You have drawn a simple random sample of 36 college students, asking each student how much rent they pay per month. You obtain a sample mean of 573 and a sample standard deviation of 22. Calculate the upper confidence limit of a 90% confidence interval for the population mean of rent paid. If necessary, round your answer to four decimal places. HINT: For a confidence interval of [a, b] we refer to "a" as the "lower confidence limit" and we refer to "b" as the "upper confidence limit". g

And the 90% confidence interval would be between [566.80 and 579.20]

Step-by-step explanation:

Information given

represent the sample mean

population mean (variable of interest)

s=22 represent the sample standard deviation

n=36 represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula:

  (1)

The degrees of freedom are given by:

The Confidence level is 0.90 or 90%, the significance would be and , and the critical value for this case would be

And replacing we got:

And the 90% confidence interval would be between [566.8033 and 579.1967]

8y^50

step-by-step explanation:

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240