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Section 11.5 Exercise 13-TQuestion Holpsurvey 0f 25 randomly sclocted customers found the ag0s shown (in years) . The mean Is 31.60 years and the standard deviation...

Question

Section 11.5 Exercise 13-TQuestion Holpsurvey 0f 25 randomly sclocted customers found the ag0s shown (in years) . The mean Is 31.60 years and the standard deviation is 91 years: a) Construct 80% confidence interval Ior the mean ag0 of all customers , assuming that tho assumptions and conditions for the confidence interval have been met; D) How large tho margin 0f emor? c) How would the confidence Interval change you had assumed that the standard deviation was krown to be 10.0 years?385] 3322 1

Section 11.5 Exercise 13-T Question Holp survey 0f 25 randomly sclocted customers found the ag0s shown (in years) . The mean Is 31.60 years and the standard deviation is 91 years: a) Construct 80% confidence interval Ior the mean ag0 of all customers , assuming that tho assumptions and conditions for the confidence interval have been met; D) How large tho margin 0f emor? c) How would the confidence Interval change you had assumed that the standard deviation was krown to be 10.0 years? 385] 3322 16 47 27 33 31 21 What ( Ine coniidence interval? (28.99 34.21 Section 11.7 Exercise 20 Question Help A random sample of the amounts for 18 purchases was taken. The mean was S42.93,the standard deviation was S22.39, and the margin of error for a 95% confidence interval was 511.13_ Assume that tn 2.0 for the 95% confidence intervals . a) To reduce the margin of error to about S5, how large would the sample size have t0 be? b) How large would the sample size have to be to reduce the margin of error to S1.1? a) The new sample size should be (Round up t0 the nearest integer:)



Answers

Determine the point estimate of the population mean and margin of error for each confidence interval. 22. A simple random sample of size $n$ is drawn. The sample mean, $\bar{x},$ is found to be 35.1 , and the sample standard deviation, $s$ is found to be 8.7 (a) Construct a $90 \%$ confidence interval for $\mu$ if the sample size, $n,$ is 40 (b) Construct a $90 \%$ confidence interval for $\mu$ if the sample size, $n,$ is $100 .$ How does increasing the sample size affect the margin of error, $E ?$ (c) Construct a $98 \%$ confidence interval for $\mu$ if the sample size, $n,$ is $40 .$ Compare the results to those obtained in part (a). How does increasing the level of confidence affect the margin of crror, $E ?$ (d) If the sample size is $n=18,$ what conditions must be satisfied to compute the confidence interval?

The following is a solution to number 20. Where were given the summary statistics X. Bar the sample mean is 100 and 23 then the population standard deviation. So we actually do know sigma this time, the population standard deviation is 17. And this allows us to use the Z interval instead of the T interval. And we're asked to find the 94% confidence interval if N is greater than area, if any is equal to 20, 94% confidence interval of in his 12 and then 85% confidence interval whenever it is back to 20. And then we're gonna compare those those three confidence intervals. Now the reason why we can have sample sizes so small is because the population of the original data was normally distributed. So the fact that it's a normal distribution that allows us to have these smaller sample sizes. So let's go and get started. I'm gonna use technology that you can use the formula if you so wish. But I'm gonna use the T. I. T. For I think it works quite well. So if you go to stat and then air over two tests and since we have the population standard deviation we can go ahead and find the Z. Interval. So there's the intervals that seventh option and here you have data or stats. Now we're not giving any data, we're just giving summary stats. So keep that stats highlighted And then the signal was 17 and the the mean 123 The sample size is 20 and the confidence level was .94 for 94%. And whenever we calculate this top line that gives us the confidence interval. So 115.85 to 130.15. I'm gonna write that down. 115 0.85 to 130 point 15 So we're 94% confident that it's between those two values. Now, let's go back to the calculator and go back to stat tests and it's the seventh option. This time we're going to decrease the sample size to 12, we're going to keep the confidence level the same, but we're going to decrease the sample size to 12 and let's see what happens. Well, that's 113.77. All the way up to 132.23 113 0.77 To 132 0.23 And let's just kind of look at this and see how these compare. So this second one, part B is quite a bit wider than Part A. The lower bound is smaller and the upper bound is larger. And so that just goes to show that as in decreases as you decrease your sample size, the margin of error increases, which means that confidence interval will be wider. Okay, so the less you know, or the smaller your sample size, that margin there will be bigger. Um And that is going to make the confidence interval wider. All right. So then last one. This is where we go to 85%, but then back to the original sample size. So if you go to stat tests and it's the seventh option again. So everything else here is the same. So we're gonna change this to 20. So n equals 20 and then the confidence level um this time is 85%. So we'll save .85. And then whenever we calculate this we get a (117.53 and 128.47 1 17.53. All the way up to 128 4 7. So if you compare this with part a. This is actually narrower so the lower bound is higher and the upper bound is lower. This means that part see is narrower. Mhm than part A. Okay, so as the confidence level decreases, the margin of error also decreases. Yeah, meaning the confidence interval is narrower or skinnier. Okay. And then part D. It says can we conduct a confidence interval if the original population is not normal and the quick answer is no. Um If population is not normal then And must be at least 30 And if you look back to the sample sizes, these sample sizes are not big enough. 2012 and 20, those are not big enough. It needs to be at least 30 if the original population is not normal or if it's skewed or have outliers of some kind, and then the last part part e if there is an outlier greater than the mean, how does that shake things up? So if there is one outlier greater than the mean, the confidence interval would be shifted up towards that outlier. Okay, so outliers tend to skew the data and excuse it towards the direction of that outlier, especially the mean. The mean is pretty non robust against outliers, so the mean is gonna get shifted up towards that outlier. And so will the confidence confidence that will follow that that print that statistic.

The following is a solution to number 11 and were given summary stats that the sample mean X bar is 18.4 and the sample standard deviation is 4.5. Now we only know the sample standard deviation, not the population standard deviation. So since we don't know the population standard deviation sigma, we have to use the tea interval instead of the z interval, so keep that in mind moving forward. So I'm gonna use technology for this problem here. We basically have three questions that are very, very similar, and then um one where we talked a little bit about conditions for inference, um now it doesn't say anything about the normality of this population, that's going to come up here in part D, But it doesn't matter because these sample sizes are big, they're greater than 30, so it doesn't matter how this population is distributed, that sample size is big enough. So we're asked to find the 95% confidence interval whenever end is 35, Then we find the 95% confidence interval will never end this 50, we're gonna compare that so the same confidence level but the sample size is a little bit bigger. Then we come down here and we increase that sample that confidence level of 99% confidence. And then we're back down to the n equals 35. So let's take a look and see what happens. So I'm gonna use the T. I. T. Four but you can use any sort of technology want or you can go and do the formula but if you go to stat and then arrow over two tests and it's this eighth option here, the T. Interval and um make sure that the summary stats is highlighted because we don't have data. We're just giving summary stats and X bar was 18.4. The s was 4.5. The end for this first one was 35 then that confidence level is 350.95 and that represents 95%. So whenever we calculate this top band here that's our confidence level. So about 16.9-19.9 is going to be our interval. So let's go and write that down. So 16 Point I'll go ahead and do what they did. So 8854 And then the upper bound is 19.946. Okay So that's our first confidence interval. Now let's see what happens when we increase that sample size from 35 to 50. So we go back to stat tests and it's this eight option here, the t interval And this time the only thing that changes is the end and that's 50 now. So we calculate And we go from about 17 to 19.7. Okay, so let's write that down. So 17 .1-1 All the way up to 19 six 79 So let's compare now. So it looks like the lower bound has actually increased by, I don't know maybe about point 2.5 4.3 and the upper bound has decreased. That means that interval is narrower. So what does that say about in? And the margin of air? Well, as in increases, which is what happened here? The margin of air decreases. Making the confidence interval more narrow. Okay, so they're inversely related. So as in increases or degrees of freedom, increase. That margin of error is going to decrease, making it less variable. And that's the reason why that interval is a little bit narrower. So now let's go back to the 35 sample size. But this time we're going to increase that confidence level to 99%. So go back to stat tests and the 8th option. Okay. And we're back down to 35 but we're going to increase that sea level of the .99. And then when we calculate, let's see what happens here. So we go from 16.3 to 20.5. Okay, so let's write that down 16 point 3 to 5, 20 point 475 So compare this to part A where we had that same sample size but different confidence levels. So the lower bound actually decreased in the upper bound increased. That means that interval is wider. Okay, So what does that mean as the confidence level increases? That margin of error is also going to increase because that that critical values actually greater. So this is larger whenever you increase that confidence level, um making the confidence interval wider. So as the confidence level increases, that confidence interval is going to increase or get wider as well. All right. And then part D. Now it says what conditions for inference are needed? If in equals 15? Well, if N equals 15, then the population from which the sample was taken must be normally distributed. Okay, Since And is less than 30. Okay. So notice in part A B and C and I already talked about this, but A B and C. Um those sample sizes are greater than equal to 30, so it doesn't matter how this population is distributed. It still works. We can still use that procedure But if N is less than 30 in this case 15, then we need that statement in there that, Hey, don't worry about the population, it's normally distributed.

Mhm Yeah. In this video we are going to look at how to form confidence intervals for the population. Mean, if we draw a simple random sample of size n from a population. Now, here we have a simple random sample of sites and was drawn from was drawn and it is found that the sample mean, remember the simple X. With a bar over is sample mean is 18.4 and the sample standard deviation. Remember the symbol for the sample standard deviation is s was equal to 4.5. And the first thing it asks us to do is to find a 95% confidence interval for μ and remember μ is the symbol for the population mean, if N is equal to 35, Well, for this, we want to recall that if we have a sample size that's 30 or bigger, 35 is greater than 30, then I don't need to have my sample drawn from a normal population. I have the central limit theorem that kicks in that says, as the sample size is larger and larger and that cut off is sample size 30 or up. Then the distribution of the sample means is approximately normal, no matter what the distribution of the individual members was. So, since I have my sample size of 35, I don't need it to say that the population was normal in the setup. But the way you find the lower bound of your confidence interval for your population menu is expe r minus t sub alpha over two times S over the square root of end. And the upper bound is expe r plus t sub alpha over two times s over the square to end. Now here for tisa Balfa over to our critical value, we take one minus the 10.95 and the 0.95 is the Confidence level in decimal form and we get 0.05, that's your alpha Alpha, divide by two is divide that number by two. So I get 0.025. And then for the student's t distribution, T.I. to find my degrees of freedom which I get from end -1. My n is 35. So my degrees of freedom is going to be 35. Subtract one or 34. Now, if you go to a student's T distribution table and you look under the column heading of point oh 25. If it's area in the right tail, that gives you those column headings. And um that's what a lot of the charts are for your student's t distribution. And then across from 34 you're going to see the tisa alpha over to value of 2.32. And I could also find this on my calculator. If I go second and the vars ski to get to the distributions and I curse her down to inverse T. The area is area to the left of the car off. So in the chart, their headings are area to the right in the calculator, they've set it up to be area to the left. So for the area for the calculator, I take one and subtract this .025. So one subtract .025 is .975. So I'm gonna put in a .975. My degrees of freedom we calculated was 34 And then go to Paste, enter to paste it and enter to have it done and you'll see that you get that 2.03 to the same number that we got from the table. So then putting these values altogether, my sample mean from the samples 18.4 Then minus the critical value of Tisa Balfe over two is the 2.03, 2 Times my sample standard deviation from the sample is 4.5 And I divide that by the square root of my sample size and my sample size is 35. So remember you put the sample size in that denominator under the square root, not the degrees of freedom. And that's the low end. And then x bar, which is my sample mean 18.4 plus My critical value of 2.03, 2 times my sample standard deviation of 4.5, Divided by the square root of my sample size. And my sample size is 35. And when you go through and calculate that out, it's a 95%. It gives you a lower value of the 16.85 for the start of your confidence interval and then your upper bound is 19.95. So if I wrote this in a sentence, we would say 95% confidence interval for the population mean is between 16.85 and 19.95. Now we can also get this on our calculator instead of having to do the formulas. So it really depends on the format of what you're allowed to use as your tools when doing this process. If you are allowed to do a graphic work with a graphing calculator and you have like a T. I 84 you're going to push the stop button and the step button is right below the delete button cursor right twice to be in tests, but we want to go the critical value is a T and we're making confidence interval. So we want to go down to where it says T interval and when we hit the T interval and push enter data would be if I had the list of numbers but they told us what the mean and standard deviation was. So we have stats, Our sample mean is 18.4. Our sample standard deviation is 4.5 And here my sample size is 35, My confidence level, they wanted me to calculate a 95% confidence interval, so .95 is my confidence level. And when I go to calculate, takes it a second, but then comes up with rounded The 16.95 is on your calculator before the comma and then that's your low. And this is just an open interval between 16.854. Which we rounded to 16.85 comma. So and the upper bound is the 19.95 rounded and that's what we put into the sentence. Okay, now let's look at that process with part B. So for part B and we still want a 95% confidence interval for population mean? But my n. is 50 this time. So when I go and look at the T symbol for over to work, I still have alpha 0.05, it's still 1 -195. My alpha over two is still divide that by two. You get 0.025. But now my degrees of freedom I get from end -1 will end is 50. So my degrees of freedom is 49. And so when we go back over to the calculator and we go second distribution to the inverse T. Do 1 -3.025. That's .975. That's the same as what I had for a previous problem. But now my degrees of freedom is 49. So I put in 49 and paste it and enter to have it calculated And I get a value of 2.0095. So 2.010 if I run that through or we can go 2.0096 and then round. Um so we don't have too much rounding error in there. So then again my ex par minus the teeth of alpha over two times s over the squirt of N. And X. Bar plus tisa alpha over two times as over the score to. And if we need to use our charts and just uh scientific calculator and aren't allowed a graphing calculator. So here we have our 18.4 -2.00096 times S. Which is the 4.5 Then divide that by the square root of my sample size here is 50 And the 18.4 plus The 2.00096 times 4.5 divided by the square root of 50. Yeah And we get here a low end of 17.12 as my lower bound. Um and My 19.68 as my upper bound. So when I compare this we again could say a 95% confidence interval for the population mean Is between 17.12 and 19.68. Now notice these this lower bound is bigger than the start of the interval I found in part a and the upper bound is smaller than the end. So the interval is tinier. So as we increase our sample size, our margin of air decreased. Okay, now part C. Oh so we can actually do this. Our margin of air decreased when the sample size increased. So now for part C let's just go ahead and look at this on the calculator. So remember stat and if you have to do the formulas, just do it from the work that we did before I go over to tests, go down to t interval, push enter to pick it. I've got stats, So I'm going to put in the 18.4 for the sample mean the 4.5 for the sample standard deviation. My end here is 35 And my confidence level is .99 Because in this situation we want a 99% confidence interval. So go down and calculate and we get A 99% confidence interval for the population, mean is between 16.33 and 20.48. Now, just a little wording with this. Um When you get to the student's T distribution table, once it gets to be larger values of n you get where they skip some values that go from 40 sample size to 50 sample size to 60 sample size. And so if you're using a table to get your critical values for those, you might have a little bit of rounding era in these. I think that with the table and working through this might have said Between 16.32 and 20.48, they didn't really changed their rounding years from what they saw calculator. They just had a value from the table that was just approximated for the critical value. And then party just asks If any equal 15, what condition would need to be met for this process to be used? Will N Equal 15 is a small sample size. That's less than 30. So if you have a sample size that small and you're trying to work with a confidence interval for the population mean using your sample statistics. Well you need to have that your random sample was drawn from a normal population. Um in order for you to use this process. So the condition that must be met is that the population would need to be normally distributed. Yeah. Yeah. With a sample size. Yeah That is less than 30. Well I hope this video has helped you put a few of those pieces together for um constructing confidence intervals for the population mean based um Using information from your sample statistics and also um giving you the way to work this process through whether you just are allowed uh scientific calculator. Where you have to use the formulas and look up the table values for your critical values or whether you're using graphing calculator and you're able to find out where on the graphing calculator. These calculations could be found

In this video, we are going to look at how to find confidence intervals using sample statistics to estimate the population mean new. So here, it says from a population that is normally distributed, a random sample of size and was drawn and it was found that in that sample, the sample mean is 50 and the sample standard deviation is eight. So one of the things also from this type of question is it could have said in words the sample mean is 50 and you would have had to recall that the notation for the sample mean is X with a bar over it. So that you had that expire is equal to 50. And then also recalled that the sample standard deviation is symbolized with the S So that you would have said S is equal to eight when they would report in words that the sample standard deviation was eight. So be sure that you're getting a good good connection between your vocabulary words and your symbols. So you're able to utilize the formulas correctly. So, this first one says find in 98% confidence interval for μ the population mean if N is equal to 20. Well, the setup for finding confidence intervals for the population means using sample statistics. When we pull our sample from a population that is normally distributed when our sample size is small or if we have a large sample size is expe r minus tisa alpha over to times S over the square root of N. Yeah, and X. Bar plus t sub alpha over two times S over the square root of N. Mhm. Now, where that tisa Alpha over two is the critical value that you can get either from your table or from the student's t distribution table or from your calculator so far tisa alpha over to work. Remember alpha Is the amount of area in the tales. So if I have a 98% confidence interval, I take one And i subtract the 0.98 and that gives me an alpha value of 0.02. Mhm. Now alpha over two then is just divide that 0.2 by two and I get 0.1. And then with the student's t distribution, remember there's a degree of freedom and your degrees of freedom is n minus one Here. In part a my N is 20, so my degrees of freedom is going to be 19, so 20 subtract one is 19. So looking that up on the table, I'm going to go underneath the .01 column and across from the 19 row and we see that that value is 2.539. Yeah. And now I have everything from my previous information to set up my bounds on my confidence interval. So X bar is the sample mean that's 50, then minus my critical value of tisa, alpha over two is the 2.539. And that's time. My s is my sample standard deviation of eight And divide that by the square root of my sample size and my sample size is 20. And then when you do the upper bound of your confidence interval, you just going to the same thing just with a plus in between so X bar is 50 then plus the critical value of 2.539 and times my s which is a sample standard deviation of eight divided by the square root of 20. Um my 20 is my sample size. And when you go through in key that through your calculator you will have a 98% confidence interval from you is between 45.5 and 54.5. And to write it out in a sentence that that's required, you would say a 98 confidence interval, Mhm, Mhm. The other one for the population mean, Mhm mhm is between 45.5 and 54.5. Now, if you're allowed to use your graphing calculator, then what you want to do here is to push the stat button and the step button is right under the delete cursor, Right to tests, cursor down until you cross from t interval, we're doing intervals instead of tests and its teeth because we again have that critical value of t stable for over two. And so then when we look at pushing enter to make that choice data is if you had the list of numbers, the sample um gathered um stats as if it you've pre done the sample mean and the sample standard deviation and that's how they gave it to us. We're going to stay on stats. Our X bar is 50 Our s is eight and our end here is 20, Then go to our confidence level of .98 and calculate It'll take just a second, but then you'll see rounded off I have 45.5 is my lower value and that's what we got. And then comma so that's the going to the other end of your confidence interval, the 54.5. So that's how you could use your calculator. Now, you can also use a calculator to get the Teesta about over two critical value. If we go to second and the virus key for the distribution and we do in verse t. The format for the inverse T. Let's clear out the stuff that I had here before. The format for the university is area, that's to the left of your critical value. And if my point no one is the alpha over to which is to the right, then one minus that, one minus the point 01 is my .99. And that's the number you want to put in for your area. So we're gonna go .99, our degrees of freedom is 19. And then when we go through and paste that and enter, We see that 2.539 number that we got for a critical value. So you can find that either on the table or from your calculator. So let's go to the next one. Find a 98% confidence interval from you if N is 15. So again recall that are mean expert is 50 and our sample standard deviation is eight And here our end is 15. So when we go to calculate artists of alpha over to We still have that alpha is 0.02. We have alpha over two is 0.01. Our degrees of freedom is N -1 here. My Ennis 15. So my degrees of freedom Is going to be 14. Yeah. And from the table or from your calculator you'll get that your critical value. Tisa Balfe over two is 2.624. And I'll verify that over here. So again, second and far is key for distribution. Go to inverse T. We're gonna put in the area of the .99. And again that was 1- the alpha over to value. And that's how we got the .99 entry. But my degrees of freedom this time is 14. And when we push enter on paste and then enter to have it done We see again the 2.6- four. So finding my confidence interval exposure minus tisa alpha over two times as over the score to end for the low end and then X bar plus T symbol for over two times s over the square to end for the upper? So I have my 50 minus the two point 6-4 times my S. Which is the eight Divided by the square root of and again this time our end is 15. Mhm. And expires 50 Then plus the 2.6-4 times eight divided by the square root of 15. And when you run that through you'll get our 44.6 And 55.4. So what happened with this is it was a 90% confidence interval in both part a and part b. Part a. Had a sample size of 20, part B had a sample size of 15. And when we look at what happened with the margin of air between them, the confidence interval got wider. I started with a lower number and ended with a higher number in part B than I did in part A. So that means my margin affair is bigger. So decreasing the sample size actually increases the margin of air. Yeah. Yeah. Mhm. Yeah. Mhm. Yeah. Mhm. Yeah. Yeah. Okay. And then over on the calculator again, if we push stat, go to tests, go down to t interval because I'm doing a confidence interval from you which enter we have stats Experts still 50 SS eight and this time we change to 15, our confidence level is still .98. And when we calculate this And just compare we see it surrounded to the 44.6 for my low and the 55.4 for number one. Okay, how about finding a 95% confidence interval from you with an equal 20. Yeah. So again we have our expire minus Teesta alpha over two times as over the score to end and X bar plus Teesta alpha over two times as over the score to end. The reason we put the and in between is because when you go to the full sentence you say a 95% confidence interval for the population mean is between the low number and the high number. So that's why it's just an abbreviation with the word and in there now with my teeth of alpha over to in this case I have 95% confidence interval. So one minus the 10.95 is 0.5. That's your alpha Alpha over two then is divide that by two, My degrees of freedom Is N -1 and is 20. So my degrees of freedom is going to be 19. So when you look up or do your university on the calculator, your teeth of alpha over to for that case is 2.093. And so I have my 50. Yeah -2.093 times The s. S. eight Divided by the square root of N is 20. And for the Top number of our interval we have 50 Then plus 2.093 times eight Divided by the square of 20. And so calculating that through, we get 46.3 and 53.7. So again wording it out, a 95% confidence interval for the population mean is between 46.3 and 53.7. And here it's with my sample size of an April 20. Now, what if it's asks me, how does the margin of air change if the confidence level is decreased? Well, if the confidence level is decreased, here's an equal 20, I'm going to compare that with part A And part A. had a low end of 45.5 and a high end of 54.5, Part C has a low end of 46.3 and a high end of 53.7. So my lower number is bigger than the start into in part A. And my higher number is smaller, so it squeezed in a bit, so my margin of air decreased when my confidence level decreased. Okay. Mhm. And then lastly, could we have used this process to construct confidence intervals in part A through C. If the population wasn't known to be normal. Well, remember with the central limit theorem, if I have a population that's normally distributed, then the sampling distribution of the sample means will also be normal. Yeah, If I don't know if my population is normally distributed, but I have sample sizes that are greater than or equal to 30. Then the the central limit theorem says that the distribution of your sample means will be approximately normal as you have that larger sample size. Well, these sample sizes were 20 And 15 smaller than 30. So I had to have the population be normally distributed in order for me to do this process to find confidence intervals. So could we have that answer is no. Well, I hope you found this video to be beneficial and keep an eye out. We will be sharing more as time goes on.


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Find the areas of the surfaces generated by revolving the curves in Exercises $31-34$ about the indicated axes. $$ x=(2 / 3) t^{3 / 2}, \quad y=2 \sqrt{t}, \quad 0 \leq t \leq \sqrt{3} ; \quad y $$
Find the areas of the surfaces generated by revolving the curves in Exercises $31-34$ about the indicated axes. $$ x=(2 / 3) t^{3 / 2}, \quad y=2 \sqrt{t}, \quad 0 \leq t \leq \sqrt{3} ; \quad y $$...
5 answers
Problem 13: Consider the basis B = {v1 = [1,2,31T = v2 = [1,2, 0JT , vg = [2, 0, 0JT} for R3 . Find the coordinate vector (v)B for the following vectors:v = [1,0, 0]T 8 U = [0,1, 0JT U = [0, 0, 1]TUse your answers for parts (a)-(c) to find the change of basis matrix AB'-B that sends a vector & (represented using the standard basis B') to (w)B.
Problem 13: Consider the basis B = {v1 = [1,2,31T = v2 = [1,2, 0JT , vg = [2, 0, 0JT} for R3 . Find the coordinate vector (v)B for the following vectors: v = [1,0, 0]T 8 U = [0,1, 0JT U = [0, 0, 1]T Use your answers for parts (a)-(c) to find the change of basis matrix AB'-B that sends a vector ...
5 answers
(4 points) Solve the equation for xx(a b) = m(xAnswer: x =This is common in algebra, but might seem weird to you: The "answer" and the value of x and the "solution" won't be number; but instead will be formula In some ways instead of asking for the answer; it is asking you tO write down the steps for the answer:If a =4b =7,c =6, and m = 3 then what is the value of x?Answer: x =Now the answer really is an answer; plain number:
(4 points) Solve the equation for x x(a b) = m(x Answer: x = This is common in algebra, but might seem weird to you: The "answer" and the value of x and the "solution" won't be number; but instead will be formula In some ways instead of asking for the answer; it is asking yo...
4 answers
Evaluate the sum f the series(-lyn+l x2n+8 7nnl n =0at x = 1 .
Evaluate the sum f the series (-lyn+l x2n+8 7nnl n =0 at x = 1 ....

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