Contributed by Jeff Hankins, 3-Oct-2008.)
Assertion
Ref Expression
fiprc Fin ∉ V
Proof of Theorem fiprc
Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 snnex 7008 . 2 {𝑥 ∣ ∃𝑦 𝑥 = {𝑦}} ∉ V
2 snfi 8079 . . . . . . . 8 {𝑦} ∈ Fin
3 eleq1 2718 . . . . . . . 8 (𝑥 = {𝑦} → (𝑥 ∈ Fin ↔ {𝑦} ∈ Fin))
42, 3mpbiri 248 . . . . . . 7 (𝑥 = {𝑦} → 𝑥 ∈ Fin)
54exlimiv 1898 . . . . . 6 (∃𝑦 𝑥 = {𝑦} → 𝑥 ∈ Fin)
65abssi 3710 . . . . 5 {𝑥 ∣ ∃𝑦 𝑥 = {𝑦}} ⊆ Fin
7 ssexg 4837 . . . . 5 (({𝑥 ∣ ∃𝑦 𝑥 = {𝑦}} ⊆ Fin ∧ Fin ∈ V) → {𝑥 ∣ ∃𝑦 𝑥 = {𝑦}} ∈ V)
86, 7mpan 706 . . . 4 (Fin ∈ V → {𝑥 ∣ ∃𝑦 𝑥 = {𝑦}} ∈ V)
98con3i 150 . . 3 (¬ {𝑥 ∣ ∃𝑦 𝑥 = {𝑦}} ∈ V → ¬ Fin ∈ V)
10 df-nel 2927 . . 3 ({𝑥 ∣ ∃𝑦 𝑥 = {𝑦}} ∉ V ↔ ¬ {𝑥 ∣ ∃𝑦 𝑥 = {𝑦}} ∈ V)
11 df-nel 2927 . . 3 (Fin ∉ V ↔ ¬ Fin ∈ V)
129, 10, 113imtr4i 281 . 2 ({𝑥 ∣ ∃𝑦 𝑥 = {𝑦}} ∉ V → Fin ∉ V)
131, 12ax-mp 5 1 Fin ∉ V
Colors of variables: wff setvar class Syntax hints: ¬ wn 3 = wceq 1523 ∃wex 1744 ∈ wcel 2030 {cab 2637 ∉ wnel 2926 Vcvv 3231 ⊆ wss 3607 {csn 4210 Fincfn 7997 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1762 ax-4 1777 ax-5 1879 ax-6 1945 ax-7 1981 ax-8 2032 ax-9 2039 ax-10 2059 ax-11 2074 ax-12 2087 ax-13 2282 ax-ext 2631 ax-sep 4814 ax-nul 4822 ax-pow 4873 ax-pr 4936 ax-un 6991 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-3or 1055 df-3an 1056 df-tru 1526 df-ex 1745 df-nf 1750 df-sb 1938 df-eu 2502 df-mo 2503 df-clab 2638 df-cleq 2644 df-clel 2647 df-nfc 2782 df-ne 2824 df-nel 2927 df-ral 2946 df-rex 2947 df-rab 2950 df-v 3233 df-sbc 3469 df-dif 3610 df-un 3612 df-in 3614 df-ss 3621 df-pss 3623 df-nul 3949 df-if 4120 df-pw 4193 df-sn 4211 df-pr 4213 df-tp 4215 df-op 4217 df-uni 4469 df-iun 4554 df-br 4686 df-opab 4746 df-tr 4786 df-id 5053 df-eprel 5058 df-po 5064 df-so 5065 df-fr 5102 df-we 5104 df-xp 5149 df-rel 5150 df-cnv 5151 df-co 5152 df-dm 5153 df-rn 5154 df-res 5155 df-ima 5156 df-ord 5764 df-on 5765 df-lim 5766 df-suc 5767 df-fun 5928 df-fn 5929 df-f 5930 df-f1 5931 df-fo 5932 df-f1o 5933 df-om 7108 df-1o 7605 df-en 7998 df-fin 8001 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator | 1,424 | 2,532 | {"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} | 3.546875 | 4 | CC-MAIN-2022-21 | latest | en | 0.144675 |
http://mathhelpforum.com/calculus/47401-huge-favour-ask-couple-graphs.html | 1,527,125,268,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865863.76/warc/CC-MAIN-20180523235059-20180524015059-00288.warc.gz | 188,331,925 | 9,958 | 1. ## Huge favour to ask, a couple of graphs.
Hi everyone.
I'm currently studying to be a maths teacher and i have to present the answer to this problem in two days to other student teachers:
manufacturer who makes replica norman windows. A norman window can be modelled as a rectangle surmounted by a semicircle. The length of framing for the heavy outside of th window is limited to 12m.
a) Find the dimensions of the window so that the maximum amount of light will be allowed into the building that has the window fitted.
b) The window maker did not use the optimum design. Why not?
So there are a couple of ways i'm solving it, algebraically and graphically. My problem is that i don't have decent software (besides excel) to get some decent graphs going on.
Would anyone please find it in their hearts to send me a couple of jpeg's of some graphs.
the two graphs i'm looking for are
Area = 12x - 2x^2 - (PI/2)x^2
and
A' = 12 - 4x - PIx
those two graphs would be great.
Also, if anyone has any interesting insights into problems like this, or even this one in particular, i'd love to hear it. It ends up being the optimum height of the rectangle is half the base of the rectangle, which i thought was interesting, but i can't explain why that is instinctively.
Thanks so much for your time
Joel
2. Originally Posted by bruxism
Hi everyone.
I'm currently studying to be a maths teacher and i have to present the answer to this problem in two days to other student teachers:
manufacturer who makes replica norman windows. A norman window can be modelled as a rectangle surmounted by a semicircle. The length of framing for the heavy outside of th window is limited to 12m.
a) Find the dimensions of the window so that the maximum amount of light will be allowed into the building that has the window fitted.
b) The window maker did not use the optimum design. Why not?
[snip]
if anyone has any interesting insights into problems like this, or even this one in particular, i'd love to hear it. It ends up being the optimum height of the rectangle is half the base of the rectangle, which i thought was interesting, but i can't explain why that is instinctively.
Thanks so much for your time
Joel
There are several salient threads to be found in these forums. Eg. http://www.mathhelpforum.com/math-he...w-problem.html | 544 | 2,336 | {"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} | 3.578125 | 4 | CC-MAIN-2018-22 | latest | en | 0.967896 |
https://www.mbatious.com/topic/511/question-bank-logical-reasoning-shashank-prabhu-cat-100-percentiler/18 | 1,597,233,966,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738892.21/warc/CC-MAIN-20200812112531-20200812142531-00537.warc.gz | 749,883,763 | 33,470 | # Question Bank - Logical Reasoning - Shashank Prabhu, CAT 100 Percentiler
• @Rajiv @Pramod-Beri
Who sat opposite to Lata? - Shankar
who sat opposite to Udit? - Kavita
• @shashank_prabhu
1.D
2.A
3.E
4.A
is it right ?
• @shashank_prabhu
Solution:
(1) L and N always together; And Quantity of N = twice as L
(2) M and O always together; And Quantity of M = O
(3) N and O NEVER together
(4) O and P NEVER together
(5) quantity of P always greater than total of others together
1. consider option (A) - 1L and 1P ; by (4) this condition is contradicting.
consider option (B) - 2M and 2L; from (1) & (2) along with M and L, O and N also should be added. but N and O can never be together from (3). so,this condition is contradicting.
consider option (C) - by (1) its contradicting
consider option (E) - by (5) its contradicting
consider option (D) - by (2) the statement satisfies the condition.
2. consider option (A) - From (1), N = twice as L. Already there is 1L,1N,5P. The condition (5) is also satisfied. by adding 1N, it becomes 2N and twice the quantity of L.
NOTE: In all other options M and O are present. With O, N can never be added.
3. from (5), P should have greater proportion than all the other elements. So, P should only be added.
option (E) is the only one that satisfies the above.
4. From the statements except L and M, all the other combinations can be added together.
• @shashank_prabhu 1-Ross 2-Monica 3-New York
• @shashank_prabhu 1. Saankar 2.Kavita
• @shashank_prabhu Why is the answer to Q4, A?
• @Tanya-Gidwani
If a perfume has L in it, then it should also contain N.
Also, if a perfume has M in it, then it should also have O in it.
But we know that N and O cannot be present together.
So L + M is not allowed.
• Q6) Each of the three logicians A, B, and C is wearing a hat. They know that the hats are either black or white and that not all the hats are white. A can see the hats of B and C; B can see the hats of A and C; C is blind. Each is asked in turn if he knows the colour of his own hat. The answers are “A : No, B : No, C : Yes” in this order only. What is the colour of C’s hat?
a. Black
b. White
c. Black if A is wearing a White hat
d. White if A is wearing a Black hat
• Q7) There are 100 people in a community. Each of these people likes one or more of the 3 characters Sheldon Cooper, Barney Stinson and Chandler Bing. Ten of them like both Sheldon Cooper and Barney Stinson. 30 of them like both Chandler Bing and Sheldon Cooper. 15 of them like both Chandler Bing and Barney Stinson. 35 of them like Chandler Bing.
(1) How many of them like all the 3 characters?
a. 5
b. 10
c. 15
d. Data insufficient
(2) How many of them like exactly one of the 3 characters?
a. 65
b. 45
c. 35
d. Data insufficient
• Q8) Six people are sitting in a row having seat numbers 1 to 6 (from left to right) such that they sit in the order of increasing age (from left to right). (A, F), (C, B ) and (E, D) are the only pairs of siblings. One pair has only boys and another pair has only girls while the third has a boy and a girl. There is only one person sitting on one seat.
I. Boys sit at the corner.
II. No siblings sit adjacent to each other.
III. The siblings A and F have more than two people between them.
IV. No two (or more) boys sit adjacent to each other.
V. Seat 4 and 5 were occupied by girls.
VI. F and E are the eldest of the boys and the girls respectively.
(1) Who occupied seat number 4?
(a) C
(b) B
(c) D
(d) Cannot be determined
(2) Which two people are definitely sitting adjacent to each other?
(a) (A, D)
(b) (C, E)
(c) (B, D)
(d) (A, B )
(3) How many people are sitting between A and E?
(a) 3
(b) 4
(c) 2
(d) Cannot be determined
(4) Who is definitely younger than E’s brother?
(a) C
(b) B
(c) A
(d) F
• Q9) 6 friends — Eddard, Joffrey, Drogo, Theon, Jon, Robb. Their wives are — Daenerys, Sansa, Catelyn, Cersei, Arya, Ygritte (not in the same order as their husbands). These 6 friends belong to Winterfell, Dragonstone, King’s Landing, The Wall, Volantis or Harrenhal (not necessarily in order). Each of them plays one of the games — Quidditch, 3-d chess, Mindgame, Whackbat, Blurnsball or Calvinball (not necessarily in that order). Each of the friends belongs to one of the places only, plays only one game and is married to one lady only.
The following information is known:
Husbands of Catelyn, Arya or Ygritte do not play 3-d chess or Mindgame.
The one who is from King’s Landing plays Quidditch.
Robb plays 3-d chess and is from Harrenhal.
Drogo and Jon are married to Daenerys and Cersei respectively but are not from King’s Landing.
The men from Dragonstone and The Wall are Blurnsball and Mindgame players respectively.
Theon is from Volantis.
Sansa is married to the man from Harrenhal.
Drogo plays Whackbat.
(1) Who is married to the man from The Wall?
(2) The person who plays Whackbat belongs to which place?
• Q10) In a target shooting competition, a person is allowed to shoot at four targets successively, and is then followed by the next competitor. After all the competitors have finished one such round, the process is repeated. If a target is hit, the shooter is awarded two points. If he misses the target, the others are awarded one point each. The first person who gets to 60 points wins the competition. In a contest between A, B and C, the final score card is A=60, B=53 and C=43. Out of a total of 78 shots fired, 43 hit the target.
(1) Who was the second person to shoot?
a) A
b) B
c) C
d) Either A or B
e) Either B or C
(2) How many targets did A hit?
a) 42
b) 34
c) 17
d) 14
e) Cannot be determined
(3) How many targets did B miss?
a) 6
b) 10
c) 12
d) 16
e) Cannot be determined
• Q11) Sanaa and Pallavi had a row over the distribution of a box of chocolates that was given to them by their friend, Danish. Pallavi yelled, "You haven’t distributed the chocolates fairly! You have kept thrice the number of chocolates that I have, for yourself. I demand a re-distribution!" To this, Sanaa calmly responded by giving Pallavi one chocolate for each year of her age, in addition to what she already had and asked, "Are you happy now?". "This is still unfair! Now, I have half as many chocolates as you do", screamed Pallavi. Sanaa lost her cool and yelled back, "Listen, I think I've given you more than enough. After all, I'm twice your age. You have to obey me." Saying this, she stormed out of the room, leaving all the chocolates, including her own, on the table. After she left, Pallavi took her own chocolates and stole chocolates from Sanaa's pile equal to Sanaa's age and ran away. Sanaa later came and collected her chocolates that were left on the table.
Who has more chocolates?
a. Sanaa
b. Pallavi
c. Both have an equal number of chocolates
• Q12) A railway company has exactly three lines: line 1, line 2, and line 3. The company prints three sets of tickets for January and three sets of tickets for February: one set for each of its lines for each of the two months. The company’s tickets are printed in a manner consistent with the following conditions:
a. Each of the six sets of tickets is exactly one of the following colors: green, purple, red, yellow.
b. For each line, the January tickets are a different color than the February tickets.
c. For each month, tickets for different lines are in different colors.
d. Exactly one set of January tickets is red.
e. For line 3, either the January tickets or the February tickets, but not both, are green.
f. The January tickets for line 2 are purple.
g. No February tickets are purple.
(1) If the line 3 tickets for January are red, then which one of the following statements must be true?
(a) The line 1 tickets for January are green.
(b) The line 1 tickets for January are yellow.
(c) The line 1 tickets for February are red.
(d) The line 2 tickets for February are yellow.
(e) The line 3 tickets for February are green.
(2) If one set of the line 2 tickets is green, then which one of the following statements must be true?
(a) The line 1 tickets for January are red.
(b) The line 3 tickets for January are red.
(c) The line 1 tickets for February are red.
(d) The line 3 tickets for February are green.
(e) The line 3 tickets for February are yellow.
(3) Which one of the following statements could be true?
(a) No January ticket is green.
(b) No February ticket is green.
(c) Only line 2 tickets are red.
(d) One set of January tickets is green and one set of January tickets is yellow.
(e) The line 2 tickets for January are the same color as the line 1 tickets for February.
(4) Which one of the following statements could be true?
(a) Both the line 1 tickets for January and the line 2 tickets for February are green.
(b) Both the line 1 tickets for January and the line 2 tickets for February are yellow.
(c) Both the line 1 tickets for January and the line 3 tickets for February are yellow.
(d) The line 1 tickets for January are green, and the line 3 tickets for February are red.
(e) The line 3 tickets for January are yellow, and the line 1 tickets for February are red.
(5) If the line 3 tickets for February are yellow, then each of the following statements must be true EXCEPT:
(a) One set of January tickets is green.
(b) One set of line 1 tickets is red.
(c) One set of line 2 tickets is red.
(d) The tickets in two of the six sets are red.
(e) The tickets in two of the six sets are yellow.
(6) Suppose that none of the ticket sets are purple. If all of the other conditions remain the same, then which one of the following statements could be true?
(a) None of the January tickets are green.
(b) None of the February tickets are green.
(c) None of the line 2 tickets are green.
(d) No line 1 or line 2 tickets are yellow.
(e) No line 2 or line 3 tickets are red.
• Q13) Six friends, Amar, Amit, Ankit, Anuj, Abhinav, and Abhijeet, went shopping and each of them purchased a different item among a pair of Shoes, a Bag, a Shirt, a Mobile Phone, a Pen, and a Watch. Further it is also known that
(i) Ankit did not purchase either a pair of Shoes or a Bag, while Amar purchased a Mobile Phone
(ii) Abhinav did not purchase either a pair of Shoes or a Pen, while Anuj Purchased a Watch
(iii) neither Amit nor Abhijeet purchased either a Pen or a Bag
(1) If Amit and Abhinav between themselves purchased a pair of Shoes and a Bag, what did Abhijeet purchase?
(a) Pen
(b) Shirt
(c) Watch
(d) Cannot be determined
(2) What did Ankit Purchase?
(a) Pen
(b) Shirt
(c) Shoes
(d) Cannot be determined
(3) Which of the following statements would be sufficient to determine which person purchased which item?
(i) Abhinav Purchased a Bag.
(ii) Abhijeet purchased a pair of Shoes
(iii) Ankit Purchased a Pen
(iv) Amit Purchased a Shirt
(a) (i), (ii) and (iii) together
(b) Both (ii) and (iv) together
(c) Both (iii) and (iv) together
(d) Either (ii) alone or (iv) alone
(4) How many combinations of persons and the items that they purchased are Possible?
(a) 1
(b) 2
(c) 3
(d) 4
• Q14) There is a colony in which each person knows how to speak in one or more of the three languages Dothraki, Elvish and Parseltongue. It is known that
1. The number of people who speak only Dothraki, the number of people who speak only Elvish and the number of people who speak only Parseltongue form an arithmetic progression.
2. Similarly, the number of people who know exactly two languages also form an arithmetic progression.
3. The number of people who know all the three languages is one-tenth of the number of people who know only Elvish, which in turn is two-thirds of the number of people who know only Parseltongue.
4. The number of people who know both Dothraki and Elvish is 15 and the number of people who know both Elvish and Parseltongue is 19.
5. The number of people who know to speak in Parseltongue is 70 and the number of people who know how to converse in Dothraki is fewer than that.
What is the sum of all the values of the total number of people in the colony that could be possible?
How many people can converse in Dothraki and Parseltongue?
What is the maximum possible number of people who can understand Dothraki?
What is the number of people who know at least two languages?
• Q15) Six dancers - Lakshmi, Madhumati, Rani, Rohini, Asha and Kriti perform six different dances: Kathakali, Kathak, Garba, Bharatnatyam, Kuchipudi and Western, not necessarily in the same order. They belong to six different states: U.P., M.P., Maharashtra, Orissa, Gujarat and Karnataka.
Following things are known about their choices:
Asha knows Kathak but does not belong to Gujarat and Maharashtra.
Rani and Kriti don’t know both Bharatnatyam and Kathakali, but belong to Karnataka and U.P. respectively.
Dancer from M.P. knows Kathak.
Lakshmi, Rani and Madhumati don’t know Garba or Western.
Rohini can perform Western while Lakshmi knows Bharatnatyam.
Lakshmi does not belong to Gujarat while Madhumati belongs to Orissa.
(1) Kuchipudi is performed by
a. Rani
b. Rohini
c. Kriti
(2) Rohini belongs to
a. Orissa
b. Maharashtra
c. Gujarat
d. None of these
(3) Garba is performed by
a. Rani
b. Kriti
c. Rohini
(4) Which of the following is definitely a false combination.
a. Rani–Kuchipudi–Karnataka
b. Lakshmi–Bharatnatyam–Maharashtra
c. Rohini–Western–Gujarat
d. Asha–Kathak–Orissa
(5) Which of the following is definitely a true combination?
a. Rani–Garba–Karnataka
b. Kriti–Garba–M.P.
c. Lakshmi–Bharatnatyam–Maharashtra
d. Rohini–Western–Orissa
• Q16) At a restaurant, a group of friends ordered four main dishes and three side dishes at a total cost of 89 Dollars. The prices of the seven items, in dollars, were all different integers, and every main dish cost more than every side dish. What was the price, in dollars, of the most expensive side dish given that the most expensive main dish cost 16 Dollars?
54
71
1
48
164
62
58 | 3,826 | 13,838 | {"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} | 3.890625 | 4 | CC-MAIN-2020-34 | latest | en | 0.939924 |
http://www.ck12.org/book/CK-12-Math-Analysis-Concepts/r4/section/5.7/ | 1,493,079,150,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917119995.14/warc/CC-MAIN-20170423031159-00295-ip-10-145-167-34.ec2.internal.warc.gz | 503,050,776 | 43,491 | # 5.7: Planes in Space
Difficulty Level: At Grade Created by: CK-12
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Practice Planes in Space
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Orlando is building a coffee table for his mother. He has the round surface cut and sanded and is ready to attach the legs. As he tries a few different heights, he notices that unless the spot on the floor is really flat, and he is extremely careful to make the legs exactly the same length, the table is unsteady and wobbles annoyingly.
How can Orlando ensure that the table remains stable, even if the floor is not entirely flat where his mom decides to put it?
Embedded Video:
### Guidance
Vectors can be used to identify directions in space and also the orientation of a plane by identifying the direction perpendicular to that plane. In this section we will look at that calculation in reverse. Rather than determining the normal vector to a plane using two vectors which lie in that plane, we will be using the normal vector to determine the equation for the plane itself. We will also use the normal vectors to determine the intersection angle between any pair of planes.
Intercept Form
The diagram below shows a plane which crosses all three coordinate axes. Points A, B, and C are the locations where the plane crosses each of the coordinate axes, called intercepts. Their locations are given by A = (a, 0, 0), B = (0, b, 0), and C = (0, 0, c). The line segments AB, BC, and CA all lie in the plane. Furthermore, segment AB is a portion of the line of intersection between this plane and the x-y axis; segment BC is a portion of the line of intersection between this plane and the y-z axis; and segment CA is a portion of the line of intersection between this plane and the z-x axis.
The intercept form of the equation for a plane is given by
\begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}\end{align*}
Using a Normal Vector
Another way to specify a plane is to know two vectors within the plane. Recall that if two vectors lie in the same plane, the normal to that plane can be found using the cross product of the two vectors. Sometimes we are given the equations of the vectors themselves. Sometimes, however, we are only given a set of points that lie on the plane. If we know three points that lie on the plane, we can use the method developed when looking at cross products to find equations for the vectors between those points and then use those vectors to identify the plane. We only need three points to accomplish this task.
Since the normal to the plane is, by definition, perpendicular to all possible vectors within a plane and since the dot product of two vectors is equal to zero for any two perpendicular vectors, we can define a plane in terms of the dot product of the normal vector with any vector, \begin{align*}\overrightarrow{v}\end{align*}, within the plane:
\begin{align*}\overrightarrow{n} \times \overrightarrow{v} = 0\end{align*}
Which we can also write as
\begin{align*}\left \langle n_x, n_y, n_z \right \rangle \times \left \langle (x - x_0), (y - y_0), (z - z_0) \right \rangle = 0\end{align*}
If we compute the dot product, we obtain another equation which specifies the plane in terms of the normal vector and two points on the plane, (x, y, z) and (x0, y0, z0).
nx (x - x0) + ny (y - y0) + nz (z - z0) = 0
This equation is frequently written as
nxx + nyy + nzz + d = 0
Where
d = -nxxo - nyyo - nzzo
and the intercepts of the plane with the x, y, and z axes are given by:
\begin{align*}a = -\frac{d}{n_x}, \ b = -\frac{d}{n_y},\end{align*} and \begin{align*}c = -\frac{d}{n_z}\end{align*}
If you only have the normal vector and one point on the plane, first determine the vector projection of the position vector of that point onto the normal vector using the dot product. Then you will have the locations of two points on the plane and can use the normal and two points method described above.
#### Example A
Find the intercepts of the plane given by the equation 3x + 5y - 2z - 4 = 0.
Solution
Rewrite the equation of the plane in the format of the intercept form of the plane equation.
3x + 5y - 2z = 4
\begin{align*}1 = \frac{3}{4} x + \frac{5}{4} y + \frac{-2}{4}z\end{align*}
\begin{align*}a = \frac{4}{3}, \ b = \frac{4}{5}\end{align*}, and \begin{align*}c = \frac{4}{-2} = -2\end{align*}. Thus the intercepts of this plane are
\begin{align*}\left (\frac{4}{3}, 0, 0 \right ), \ \left (0, \frac {4}{5}, 0 \right )\end{align*}, and \begin{align*}(0, 0, -2)\end{align*}.
#### Example B
A plane has intersections at (12, 0, 0), (0, 6, 0), and (0, 0, 4). Write the equation of the plane.
Solution
\begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = \frac{x}{12} + \frac{y}{6} + \frac{z}{4}\end{align*} or \begin{align*}12 = x + 2y + 3z\end{align*}
#### Example C
Use the equation 3x + 5y - 2z - 4 = 0 to determine the normal unit-vector to that plane.
Solution
Comparing this equation to nxx + nyy + nzz + d = 0, we can see that \begin{align*}\overrightarrow{n} = \left \langle 3, 5, -2\right \rangle\end{align*}. To find the unit normal vector, find the magnitude of this normal vector and divide each component by the magnitude.
\begin{align*}|\overrightarrow{n}| = \sqrt{n_x^2 + n_y^2 + n_z^2} = \sqrt{3^2 + 5^2 + (-2)^2} = \sqrt{38}\end{align*}
\begin{align*}\hat{n} = \left \langle \frac{3}{\sqrt{38}}, \frac{5}{\sqrt{38}}, \frac{-2}{\sqrt{38}} \right \rangle\end{align*}
#### Example D
The three points P = (3, 7, 2), Q = (1, 4, 3), and R = (2, 3, 4) define a plane. Determine the equation of the plane.
Solution
First find the vectors between two pairs of the points.
\begin{align*}\overrightarrow{PQ} = \left \langle (Q_x - P_x), (Q_y - P_y), (Q_z - P_z) \right \rangle = \left \langle (1 - 3), (4 - 7), (3 - 2) \right \rangle =\end{align*} \begin{align*}\left \langle -2, -3, 1 \right \rangle\end{align*}
\begin{align*}\overrightarrow{PR} = \left \langle (R_x - P_x), (R_y - P_y), (R_z - P_z) \right \rangle = \left \langle (2 - 3), (3 - 7), (4 - 2) \right \rangle =\end{align*} \begin{align*}\left \langle -1, -4, 2 \right \rangle\end{align*}
The cross product of these two vectors is normal to the plane.
\begin{align*}\overrightarrow{PQ} \times \overrightarrow{PR} = \left \langle (PQ_yPR_z - PQ_zPR_y), (PQ_zPR_x - PQ_xPR_z), (PQ_xPR_y - PQ_yPR_x)\right \rangle\end{align*}
\begin{align*}\overrightarrow{PQ} \times \overrightarrow{PR} = \left \langle [(-3 \cdot 2) - (1 \cdot -4)], [(1 \cdot -1) - (-2 \cdot 2)], [(-2 \cdot -4) - (-1 \cdot -3)] \right \rangle\end{align*}
\begin{align*}\overrightarrow{n} = \overrightarrow{PQ} \times \overrightarrow{PR} = \left \langle [(-6) - (-4)], [(-1) - (-4)], [(8) - (3)] \right \rangle = \left \langle -2, 3, 5 \right \rangle\end{align*}
This normal vector and one of the points will give an equation for the plane.
nx (x - Px) + ny (y - Py) + nz (z - Pz) = 0
-2(x - 3) + 3(y - 7) + 5(z - 2) = 0
-2x + 3y + 5z + (6 - 21 - 10) = 0
-2x + 3y + 5z - 25 = 0
Concept question wrap-upHow can Orlando ensure that the table he is making for his mother remains stable, even if the floor is not entirely flat where his mom decides to put it?
Recall from the lesson that any plane can be defined by three points. If Orlando makes the table with only three legs, then the endpoints of each leg will define a stable plane for the table to sit on. Even if the plane defined by the leg endpoints is not exactly parallel to the average plane of the floor across the entire room, the table itself will remain stable.
### Vocabulary
A plane is the 3 dimensional equivalent of a line on a standard rectangular graph. It can be conceptualized as a sheet of paper of infinite area.
The intercept form of the equation for a plane is given by: \begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}\end{align*}
The normal vector to a plane is perpendicular to all possible vectors within that plane
### Guided Practice
Questions
1) Rewrite the equation of the plane 7x + 3y + z + 12 = 0 in intercept form.
2) Determine the equation for the unit vector which is perpendicular to the plane, 7x + 3y + z + 12 = 0.
3) Find the intercepts of the plane described by the equation 2.4x + 3.6y - 4.8z - 5.9 = 0.
4) A plane is defined by three points having position vectors \begin{align*}\overrightarrow{r_1} = \left \langle 1, 0, -1 \right \rangle, \ \overrightarrow{r_2} = \left \langle 2, 4, 6 \right \rangle\end{align*}, and \begin{align*}\overrightarrow{r_3} = \left \langle -3, 7, 5 \right \rangle\end{align*}. Determine the components of the unit vector which is perpendicular to the plane passing through those points.
5) Determine the components of the unit vector which is perpendicular to the plane 12x + 23y + 14z - 5 = 0.
Solutions
1) The equation \begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}\end{align*} must be true for all points on a plane. Therefore, we should first rearrange 7x + 3y + z + 12 = 0 into the form \begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}\end{align*}.
7x + 3y + z = -12
\begin{align*}\frac{7}{-12}x + \frac{3}{-12}y + \frac{1}{-12}z = 1\end{align*}
Therefore, \begin{align*}a = \frac{-12}{7}, \ b = \frac{-12}{3} = -4\end{align*}, and \begin{align*}c = \frac{-12}{1} = -12\end{align*} and the position vectors of the three intercepts are \begin{align*}\overrightarrow{A} = \left \langle -1.714, 0, 0 \right \rangle, \ \overrightarrow{B} = \left \langle 0, -4, 0 \right \rangle\end{align*}, and \begin{align*}\overrightarrow{C} = \left \langle 0, 0, -12 \right \rangle\end{align*}
2) Comparing this equation to \begin{align*}n_xx + n_yy + n_zz + d = 0\end{align*}, we can see that \begin{align*}\overrightarrow{n} = \left \langle 7, 3, 1 \right \rangle\end{align*}
\begin{align*}\hat{n} = \frac{\overrightarrow{n}}{|\overrightarrow{n}|} = \frac{\left \langle n_x, n_y, n_z \right \rangle}{\sqrt{n_x^2 + n_y^2 + n_z^2}} = \frac{\left \langle 7, 3, 1 \right \rangle}{\sqrt{(7)^2 + (3)^2 + (1)^2}} = \frac{\left \langle 7, 3, 1 \right \rangle}{\sqrt{49 + 9 + 1}} = \frac{\left \langle 7, 3, 1 \right \rangle}{\sqrt{59}} = \left \langle \frac{7}{\sqrt{59}}, \frac{3}{\sqrt{59}}, \frac{1}{\sqrt{59}} \right \rangle\end{align*}
3) First write the equation of the plane in intercept form, \begin{align*}1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}\end{align*}.
2.4x + 3.6y - 4.8z = 5.9
\begin{align*}\frac{2.4}{5.9}x + \frac{3.6}{5.9}y - \frac{4.8}{5.9}z = 1\end{align*}
Therefore the x-intercept is \begin{align*}\left \langle \frac{5.9}{2.4}, 0, 0 \right \rangle\end{align*}, the y-intercept is \begin{align*}\left \langle 0, \frac{5.9}{3.6}, 0 \right \rangle\end{align*}, and the z-intercept is \begin{align*}\left \langle 0, 0, \frac{5.9}{4.8} \right \rangle\end{align*}.
4) The cross-product determines the direction perpendicular to a pair of vectors. Therefore we can use these three points to define two vectors in the same plane. The vector from point 1 to point 2 is given by subtracting vector 2 from vector 1:
\begin{align*}\overrightarrow{r_{1-2}} = \overrightarrow{r_1} - \overrightarrow{r_2} = \left \langle 1, 0, -1\right \rangle - \left \langle 2, 4, 6 \right \rangle = \left \langle 1 - 2, 0 - 4, -1 - 6 \right \rangle = \left \langle -1, -4, -7 \right \rangle\end{align*}
Likewise, the vector from point 1 to point 3 is given by subtracting vector 3 from vector 1:
\begin{align*}\overrightarrow{r_{1-3}} = \overrightarrow{r_1} - \overrightarrow{r_3} = \left \langle 1, 0, -1\right \rangle - \left \langle -3, 7, 5 \right \rangle = \left \langle 1 - (-3), 0 - 7, -1 - 5 \right \rangle =\end{align*} \begin{align*}\left \langle 4, -7, -6 \right \rangle\end{align*}
Now we can use the cross-product of the two vectors in the plane to determine a vector which is perpendicular to that plane,
\begin{align*}\overrightarrow{n} = \overrightarrow{r_{1-2}} \times \overrightarrow{r_{1-3}} = \left \langle (r_{1-2y}r_{1-3z} - r_{1-2z}r_{1-3y}), (r_{1-2z}r_{1-3x} - r_{1-2x}r_{1-3z}), (r_{1-2x}r_{1-3y} - r_{1-2y}r_{1-3x}) \right \rangle\end{align*}
\begin{align*}\overrightarrow{n} = \overrightarrow{r_{1-2}} \times \overrightarrow{r_{1-3}} = \left \langle ((-4)(-6) - (-7)(-7)), ((-7)(4) - (-1)(-6)), ((-1)(-7) - (-4)(4)) \right \rangle\end{align*}
\begin{align*}\overrightarrow{n} = \overrightarrow{r_{1-2}} \times \overrightarrow{r_{1-3}} = \left \langle ((24) - (49), ((-28) - (6)), ((7) - (-16)) \right \rangle = \left \langle -25, -34, 23 \right \rangle\end{align*}
Now use the definition of the unit vector to complete the problem.
\begin{align*}\hat{n} = \frac{\overrightarrow{n}}{|\overrightarrow{n}|} = \frac{\left \langle n_x, n_y, n_z \right \rangle}{\sqrt{n_x^2 + n_y^2 + n_z^2}} = \frac{\left \langle -25, -34, 23 \right \rangle}{\sqrt{(-25)^2 + (-34)^2 + (23)^2}} = \frac{\left \langle -25, -34, 23 \right \rangle}{\sqrt{625 + 1156 + 529}} = \frac{\left \langle -25, -34, 23 \right \rangle}{\sqrt{2310}} \end{align*}
\begin{align*}\hat{n} = \frac{\overrightarrow{n}}{|\overrightarrow{n}|} = \frac{\left \langle {-25, -34, 23}\right \rangle}{48.06} = \left \langle -0.5202, -0.7074, 0.4785 \right \rangle\end{align*}
5) Comparing this equation to \begin{align*}n_xx + n_yy + n_zz + d = 0\end{align*}, we can see that \begin{align*}\overrightarrow{n} = \left \langle 12, 23, 14 \right \rangle\end{align*}.
Now we can use the definition of the unit vector to complete the problem.
\begin{align*}\hat{n} = \frac{\overrightarrow{n}}{|\overrightarrow{n}|} = \frac{\left \langle n_x, n_y, n_z \right \rangle}{\sqrt{n_x^2 + n_y^2 + n_z^2}} = \frac{\left \langle 12, 23, 14 \right \rangle}{\sqrt{12^2 + 23^2 + 14^2}} = \frac{\left \langle 12, 23, 14 \right \rangle}{\sqrt{869}} = \frac{\left \langle 12, 23, 14 \right \rangle}{29.5} = \left \langle \frac{12}{29.5}, \frac{23}{29.5}, \frac{14}{29.5} \right \rangle\end{align*}
### Practice
Given the following intersections, write the equation of the plane.
1. \begin{align*}(7, 0, 0), (0, 3, 0)\end{align*} and \begin{align*}(0, 0, 19)\end{align*}
2. \begin{align*}(2, 0, 0), (0, 8, 0)\end{align*} and \begin{align*}(0, 0, 5)\end{align*}
3. \begin{align*}(13, 0, 0), (0, 21, 0)\end{align*} and \begin{align*}(0, 0, 17)\end{align*}
4. \begin{align*}(5, 0, 0), (0, 1, 0)\end{align*} and \begin{align*}(0, 0, 2)\end{align*}
5. \begin{align*}(27, 0, 0), (0, 12, 0)\end{align*} and \begin{align*}(0, 0, 18)\end{align*}
Find the intercepts of the plane given the following equations:
1. \begin{align*}3x + 2y + z - 6 = 0\end{align*}
2. \begin{align*}1x - 7y - z + 10 = 0\end{align*}
3. \begin{align*}-2x + 9y + 4z - 1 = 0\end{align*}
4. \begin{align*}6x - 11y + 2z + 3 = 0\end{align*}
5. \begin{align*}-2x + 5y + 5z + 6 = 0\end{align*}
Use the given equations to determine the normal unit-vector to that plane
1. \begin{align*}7x + 5y - 1z - 10 = 0\end{align*}
2. \begin{align*}4x - 13y + 5z - 3 = 0\end{align*}
3. \begin{align*}-8x + 7y + 2z + 5 = 0\end{align*}
4. \begin{align*}10x + 3y - z - 2 = 0\end{align*}
5. \begin{align*}-1x - 2y + 7z + 16 = 0\end{align*}
Determine the equation of the planes below using the three points given:
1. \begin{align*}P = (3, 6, 9), Q = (9, 6, 3)\end{align*} and \begin{align*}R = (6, -9, 9)\end{align*}
2. \begin{align*}P = (1, -7, 2), Q = (4, 2, 9)\end{align*} and \begin{align*}R = (3, -5, 1)\end{align*}
3. \begin{align*}P = (3, 8, 10), Q = (-2, 5, 8)\end{align*} and \begin{align*}R = (7, 4, 8)\end{align*}
4. \begin{align*}P = (9, -1, 4), Q = (6, 2, -8)\end{align*} and \begin{align*}R = (12 , 9, 10)\end{align*}
5. \begin{align*}P = (5, 8,-9), Q = ( -5, 3, 9)\end{align*} and \begin{align*}R = (10, 4, -6)\end{align*}
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Cross product The cross product of two vectors is a third vector that is perpendicular to both of the original vectors.
dot product The dot product is also known as inner product or scalar product. The two forms of the dot product are $\vec{a} \cdot \vec{b} = \Big \| \vec{a}\Big \| \ \Big \| \vec{b}\Big \| \cos \theta$ and $\vec{a} \cdot \vec{b} = x_a x_b + y_a y_b$.
intercept form The intercept form of the equation for a plane is given by $1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}$.
Intercepts The intercepts of a curve are the locations where the curve intersects the $x$ and $y$ axes. An $x$ intercept is a point at which the curve intersects the $x$-axis. A $y$ intercept is a point at which the curve intersects the $y$-axis.
normal vector A normal vector is a vector that is perpendicular to a given surface or plane. A unit normal vector is a normal vector with a magnitude of one.
plane A plane is a flat, two-dimensional surface. It can be conceptualized as a sheet of paper of infinite area.
unit vector A unit vector is a vector with a magnitude of one.
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Subjects: | 5,985 | 17,066 | {"found_math": true, "script_math_tex": 82, "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": 9, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2017-17 | latest | en | 0.935151 |
https://web2.0calc.com/questions/please-help-me_83398 | 1,627,788,035,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154158.4/warc/CC-MAIN-20210801030158-20210801060158-00498.warc.gz | 611,232,356 | 5,595 | +0
+1
112
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Two liters of a 30% acid solution are mixed with three liters of a 20% acid solution. What percent of the mixture is acid?
Apr 7, 2021
#1
0
The mixture is (30% + 20%)/2 = 25% acid.
Apr 7, 2021
#2
+2122
0
Not quite exactly, the amounts of the solutions are different.
30% of 2 liters is .6 liters.
20% of 3 liters is .6 liters.
Total 1.2 liters of acid and 5 liters of total mixture
1.2/5 = 24%
I hope this helped. :))
=^._.^=
catmg Apr 7, 2021
#3
0
.30 (2) + .20 (3) = x ( 2+3)
1.2 / 5 = .24 = x or 24 % final solution
Apr 7, 2021 | 232 | 583 | {"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} | 3.78125 | 4 | CC-MAIN-2021-31 | latest | en | 0.876908 |
https://clrs.skanev.com/04/problems/01.html | 1,713,012,642,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816734.69/warc/CC-MAIN-20240413114018-20240413144018-00358.warc.gz | 164,488,642 | 2,269 | # Problem 4.1
## Recurrence examples
Give asymptotic upper and lower bound for $T(n)$ in each of the following recurrences. Assume that $T(n)$ is constant for $n \le 2$. Make your bounds as tight as possible, and justify your answers.
1. $T(n) = 2T(n/2) + n^4$
2. $T(n) = T(7n/10) + n$
3. $T(n) = 16T(n/4) + n^2$
4. $T(n) = 7T(n/3) + n^2$
5. $T(n) = 7T(n/2) + n^2$
6. $T(n) = 2T(n/4) + \sqrt{n}$
7. $T(n) = T(n - 2) + n^2$
1. $\Theta(n^4)$ (master method)
2. $\Theta(n)$ (master method, $\log_{10/7}1 = 0$)
3. $\Theta(n^2\lg{n})$ (master method)
4. $\Theta(n^2)$ (master method)
5. $\Theta(n^{\log_2{7}})$ (master method)
6. $\Theta(\sqrt{n}\lg_{n})$ (master method)
7. $\Theta(n^3)$ by the following:
\begin{aligned} T(n) &= n^2 + T(n - 2) \\ &= n^2 + (n - 2)^2 + T(n - 4) \\ &= \sum_{i=0}^{n/2}(n -2i)^2 \\ &= n^2 \sum_{i=0}^{n/2} 1 + 4 \sum_{i=0}^{n/2} i^2 - 4n \sum_{i=0}^{n/2} i \\ \end{aligned}
For the three sums above:
$$n^2 \sum_{i=0}^{n/2} 1 = \frac{n^3}{2}$$
$$4 \sum_{i=0}^{n/2} i^2 = \frac{4}{6} \left[\frac{n}{2}\left(\frac{n+2}{2}(n+1)\right)\right] = \frac{1}{3}(2n^3 + 6n^2 + 4n)$$
$$4n \sum_{i=0}^{n/2} i = 4n\left[\frac{1}{2}\frac{n}{2}\left(\frac{n+2}{2}\right)\right] = \frac{1}{2}(n^3 + 2n^2)$$
Now substitute back in these 3 simplifications:
\begin{aligned} T(n) &= n^2 \sum_{i=0}^{n/2} 1 + 4 \sum_{i=0}^{n/2} i^2 - 4n \sum_{i=0}^{n/2} i \\ &= \frac{n^3}{2} + \frac{1}{3}(2n^3 + 6n^2 + 4n) - \frac{1}{2}(n^3 + 2n^2) \\ &= \frac{2}{3}n^3 + n^2 + \frac{4}{3}n \\ &= \Theta(n^3) \end{aligned} | 759 | 1,522 | {"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} | 4.5625 | 5 | CC-MAIN-2024-18 | latest | en | 0.512245 |
null | null | null | null | null | null | Remember the Drunk Chrysler Workers? Car Pro News
19 Dec
Chrysler took quick action two years ago after television news reports of workers at its Jefferson North plant in Detroit who were apparently drinking beer or smoking marijuana on lunch breaks against factory policies.
It fired 13 of them. But now they’re back on the job, having won an arbitration decision that reinstated them to their union jobs.
Chrysler officials aren’t happy about the decision.
“An arbitrator decided in the workers’ favor, citing insufficient conclusive evidence to uphold the dismissals. This was a decision that Chrysler Group does not agree with,” says Scott Garberding, senior vice president of manufacturing, in a statement.
He notes that the company hasn’t forgotten that it only stayed in business because of a federal bailout and having to reinstate the drinking workers is “an unfortunate aberration.”
Fox 2 News’ report caused quite a stir at the time. News crews followed workers to a park where they were seen drinking or lighting up marijuana on their breaks, then heading back to the plant. It went back a year later and found more workers engaged in the same behavior on breaks.
1. Arnie says:
I think Chrysler should sue the union . The union is supposed to supply labor to the companyand is responsible for them .
2. Thomas Drapela says:
That is the bad part of unions and they see nothing wrong with it!
3. Nick Stumpf says:
These worker’s behavior, the union position, and the courts decision are absolutely unacceptable. There are worker safety, product quality, production efficiency, public image, and product cost implications that ultimately affect Chysler’s competitiveness and thus those worker’s jobs. Both the workers and the union should appreciate those jobs and do their best help Chrysler compete successfully in the automotive business.
4. edd altieri says:
This is another case where the union thugs call the shots. I’am sure they wanted them back to work so they can get there union dues from these jerks. I would transfer these guys to restroom clean up.
I wonder how much the arbitrator got payed? Now I know why our P. T. cruiser conv. was in the shop more than out . We traded it in for a ford explorer
Speak Your Mind | null | null | null | null | null | null | null | null | null |
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# find sin theta
0
370
1
If cos(θ) = − 10/11 with θ in Quadrant III, what is sin(θ)?
Oct 10, 2017
#1
+7352
+2
From the Pythagorean identity,
sin2 θ + cos2 θ = 1 Plug in -10/11 for cos θ .
sin2 θ + (-10/11)2 = 1 $$(-\frac{10}{11})^2=(-\frac{10}{11})(-\frac{10}{11})=\frac{100}{121}$$
sin2 θ + 100/121 = 1 Subtract 100 / 121 from both sides.
sin2 θ = 21/121 Since sin is negative in Quad III, take the negative sqrt of both sides.
sin θ = $$-\sqrt{\frac{21}{121}}$$
sin θ = $$-\frac{\sqrt{21}}{11}$$
.
Oct 10, 2017
edited by hectictar Oct 10, 2017
#1
+7352
+2
From the Pythagorean identity,
sin2 θ + cos2 θ = 1 Plug in -10/11 for cos θ .
sin2 θ + (-10/11)2 = 1 $$(-\frac{10}{11})^2=(-\frac{10}{11})(-\frac{10}{11})=\frac{100}{121}$$
sin2 θ + 100/121 = 1 Subtract 100 / 121 from both sides.
sin2 θ = 21/121 Since sin is negative in Quad III, take the negative sqrt of both sides.
sin θ = $$-\sqrt{\frac{21}{121}}$$
sin θ = $$-\frac{\sqrt{21}}{11}$$
hectictar Oct 10, 2017
edited by hectictar Oct 10, 2017 | 538 | 1,170 | {"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} | 4.0625 | 4 | CC-MAIN-2019-13 | latest | en | 0.192311 |
null | null | null | null | null | null | A new beginning will come after the world's destruction. The earth will emerge out of the sea and flourish vigorously. The sons and daughters of the Aesir and Vanir will survive Ragnarok and meet in council on the plain of Ida, where Asgard used to be.
Odin's sons, Vidar and Vali will meet there and the sons of Thor will join them and be heirs of their fathers hammer, Mjollnir. The beloved god Baldr and his brother Hod will return fron Hel and join the rest, while Hoenir will predict what is to become of the new world.
Bor's sons, Vili and Ve, will be sent to the heavens to rule with the rest. The new ruling gods will congregate and recall past memories of Ragnarok. Treasures that once belonged to the Aesir will be found scattered on grassy plains and will be looked upon with great amazement.
Gimle will once more house the gods in peace and bounty. Good and evil, however, will not cease to exist, for there will be a region in Hel called Nastrond, the shore of the dead. The dragon Nidhogg will survive the fiery doom and continue to gnaw on the bodies of the deceased.
The man and woman who sought shelter under the branches of Yggdrasil will be called Lif and Lifthrasir. They will nourish themselves with dew drops and give birth to many children to repopulate the earth.
From the great Ash new rays of light will come from the skies above, for, a daughter was born by the sun before the wolf swallowed her in the dawn of Ragnarok. This is how it all has ended; and this is how the world begins. | null | null | null | null | null | null | null | null | null |
http://www.algebra.com/algebra/homework/Linear-equations/Linear-equations.faq.question.2354.html | 1,369,400,121,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704658856/warc/CC-MAIN-20130516114418-00036-ip-10-60-113-184.ec2.internal.warc.gz | 310,294,249 | 4,612 | # SOLUTION: I'm confused about the whole Slope deal. I dont really understand the equations for it. I know that the equation is y=mx+b but how do you solve a problem that is like 10y - 12x = 8
Algebra -> Algebra -> Linear-equations -> SOLUTION: I'm confused about the whole Slope deal. I dont really understand the equations for it. I know that the equation is y=mx+b but how do you solve a problem that is like 10y - 12x = 8 Log On
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Algebra: Linear Equations, Graphs, Slope Solvers Lessons Answers archive Quiz In Depth
Click here to see ALL problems on Linear-equations Question 2354: I'm confused about the whole Slope deal. I dont really understand the equations for it. I know that the equation is y=mx+b but how do you solve a problem that is like 10y - 12x = 8? Answer by longjonsilver(2297) (Show Source): You can put this solution on YOUR website!re-arrange it to the form you need, ie 1y equals, like so... 10y-12x=8 10y = 12x + 8 y = 1.2x + 0.8 so now, m=1.2 ie the gradient is +1.2 (just over 1). A gradient of 1 is a line at 45 degrees (for equal scales on the 2 axes) jon | 365 | 1,281 | {"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-2013-20 | latest | en | 0.911418 |
https://hr.mathigon.org/course/triangles/properties | 1,600,995,140,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400221382.33/warc/CC-MAIN-20200924230319-20200925020319-00395.warc.gz | 413,034,541 | 16,729 | ???
Well done!
Archie
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# Triangles and TrigonometryProperties of Triangles
Vrijeme čitanja: ~30 min
Let’s start simple: a triangle is a closed shape that has three sides (which are line segments) and three vertices (the points where the sides meet). It also has three internal angles, and we already know that the sum of them is always °.
We can classify triangles by the size of their angles:
A right-angled triangle
has one right angle.
An obtuse triangle
has one obtuse angle.
An acute triangle
has acute angles.
For convenience, we always label triangles in the same way. The vertices are labelled with capital letters A, B and C, the sides are labelled with lowercase letters a, b and c, and the angles are labelled with Greek letters α, β and γ (“alpha”, “beta” and “gamma”).
The side that lies opposite vertex A is labeled a, and the angle that lies right next to A is labelled α. The same pattern works for B/b/β and for C/c/γ.
## Medians
Here you can see a triangle as well as the midpoints of its three sides.
A median of a triangle is a line segment that joins a vertex and the midpoint of the opposite side. Draw the three medians of this triangle. What happens as you move the vertices of the triangle?
It seems like the medians always . This point is called the centroid.
Medians always divide each other in the ratio 2:1. For each of the three medians, the distance from the vertex to the centroid is always as long as the distance from the centroid to the midpoint.
The centroid is also the “balancing point” of a triangle. Draw a triangle on some cardboard, cut it out, and find the three medians. If you were accurate, you can now balance the triangle on the tip of a pencil, or hang it perfectly level from a piece of string that’s attached to its centroid:
This works because the weight of the triangle is evenly distributed around the centroid. In physics, this point is often called the center of mass.
## Perpendicular Bisectors and Circumcircle
Recall that the perpendicular bisector of a line is the perpendicular line that goes through its .
Draw the perpendicular bisector of all three sides of this triangle. To draw the perpendicular bisector of a side of the triangle, simply click and drag from one of its endpoints to the other.
Like before, the three perpendicular bisectors meet in a single point. And again, this point has a special property.
Any point on a perpendicular bisector has the same distance from the two endpoints of the lines it bisects. For example, any point on the blue bisector has the same distance from points A and C and any point on the red bisector has the same distance from points .
The intersection point lies on all three perpendicular bisectors, so it must have the same distance from all three of the triangle.
This means we can draw a circle around it that perfectly touches all the vertices. This circle is called the circumcircle of the triangle, and the center is called the circumcenter.
In fact, this means that if you are given any three points, you can use the circumcenter to find a circle that goes through all three of them. (Unless the points are , in which case they all lie on a straight line.)
## Angle Bisectors and Incircle
You’re probably getting the hang of this now: we pick a certain construction, do it three times for all sides/angles of the triangles, and then we work out what’s special about their intersection.
Recall that the angle bisector divides an angle exactly in the middle. Draw the angle bisector of the three angles in this triangle. To draw an angle bisector, you have to click on three points that form the angle you want to bisect.
Once again, all three lines intersect at one point. You probably expected something like this, but it is important to notice that there is no obvious reason why this should happen – triangles are just very special shapes!
Points that lie on an angle bisector have the same distance from the two lines that form the angle. For example any point on the blue bisector has the same distance from side a and side c, and any point on the red bisector has the same distance from sides .
The intersection point lies on all three bisectors. Therefore it must have the same distance from all three of the triangle.
This means we can draw a circle around it, that lies inside the triangle and just touches its three sides. This circle is called the incircle of the triangle, and the center is called the incenter.
## Area and Altitudes
Finding the area of a rectangle is easy: you simply multiply its width by its height. Finding the area of a triangle is a bit less obvious. Let’s start by “trapping” a triangle inside a rectangle.
The width of the rectangle is the length of the bottom side of the triangle (which is called the base). The height of the rectangle is the perpendicular distance from the base to the opposite vertex.
The height divides the triangle into two parts. Notice how the two gaps in the rectangle are exactly as big as the two parts of the triangle. This means that the rectangle is large as the triangle.
We can easily work out the area of the rectangle, so the area of the triangle must be half that:
A=12× base × height
To calculate the area of a triangle, you can pick any of the three sides as base, and then find the corresponding height, which is the line that is to the base and goes through the opposite vertex.
In triangles, these heights are often called altitudes. Every triangle has altitudes.
Like the medians, perpendicular bisectors and angle bisectors, the three altitudes of a triangle intersect in a single point. This is called the orthocenter of the triangle.
In acute triangles, the orthocenter the triangle.
In obtuse triangles, the orthocenter the triangle.
In right-angled triangles, the orthocenter the triangle. Two of the altitudes are actually just sides of the triangle. | 1,295 | 5,960 | {"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-2020-40 | longest | en | 0.949156 |
https://physics.stackexchange.com/questions/684648/how-to-represent-orbital-velocity-of-a-planet-on-an-elliptical-path-in-terms-of | 1,716,617,389,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058773.28/warc/CC-MAIN-20240525035213-20240525065213-00277.warc.gz | 383,151,762 | 41,786 | # How to represent orbital velocity of a planet on an elliptical path in terms of time?
Before reading this long message, please know that this is question is posted to know whether there is a method to find the relation between between the orbital velocity of a planet on an elliptical orbit and time.
I am trying to model the orbital velocity of Mercury orbiting the Sun on an elliptical orbit,that is the path of the orbit cannot be assumed to be circular as one might do at a high school level.
I have visited several sources online, I have tried to derive the equation myself and no matter what I do, I cannot seem to come up with an equation for this. I did come across the formula for orbital velocity in terms of the radius of orbit at any given instant; however, I could not use this to figure out a relationship between velocity and time either. The equation is as follows: $$v=\sqrt {\frac{2GM}{r}-\frac{GM}{a}}$$ Where, $$v$$ is the orbital velocity, $$G$$ is Newton's gravitational constant, $$M$$ is the mass of Sun, $$r$$ is the orbital radius of a planet around the sun at any given instant (not constant on an elliptical path), and $$a$$ is the semimajor axis of the elliptical orbit.
On the other hand, when I tried deriving the equation, I got this differential equation:
Now, I tried solving this differential equation and I believe this is not solvable since the integral for the velocity terms is unsolvable (if I am correct). So, for this, I reached out to my Math teacher and asked her if I there was some other method to solve the differential equation and she suggested I use the slope fields method. I tried it out but for some reason, this specific differential equation is not plotting out on any of the technology I used for plotting slope fields. Next I tried to plot these slope fields manually by the use of a graphing calculator and realized that the differential equation that for a particular range of $$v$$, $$\frac{dv}{dt}$$ is coming out to be complex.
So from here, I tried to find the range for which the value of $$\frac{dv}{dt}$$ is real; however, so far I have been unsuccessful. If someone can provide me with a pre-existing equation which describes this relation or knows a method derive the equation, or knows how to solve the differential equation mentioned above, it would be of great help to me.
• BTW, it's not a good idea to plug all those raw numbers into your equations. It's much better to use symbolic constants, and only plug in actual numbers after you've done all the algebra & calculus. Dec 22, 2021 at 10:13
The algorithm for the calculation is as follows:
1. We call $$t_0$$ the time of passage through the perihelion and $$T$$ the orbital period. We first calculate the mean anomaly $$M$$ at the instant $$t$$ we want to know the velocity
$$M=\frac{2\pi}T \ (t-t_0)$$
1. From the mean anomaly we calculate the eccentric anomaly $$E$$ using Kepler's Equation. Because it is a transcendental equation, the calculation must be done by numerical methods, but it is a very simple calculation. Look, for example, Kepler's Equation Fixed-point iteration
$$E=M+e \sin E$$
1. Once the eccentric anomaly is known, we calculate the **true anomaly $$\theta$$ :
$$\tan \frac{\theta}2=\sqrt{\frac{1+e}{1-e}} \ \tan \frac E 2$$
1. From the true anomaly we calculate the distance $$r$$ to the Sun by means of:
$$r=\frac{a \ (1-e^2)}{1+e\cos \theta}$$
1. Finally, knowing the mass of the sun $$M_s$$, we calculate the velocity by means of:
$$v=\sqrt{2 G M_s \left (\frac 1 r - \frac 1{2a} \right )}$$
Best regards.
** If desired, it is possible "to skip" the calculation of the true anomaly by using directly the expression of the radius vector as a function of the eccentric anomaly:
$$r=a \ (1-e \cos E)$$
If you can use numeric values for position and velocity as a function of time, you might go for a computer generated simulation of the orbit. There are various techniques for improving accuracy. You might aim for an orbit that ends where it started.
• Hello! Thank you for the response; however, I had already tried what you have suggested. I picked up data from the NASA Horizons System and plotted the graph as well. The problem with this method is that even though the final function (v-t function) is periodic, it cannot be modelled by the usual periodic functions (trigonometric functions). I am not able to figure out what model suits those graphs best and therefore I am trying to derive the models on my own. Look at this question I asked on Math stack exchange: math.stackexchange.com/q/4337550/950470. Dec 27, 2021 at 18:48
• I like to do simulations on a spreadsheet. It's easy to plot x and y coordinates using an orbit equation. Then use conservation of energy to get the speed at each point. With the speed, you can estimate the time to reach the next point. Dec 28, 2021 at 14:44 | 1,164 | 4,871 | {"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": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 23, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-22 | latest | en | 0.963119 |
https://www.easyerra.com/confidence-interval-calculator/ | 1,701,468,461,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100308.37/warc/CC-MAIN-20231201215122-20231202005122-00321.warc.gz | 838,453,838 | 16,892 | # Confidence Interval Calculator
Calculate confidence intervals efficiently with our Confidence Interval Calculator.
Confidence Level 909599 % Sample Size % Population Percentage %
Result: Confidence Interval %
## Similar Calculators:
A confidence interval is a range of values that provides a statistical measure of the uncertainty or margin of error associated with a particular estimate, typically the mean, proportion, or other population parameter. It’s a valuable tool for expressing the reliability of sample-based estimates and is often used in research, surveys, and statistical analysis.
## How to Calculate a Confidence Interval
Calculating a confidence interval involves several steps, and the exact formula or method used may vary depending on the type of data and the desired confidence level. Here’s a general overview of the process:
### Step 1: Gather Data
Collect the data for which you want to calculate a confidence interval. This could be data related to a population’s mean, proportion, or other parameters.
### Step 2: Choose a Confidence Level
Select a desired confidence level, typically expressed as a percentage. Common confidence levels include 95%, 99%, or other values depending on your specific requirements.
### Step 3: Calculate the Sample Mean and Standard Error
For a confidence interval related to a population mean, calculate the sample mean and standard error. The standard error is a measure of the uncertainty associated with the sample estimate. The formula for standard error may vary depending on the data and sample size.
### Step 4: Determine the Appropriate Z-Score or T-Score
Choose the corresponding z-score or t-score from a standard normal distribution table based on your chosen confidence level. If the sample size is large (typically over 30), you can use z-scores. For smaller sample sizes, you should use t-scores from a t-distribution table. The z-score or t-score depends on the confidence level and the number of degrees of freedom.
### Step 5: Calculate the Margin of Error
Determine the margin of error by multiplying the standard error by the selected z-score or t-score. This accounts for the uncertainty and variability in the data.
### Step 6: Calculate the Confidence Interval
Finally, calculate the confidence interval by adding and subtracting the margin of error from the sample mean. This yields a range of values within which you can be confident the population parameter falls.
## Interpreting the Confidence Interval
The confidence interval provides a range of values within which the true population parameter is likely to be found with the chosen confidence level. For example, if you calculate a 95% confidence interval for a population mean, it means that you can be 95% confident that the true population mean falls within that interval. The narrower the confidence interval, the more precise the estimate, while wider intervals indicate greater uncertainty.
Confidence intervals are essential in scientific research, surveys, quality control, and decision-making processes, as they offer insights into the reliability and uncertainty of statistical estimates. | 592 | 3,160 | {"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} | 3.625 | 4 | CC-MAIN-2023-50 | latest | en | 0.802421 |
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Rediff News All News » Sports » Suarez grateful for Liverpool support: Dalglish
Suarez grateful for Liverpool support: Dalglish
Last updated on: December 24, 2011 09:32 IST
'He was quite emotional and very grateful'
Liverpool manager Kenny Dalglish has defended the club's support of Luis Suarez following criticism for their reaction to the eight-match ban given to Uruguayan who was found guilty of racially abusing Manchester United's Patrice Evra.
Suarez is free to play for the club until an appeal is heard and his team mates showed they were fully behind the striker when they all wore T-shirts emblazoned with his picture before the Premier League match at Wigan Athletic on Wednesday.
"He was quite emotional and very grateful," Dalglish said, on Friday, as they prepare for a hectic holiday fixture list.
"I don't think it's ever a disappointment when your team-mates and the people you work for give you their undivided support. For me, that's the least he deserves."
Liverpool are waiting for full documentation of the case from the Football Association and Dalglish was reluctant to go into detail about what it might contain.
Image: Luis Suarez
'Players have done nothing wrong'
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The club was criticised by anti-racist organisations and some former players on Thursday for their stance over Suarez, but Dalglish said his players had done nothing wrong.
"We've certainly said what we have to say. We'll wait until we get the judgement and take it from there," Dalglish said.
"The players have made their statement both visually and verbally. It's best for everybody to wait and see what the written retort is because nobody knows - we don't even know.
"At this moment in time the club aren't permitted to go into any further detail than what they have done in their statement, which couldn't have caused anyone any trouble.
"I don't think the players have caused any trouble with the FA either with their statement or the t-shirts, so if we're not in any trouble we'll just leave it at that before we do give ourselves some trouble."
Liverpool face Dalglish's former club Blackburn Rovers on December 26 with the Scot offering some sympathy to Rovers' beleaguered boss Steve Kean who has come under fire with his side propping up the table.
Image: Kenny Dalglish
Photographs: Getty Images
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http://slideplayer.com/slide/3306665/ | 1,529,599,700,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864191.74/warc/CC-MAIN-20180621153153-20180621173153-00377.warc.gz | 295,442,992 | 22,495 | # 1 1 Slide © 2005 Thomson/South-Western Slides Prepared by JOHN S. LOUCKS ST. EDWARD’S UNIVERSITY.
## Presentation on theme: "1 1 Slide © 2005 Thomson/South-Western Slides Prepared by JOHN S. LOUCKS ST. EDWARD’S UNIVERSITY."— Presentation transcript:
1 1 Slide © 2005 Thomson/South-Western Slides Prepared by JOHN S. LOUCKS ST. EDWARD’S UNIVERSITY
2 2 Slide © 2005 Thomson/South-Western Chapter 4 Linear Programming Applications n Blending Problem n Portfolio Planning Problem n Product Mix Problem
3 3 Slide © 2005 Thomson/South-Western Blending Problem Ferdinand Feed Company receives four raw grains from which it blends its dry pet food. The pet food advertises that each 8-ounce packet meets the minimum daily requirements for vitamin C, protein and iron. The cost of each raw grain as well as the vitamin C, protein, and iron units per pound of each grain are summarized on the next slide.
4 4 Slide © 2005 Thomson/South-Western Blending Problem Vitamin C Protein Iron Vitamin C Protein Iron Grain Units/lb Units/lb Units/lb Cost/lb Grain Units/lb Units/lb Units/lb Cost/lb 1 9 12 0.75 1 9 12 0.75 2 16 10 14.90 2 16 10 14.90 3 8 10 15.80 3 8 10 15.80 4 10 8 7.70 Ferdinand is interested in producing the 8-ounce mixture at minimum cost while meeting the minimum daily requirements of 6 units of vitamin C, 5 units of protein, and 5 units of iron.
5 5 Slide © 2005 Thomson/South-Western Blending Problem n Define the decision variables x j = the pounds of grain j ( j = 1,2,3,4) x j = the pounds of grain j ( j = 1,2,3,4) used in the 8-ounce mixture used in the 8-ounce mixture n Define the objective function Minimize the total cost for an 8-ounce mixture: Minimize the total cost for an 8-ounce mixture: MIN.75 x 1 +.90 x 2 +.80 x 3 +.70 x 4 MIN.75 x 1 +.90 x 2 +.80 x 3 +.70 x 4
6 6 Slide © 2005 Thomson/South-Western Blending Problem Define the constraints Define the constraints Total weight of the mix is 8-ounces (.5 pounds): (1) x 1 + x 2 + x 3 + x 4 =.5 (1) x 1 + x 2 + x 3 + x 4 =.5 Total amount of Vitamin C in the mix is at least 6 units: (2) 9 x 1 + 16 x 2 + 8 x 3 + 10 x 4 > 6 (2) 9 x 1 + 16 x 2 + 8 x 3 + 10 x 4 > 6 Total amount of protein in the mix is at least 5 units: (3) 12 x 1 + 10 x 2 + 10 x 3 + 8 x 4 > 5 (3) 12 x 1 + 10 x 2 + 10 x 3 + 8 x 4 > 5 Total amount of iron in the mix is at least 5 units: (4) 14 x 2 + 15 x 3 + 7 x 4 > 5 (4) 14 x 2 + 15 x 3 + 7 x 4 > 5 Nonnegativity of variables: x j > 0 for all j
7 7 Slide © 2005 Thomson/South-Western n The Management Scientist Output OBJECTIVE FUNCTION VALUE = 0.406 OBJECTIVE FUNCTION VALUE = 0.406 VARIABLE VALUE REDUCED COSTS VARIABLE VALUE REDUCED COSTS X1 0.099 0.000 X1 0.099 0.000 X2 0.213 0.000 X2 0.213 0.000 X3 0.088 0.000 X3 0.088 0.000 X4 0.099 0.000 X4 0.099 0.000 Thus, the optimal blend is about.10 lb. of grain 1,.21 lb. of grain 2,.09 lb. of grain 3, and.10 lb. of grain 4. The mixture costs Frederick’s 40.6 cents. Blending Problem
8 8 Slide © 2005 Thomson/South-Western Portfolio Planning Problem Winslow Savings has \$20 million available for investment. It wishes to invest over the next four months in such a way that it will maximize the total interest earned over the four month period as well as have at least \$10 million available at the start of the fifth month for a high rise building venture in which it will be participating.
9 9 Slide © 2005 Thomson/South-Western Portfolio Planning Problem For the time being, Winslow wishes to invest only in 2-month government bonds (earning 2% over the 2-month period) and 3-month construction loans (earning 6% over the 3-month period). Each of these is available each month for investment. Funds not invested in these two investments are liquid and earn 3/4 of 1% per month when invested locally.
10 Slide © 2005 Thomson/South-Western Portfolio Planning Problem Formulate a linear program that will help Winslow Savings determine how to invest over the next four months if at no time does it wish to have more than \$8 million in either government bonds or construction loans.
11 Slide © 2005 Thomson/South-Western Portfolio Planning Problem n Define the decision variables g j = amount of new investment in g j = amount of new investment in government bonds in month j government bonds in month j c j = amount of new investment in construction loans in month j c j = amount of new investment in construction loans in month j l j = amount invested locally in month j, l j = amount invested locally in month j, where j = 1,2,3,4 where j = 1,2,3,4
12 Slide © 2005 Thomson/South-Western Portfolio Planning Problem n Define the objective function Maximize total interest earned over the 4-month period. Maximize total interest earned over the 4-month period. MAX (interest rate on investment)(amount invested) MAX (interest rate on investment)(amount invested) MAX.02 g 1 +.02 g 2 +.02 g 3 +.02 g 4 MAX.02 g 1 +.02 g 2 +.02 g 3 +.02 g 4 +.06 c 1 +.06 c 2 +.06 c 3 +.06 c 4 +.0075 l 1 +.0075 l 2 +.0075 l 3 +.0075 l 4 +.0075 l 1 +.0075 l 2 +.0075 l 3 +.0075 l 4
13 Slide © 2005 Thomson/South-Western Portfolio Planning Problem n Define the constraints Month 1's total investment limited to \$20 million: Month 1's total investment limited to \$20 million: (1) g 1 + c 1 + l 1 = 20,000,000 (1) g 1 + c 1 + l 1 = 20,000,000 Month 2's total investment limited to principle and interest invested locally in Month 1: Month 2's total investment limited to principle and interest invested locally in Month 1: (2) g 2 + c 2 + l 2 = 1.0075 l 1 (2) g 2 + c 2 + l 2 = 1.0075 l 1 or g 2 + c 2 - 1.0075 l 1 + l 2 = 0
14 Slide © 2005 Thomson/South-Western Portfolio Planning Problem n Define the constraints (continued) Month 3's total investment amount limited to principle and interest invested in government bonds in Month 1 and locally invested in Month 2: (3) g 3 + c 3 + l 3 = 1.02 g 1 + 1.0075 l 2 (3) g 3 + c 3 + l 3 = 1.02 g 1 + 1.0075 l 2 or - 1.02 g 1 + g 3 + c 3 - 1.0075 l 2 + l 3 = 0
15 Slide © 2005 Thomson/South-Western Portfolio Planning Problem n Define the constraints (continued) Month 4's total investment limited to principle and interest invested in construction loans in Month 1, goverment bonds in Month 2, and locally invested in Month 3: (4) g 4 + c 4 + l 4 = 1.06 c 1 + 1.02 g 2 + 1.0075 l 3 (4) g 4 + c 4 + l 4 = 1.06 c 1 + 1.02 g 2 + 1.0075 l 3 or - 1.02 g 2 + g 4 - 1.06 c 1 + c 4 - 1.0075 l 3 + l 4 = 0 \$10 million must be available at start of Month 5: (5) 1.06 c 2 + 1.02 g 3 + 1.0075 l 4 > 10,000,000 (5) 1.06 c 2 + 1.02 g 3 + 1.0075 l 4 > 10,000,000
16 Slide © 2005 Thomson/South-Western Portfolio Planning Problem n Define the constraints (continued) No more than \$8 million in government bonds at any time: (6) g 1 < 8,000,000 (6) g 1 < 8,000,000 (7) g 1 + g 2 < 8,000,000 (7) g 1 + g 2 < 8,000,000 (8) g 2 + g 3 < 8,000,000 (8) g 2 + g 3 < 8,000,000 (9) g 3 + g 4 < 8,000,000 (9) g 3 + g 4 < 8,000,000
17 Slide © 2005 Thomson/South-Western Portfolio Planning Problem n Define the constraints (continued) No more than \$8 million in construction loans at any time: (10) c 1 < 8,000,000 (10) c 1 < 8,000,000 (11) c 1 + c 2 < 8,000,000 (11) c 1 + c 2 < 8,000,000 (12) c 1 + c 2 + c 3 < 8,000,000 (12) c 1 + c 2 + c 3 < 8,000,000 (13) c 2 + c 3 + c 4 < 8,000,000 (13) c 2 + c 3 + c 4 < 8,000,000 Nonnegativity: g j, c j, l j > 0 for j = 1,2,3,4
18 Slide © 2005 Thomson/South-Western Product Mix Problem Floataway Tours has \$420,000 that can be used to purchase new rental boats for hire during the summer. The boats can be purchased from two different manufacturers. Floataway Tours would Floataway Tours would like to purchase at least 50 boats and would like to purchase the same number from Sleekboat as from Racer to maintain goodwill. At the same time, Floataway Tours wishes to have a total seating capacity of at least 200.
19 Slide © 2005 Thomson/South-Western Formulate this problem as a linear program. Maximum Expected Maximum Expected Boat Builder Cost Seating Daily Profit Boat Builder Cost Seating Daily Profit Speedhawk Sleekboat \$6000 3 \$ 70 Silverbird Sleekboat \$7000 5 \$ 80 Catman Racer \$5000 2 \$ 50 Classy Racer \$9000 6 \$110 Product Mix Problem
20 Slide © 2005 Thomson/South-Western n Define the decision variables x 1 = number of Speedhawks ordered x 1 = number of Speedhawks ordered x 2 = number of Silverbirds ordered x 2 = number of Silverbirds ordered x 3 = number of Catmans ordered x 3 = number of Catmans ordered x 4 = number of Classys ordered x 4 = number of Classys ordered n Define the objective function Maximize total expected daily profit: Maximize total expected daily profit: Max: (Expected daily profit per unit) Max: (Expected daily profit per unit) x (Number of units) x (Number of units) Max: 70 x 1 + 80 x 2 + 50 x 3 + 110 x 4 Product Mix Problem
21 Slide © 2005 Thomson/South-Western n Define the constraints (1) Spend no more than \$420,000: 6000 x 1 + 7000 x 2 + 5000 x 3 + 9000 x 4 < 420,000 6000 x 1 + 7000 x 2 + 5000 x 3 + 9000 x 4 < 420,000 (2) Purchase at least 50 boats: (2) Purchase at least 50 boats: x 1 + x 2 + x 3 + x 4 > 50 x 1 + x 2 + x 3 + x 4 > 50 (3) Number of boats from Sleekboat equals number of boats from Racer: (3) Number of boats from Sleekboat equals number of boats from Racer: x 1 + x 2 = x 3 + x 4 or x 1 + x 2 - x 3 - x 4 = 0 x 1 + x 2 = x 3 + x 4 or x 1 + x 2 - x 3 - x 4 = 0 Product Mix Problem
22 Slide © 2005 Thomson/South-Western n Define the constraints (continued) (4) Capacity at least 200: 3 x 1 + 5 x 2 + 2 x 3 + 6 x 4 > 200 3 x 1 + 5 x 2 + 2 x 3 + 6 x 4 > 200 Nonnegativity of variables: Nonnegativity of variables: x j > 0, for j = 1,2,3,4 x j > 0, for j = 1,2,3,4 Product Mix Problem
23 Slide © 2005 Thomson/South-Western n Complete Formulation Max 70 x 1 + 80 x 2 + 50 x 3 + 110 x 4 s.t. 6000 x 1 + 7000 x 2 + 5000 x 3 + 9000 x 4 < 420,000 6000 x 1 + 7000 x 2 + 5000 x 3 + 9000 x 4 < 420,000 x 1 + x 2 + x 3 + x 4 > 50 x 1 + x 2 + x 3 + x 4 > 50 x 1 + x 2 - x 3 - x 4 = 0 x 1 + x 2 - x 3 - x 4 = 0 3 x 1 + 5 x 2 + 2 x 3 + 6 x 4 > 200 3 x 1 + 5 x 2 + 2 x 3 + 6 x 4 > 200 x 1, x 2, x 3, x 4 > 0 x 1, x 2, x 3, x 4 > 0 Product Mix Problem
24 Slide © 2005 Thomson/South-Western n Partial Spreadsheet Showing Problem Data Product Mix Problem
25 Slide © 2005 Thomson/South-Western n Partial Spreadsheet Showing Solution Product Mix Problem
26 Slide © 2005 Thomson/South-Western n Solution Summary Purchase 28 Speedhawks from Sleekboat. Purchase 28 Speedhawks from Sleekboat. Purchase 28 Classy’s from Racer. Purchase 28 Classy’s from Racer. Total expected daily profit is \$5,040.00. Total expected daily profit is \$5,040.00. The minimum number of boats was exceeded by 6 (surplus for constraint #2). The minimum number of boats was exceeded by 6 (surplus for constraint #2). The minimum seating capacity was exceeded by 52 (surplus for constraint #4). The minimum seating capacity was exceeded by 52 (surplus for constraint #4). Product Mix Problem
27 Slide © 2005 Thomson/South-Western n Sensitivity Report Product Mix Problem
28 Slide © 2005 Thomson/South-Western n Sensitivity Report Product Mix Problem
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## Kirchhoff’s Law Writing Service
Introduction
Kirchhoff’s laws, pair of laws mentioning basic constraints on the present and voltage in an electrical circuit. The very first of these states that at any provided immediate the amount of the voltages around any closed course, or loop, in the network is no. The 2nd states that at any junction of courses, or node, in a network the amount of the currents reaching any immediate amounts to the amount of the currents streaming away.
In examining circuits utilizing Kirchhoff’s law, it is helpful to bear in mind the following standards.
Kirchhoff’s Law Writing Service
1. Draw the circuit and appoint labels to the known and unknown amounts, including currents in each branch. You must designate instructions to currents; do not stress if you think improperly the instructions of a particular unknown present, as the response resulting from the analysis in this case will just come out damaging, but with the ideal magnitude.
2. Apply the junction rule to as numerous junctions in the circuit as possible to acquire the maximum number of independent relations.
3. Apply the loop guideline to as numerous loops in the circuit as needed in order to solve for the unknowns. If one has n unknowns in a circuit one will need n independent formulas, note that. In basic there will be more loops present in a circuit than one needs to resolve for all the unknowns; the relations resulting from these “extra” loops can be utilized as a consistency look at your last answers.
4. Solve the resulting set of simultaneous formulas for the unknown quantities.
Kirchhoff’s Law, Expect the point is on the conductor through which the current is flowing, then the exact same present crosses the point which can alternatively stated that the present gets in at the point and same will leave the point. Overall amount of current gets in at the junction point have to be exactly equal to total amount of present that leaves the junction. If we consider all the currents go into in the junction are considered as favorable current, then convention of all the branch currents leaving the junction are negative.
Believe about one point on a closed loop in an electrical circuit. Kirchhoff’s Law, If someone goes to any other point on the same loop, he or she will discover that the capacity at that 2nd point might be different from very first point. That is what Kirchhoff’s Law Voltage law states.
The overall current streaming into any DC circuit node, likewise called a branch point, is always the exact same as the total current streaming out of the node. The total current passing into the node is a + b + c + d, and the overall current flowing out is e + f. Suppose the point is on the conductor through which the existing is flowing, then the exact same present crosses the point which can additionally said that the existing enters at the point and same will leave the point. Kirchhoff’s Law, Total quantity of current gets in at the junction point have to be exactly equivalent to total quantity of present that leaves the junction. If we think about all the currents go into in the junction are considered as favorable present, then convention of all the branch currents leaving the junction are unfavorable.
Kirchhoff’s Law offer an easy approximate description of how voltage and existing act in electrical circuits. Individuals truly like this easy method of taking a look at things, and they frequently take discomforts to develop their circuits so about decrease the significance of the different offenses of Kirchhoff’s law. The direct effectiveness of these laws is far less than you may have thought of.
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# ps combinatorics
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ps combinatorics [#permalink] 01 Sep 2004, 11:15
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6 ppl need to divide into 3 groups of 2
is it 6c2 * 4c2 * 2c2*??
another one... 5 letter word
DEFGG
how mnay words can you make where the two G's are not together!
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Your first question needs to be divided by 3! because order of group does not count. Hence, answer should be 6c2*4c2/3!
Second question is 5! - 4! = 120 - 24 = 96
4! is total number of unfavorable outcomes when 2 Gs(as a single entity) are next to each other:
(GG) - D - E - F --> 4! ways of sorting these differently
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Paul
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You are right on the first combination equation.
As for the word, my guess is that there are 5! total words that can be made....120. There are 5C2 unfavorable combinations of having the two G's together. So 120-5C2=110.
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I am wrong for the second question:
Total outcome should be 5!/2! because there are 2 G's which can be interchanged. --> 60
4! would be all unfavorable outcomes as previously explained
Answer should be: 60 - 24 = 36
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I would like to underline one point and gather your comments guys :
About the 2nd question, the 96 answer considers that all words are 5 letters long and the two G are distinct.
When I first tried to answer I was thinking about 5 ar less characters words...
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thanks guys
36 is the caorrect answer for the second problem
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6 ppl need to divide into 3 groups of 2
is it 6c2 * 4c2 * 2c2*??
Hah ! Let me try something Ian taught me yesterday and see if I get this right !
6C2*4C2*2C2/3! = 15*6*1/6 = 15 ways
-----------------------------------------------------------------
another one... 5 letter word
DEFGG
how mnay words can you make where the two G's are not together!
Do the words have to be real words ? Or just any combination of letters as long as the two Gs are not together.
If the words have to be real:
Then the number of words = 0 (The only word that can be made is egg, and it's out because the two g's are together)
If it's any letters and the words do not need to make sense:
Then , we can either has 1 letter word, 2 letter words, 3 letter words, 4 letter words, can't be 5 as that will give 2 g's
So 1 letter word - 5
2 letter words - 4P2 (remove 1 G) = 12 (use permutation assuming we can swtich the letters around and form another word)
3 letter words - 4P3 = 24
4 letter words - 4P4 = 16
So total - 4 + 12 + 24 + 16 = 56
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ywilfred, I believe that you have to use all the letters to form a 5 letter word. Also, why did you omit to account for 5 letter words in your calculation? Although there are 2 G's, you should still account for words that can be formed when the 2 G's are apart.
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The answer to the first problem according to my working is
6C2*4C2*2C2/3!
=15*6/6=15
In the word problem, the number of different words possible is 5!/2! ( as G is coming twice)
=60
If G and G is treated as one unit, number of ways that the rest of the letters can be arranged= 4!=24
The number of ways in which the two Gs will be apart
=60-24
=36
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humakhan
Junior Member
A total of 52 dogs and cats were on the ranch. if three times the number of dogs exceeded the number of cats by four , how many of each were on the ranch?
3D + 4C = 52
subtract 4C from both sides
3D = 48
48/3
D=16
3(16) + 4C = 52
48 + 4C = 52
sybtract 48 from both sides
4C = 4
4/4
C = 1
check
3(16dogs) + 4(1cat) = 52 on the ranch
48dogs + 4cats = 52
52dogs&cats= 52 on the ranch
wrong
stapel
Super Moderator
Staff member
Re: [MOVED] 52 dogs and cats; three times the number of dogs
humakhan said:
A total of 52 dogs and cats were on the ranch. if three times the number of dogs exceeded the number of cats by four, how many of each were on the ranch?
3D + 4C = 52
Hint: Look up the word "exceeded". Once you know what it means, you'll know what operation to use.
humakhan said:
cats = 1
dogs = 16
Since 1 + 16 does not equal 52, the above cannot be the correct solution.
Eliz.
humakhan
Junior Member
exceeded means :To be greater than
so here is how you write the problem out
3D > 4C = 52
Denis
Senior Member
"exceeds by 4" simply means "4 more"; like 10 is 4 more than 6 :idea: | 394 | 1,182 | {"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-2019-13 | latest | en | 0.955492 |
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### Learning Objectives
• Use mean and standard deviation to describe a distribution.
### Introduction
In the section “Distributions for Quantitative Data,” we discussed the spread of a distribution in terms of a typical range of values. In “Quantifying Variability Relative to the Median,” we made this idea more precise with the interquartile range, IQR. The IQR gives us a measure of spread about the median. We defined a typical range of values about the median as the values between the first and third quartiles.
Now we want to develop a numerical measure of spread that we can use with the mean. In constructing a measure of spread about the mean, we want to compute how far a “typical” number is away from the mean.
Let’s consider the sample data set 2, 2, 4, 5, 6, 7, 9. The mean of this data set is
$\stackrel{¯}{x}=\frac{2+2+4+5+6+7+9\text{}}{7}\text{}=\text{}\frac{35}{7}=5$
Here is a dotplot of this data set with the mean marked by the vertical blue line.
We can see that some data is close to the mean and some data is further from the mean.
Since we want to see how the data points deviate from the mean, we determine how far each point is from the mean. We compute the difference between each of these values and the mean. These differences are called the deviations from the mean for each point.
2 − 5 = −3 2 − 5 = −3 4 − 5 = −1 5 − 5 = 0 6 − 5 = 1 7 − 5 = 2 9 − 5 = 4
When visualized on a dotplot, these differences are viewed as distances between each point and the mean. A negative difference indicates that the data point is to the left of the mean (shown in blue on the graph below). A positive difference indicates that the data point is to the right of the mean (shown in green on the graph below).
Our goal is to develop a single measurement that summarizes a typical distance from the mean. Before we continue, let’s practice determining the distance of a single data point from the mean.
### Learn By Doing
The two questions below refer to the following dotplot. The mean is 9 and it is marked by the vertical blue line.
Since we want to determine how far a typical number is away from the mean, we might try to average these numbers. However, if we add them all up, we will get 0 (try it). Getting 0 with this procedure (finding differences from mean and adding them all together) is no accident – it always produces 0. We have to overcome this problem.
Recall that we are trying to find the typical distance between data points and the mean. It therefore makes sense to take the absolute value of each of these differences.
| 2 − 5 | = | −3 | = 3 | 2 − 5 | = | −3 | = 3 | 4 − 5 | = | −1 | = 1 | 5 − 5 | = | 0 | = 0 | 6 − 5 | = | 1 | = 1 | 7 − 5 | = | 2 | = 2 | 9 − 5 | = | 4 | = 4
Now we can compute the average of these deviations. There are seven data points, so we add these seven distances and divide by 7. The result is a measure of spread about the mean called the average deviation from the mean (ADM).
$\frac{3+3+1+0+1+2+4\text{}}{7}\text{}=\text{}\frac{14}{7}=2$
We can indicate this average deviation on a dotplot with a graphic similar to a boxplot as follows.
The shaded box in the middle is centered at the mean. It extends left and right a distance of 1 average deviation from the mean. Because the average deviation about the mean for this data set is 2, the box starts at 3 (because 5 − 2 = 3) and ends at 7 (because 5 + 2 = 7). In this way, we can use the ADM to define a typical range of values about the mean. Notice that this typical range of values (within 1 ADM of the mean) contains more than half of the values in the data set.
The goal of the next Learn By Doing exercise is to improve our intuition of what the ADM measures. We use the following simulation to investigate how the ADM responds to changes in a data set.
Instructions for adding or removing data points:
• To add a point, move the slider to the value you want, then click the + sign.
• To remove a point, move the slider to the value you want, then click the – sign.
• To reset the simulation to a blank screen, click the button in the upper left corner that says Reset.
### Learn By Doing
Before we continue, let’s summarize our main points:
• The ADM (average distance from the mean) is a measurement of spread about the mean. More precisely, ADM measures the average distance of the data from the mean.
• We can use the ADM to define a typical range of values about the mean. We mark the mean, then we mark 1 ADM below the mean and 1 ADM above the mean. This interval is centered at the mean and captures typical values about the mean.
Using these two ideas, we can estimate the ADM by looking at a graph of the distribution of data. We practice this important skill in the next Learn By Doing.
### Learn By Doing
In the next example, we compare the ADM as a measure of spread to the other ways we have measured spread.
## Measuring Variability in Different Ways
The following dotplots show the potassium content in 76 cereals. Compare children’s cereals to adult cereals. Which type of cereal has more variability in potassium content?
We can visually see that there is more variability in the potassium content of the adult cereals than in the children’s cereals. We can measure this spread in three ways:
• Using overall range: The range of potassium content is larger for the adult cereals than for the children’s cereals. The children’s cereal set has a range of 90 (because 110 − 20 = 90), whereas the adult cereal set has a range of 315 (because 330 − 15 = 315).
• Using IQR: The IQR of the adult cereal set is larger than the IQR of the children’s cereal set. The adult cereal set has an IQR of 80, since for that set Q1 = 80 and Q3 = 160. The children’s cereal set has an IQR of 30, since for that set Q1 = 35 and Q3 = 65. Notice here we use the median as a measure of center. The median is marked with a red line. IQR measures spread about the median.
• Using ADM: The children’s cereal data set has an ADM of 22. The adult cereal data set has an ADM of 55. Notice here we use the mean as a measure of center. The mean is marked with a blue line. ADM measures spread about the mean.
Based on the preceding example, we might expect the data set with the larger range to also have the larger ADM. This is not true, as we illustrate in the next example.
### Example
Which data set has more variability? Our answer to this question depends on how we measure variability.
We can see that the overall range is larger for data set b.
• Data set a: range = 100 − 0 = 100
• Data set b: range = 200 − 0 = 200
If we use overall range to measure spread, we will say that data set b has more variability.
• Data set a: ADM = 41
• Data set b: ADM = 23
Most of the data in data set a is located away from the mean, so the ADM is large: 41. Compare this to data set b. Most of the data in data set b is located close to the mean, so the ADM is small: 23.
If we use ADM as a measure of spread, we will say that data set a has more variability.
The ADM is a reasonable measure of spread about the mean, but there is another measure that is used much more often: the standard deviation (SD). The standard deviation behaves very much like the average deviation. So all of the work we have done on this page is useful in understanding standard deviation. We discuss standard deviation next. | 1,850 | 7,380 | {"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} | 4.9375 | 5 | CC-MAIN-2024-26 | latest | en | 0.909137 |
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# New way to find vertex ok if I have y = x^2+6x+8 so 9+y = (x^2+6x+9) + 8 we subtract 8 so we have 9-8+y = (x+3)^2 we have (1+y) = (x+3)^2 right and 1+y=0 y is equal to -1 and x is x+3= 0 x is -3 :) ⚑ Flag
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A General Solution Calculator is an online calculator that helps you solve complex differential equations. The General Solution Calculator needs a single input, a differential equation you
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## Differential Equations
The general solution of the differential equation is one that comprises as many arbitrary constants as the order of the differential equation. Thus, a general solution of $$ntℎ$$ order differential equation has $$n$$ arbitrary constants. For example, $$y = A\,\cos \,x + B\,\sin \,x$$ is the general solution of the second order diffe See more
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## Solution Of A Differential Equation -General and Particular
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## What is the general solution of the differential equation?
Solve ordinary differential equations (ODE) step-by-step. full pad ». x^2. x^ {\msquare} \log_ {\msquare} \sqrt {\square} \nthroot [\msquare] {\square} \le. \ge. | 440 | 2,000 | {"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} | 3.75 | 4 | CC-MAIN-2023-06 | latest | en | 0.908852 |
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