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Score: 0 of 1 ptof 15 (13 complete)HW Score: 63.33% _5.5.688 Skill BuilderQuestioThe population of Adamsville grew from 10,000 to 16,000 in 5 years. Assuming uninhi...

Question

Score: 0 of 1 ptof 15 (13 complete)HW Score: 63.33% _5.5.688 Skill BuilderQuestioThe population of Adamsville grew from 10,000 to 16,000 in 5 years. Assuming uninhibited exponential growth , what is the expected population in an additional years?The expected population is (Do not round until the final answer: Then round to the nearest whole number as needed:)

Score: 0 of 1 pt of 15 (13 complete) HW Score: 63.33% _ 5.5.6 88 Skill Builder Questio The population of Adamsville grew from 10,000 to 16,000 in 5 years. Assuming uninhibited exponential growth , what is the expected population in an additional years? The expected population is (Do not round until the final answer: Then round to the nearest whole number as needed:)



Answers

Use the normal distribution in Exercise 13. (a) What percent of the SAT writing scores are less than $600 ?$ (b) Out of 1000 randomly selected SAT writing scores, about how many would you expect to be greater than $500 ?$

In question Number nine were given five values in the population. It does. Population is five. The values are to 13, 4, 10 and six and were asked to calculate a few things first, the sum of the square. Um, some of the squares of the differences mean so basically, we're gonna take the summation of all of the values minus the mean and square that do that for all of these values. So in order to do this, we need to know. I mean, eso to calculate the mean we're gonna add those phase together to buy by five. We get 35 divided by five and the mean is seven. So we do that. We'll play these in as to minus seven squared plus 13 minus seven square. Do this for all the values. Four minus seven squared plus 10 minus seven squared plus six months. Seven squared. This produces 25 plus 36 plus nine plus nine plus one, and we get 80. We also asked to calculate the variance and then the standard deviation. So the variance in this case is the value that we just found 80 divided by in because this is the population. We don't have to subtract by one. Uh, that's 80 divided about five and we get 16 standard deviation. In this case, standard deviation is the same thing as the square root of the variance. So we know the variances. 16 Take the square root of that. We get a standard deviation of four.

So formal. 100 published beauty. They called you too bizarre terms with the the right here will be to bunk to five. So we, uh, won't plus the really bizarre Bonzo Ju ju five seems to be here. So from here, we want to know when it would be the bubbles and reached a 1,000,000 on we have is gonna be de Maria. 123 It could be easily could your trees 75 You wanted to dream? Well, now this in terms with, uh, one Bonzo When Tom Izzo Jew do five times a day. So we have 1000 one million. Do you value guy trees? 75123 So equal Ju ju pawn 667 We go to one point you 2 to 5 day. So we turned an animal science. So we got on and off 2.667 in code to the Times. And then I went from zero juju. Fact that isn't the agriculture. And after pawn 667 grinding the Elena one bones of juju five. And then what? You go, Jew. But before Yes. Every speaking

In Problem 58. We have an equation for the population off the United States after the year off two thousands, where T is number of years after 20 thousands and these population is in millions. We want to find the average value of the baby relation from the year 2000 and one to the year 2000 and five. This means in our equation, T equals one, 23 equals five. And to get the average of regulation through this period, the average equals the definite integral for the equation that represents the population. This equation. 282.3 he is a lot of open Oh won t d t through the interval from 1 to 5, divided by the length off. The interval, which is 45 minus one, equals 1/4 multiplied boy 200 82.3 out off the integral supplied by the integration off E to the bar off. Any function is it was about dysfunction divided by the differentiation of this function. Divided by opening all one. Then we substitute from 1 to 5 equals 200. The 82.3 divided by four by going. Four point offer multiplied boy, we substitute boy T equals five e to the bottle 0.5 minus mhm to the power off. All 0.1 We can evaluate this expression using calculator multiplied by E. There's a lot of a 0.45 minus e Sabbar or going to a one equals 200 and 20 0.9 167 Well, 85 We can multiply it by one million to get the number of population. The average number, the average value off the regulation through 21 to 20 or five it equals 29 220 millions. 900 16 thousands and 700 it 0.5 there's and this is the final answer off our problem.

So we're assuming we have a normal distribution and we are given that they mean it's supposed to be 498 with a standard deviation of 100. So this would be a score of about 5 98 right here, and this would be a score of about 3 98 there. And we want to find what's the likelihood of having a score for this particular type of? I think it was an exam between 405 100. We're supposed to assume a normal population again, so we would convert both these into Z values. We take the score minus the mean yeah, divided by the standard deviation and the score minus the main, divided by the standard deviation. And that will give us our Z values and that differences negative. 98 negative night. Excuse me. Negative 98 divided by 100. So that's going to be a Z score of negative 1000.98 and this is the score is only two divided by 100. So 1000.2 and picture wise that 500 would be about here. So it's again very close to a zero Z value and the 400 is going to be about here. So we're finding that area, really? And I'm gonna use my calculator. I could use a table as well, but I'm going to use my normal CTF. And when I hit on my calculator second and distribution normal CDF, I'm going to use this as my low number negative 0.98 My upper number is going to be 0.2 and I'm leaving the mean for the standard normal distribution at zero and standard deviation of one. And when I do that, I find out that that probability is 0.34444 So I don't recall if it said percent. If it said percent, we'd get at about 34% chance of that happening or the probability is about 340.34 Then we had to look at Part B, and it said if you had a total of 300 people, how many of those would you find would have a score that's over 700? So let's find the probability of getting a score over 701st, and we can see that it's not real likely. Here's at 600 roughly and 700 would be about here. This would be at 6 98 so I can see that that probability I'll kind of color in blue is not going to be very big, which means you're not going to have a huge number here. So let's convert this into a Z score again and our Z score. What we got minus what we're assuming and think that was 4 98. It wasn't it. Yep, 4 98 and then divided by 100 and so 700 minus the 4 98. That difference is two oh two. And then we're dividing that by 100. So it's going to be two point Oh two. So that's our Z value, or about two standard deviations higher than the mean. And again I'm going to use my normal CDF here. Second and distribution go to normal CDF, and I'm going to use this as the low number 2.2 You could also look this up in a Z table and then for the upper value. I'm just going to put in 1000 and leave the mean and standard deviation at zero and one. And when I type that in. I find out that that percent is 0.217 approximately. And so now I'm going to multiply that by 300 to find out what 2.17% is of 300 times 300 and I find out that that comes out to be 6.5, so there would be probably six or seven, six or seven people I would anticipate to have a score that is greater than, uh, 700.


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