Asupervisor records the repair cost for 11 randomly selected refrigerators. a sample mean of $82.43 and standard deviation of $13.96 are subsequently computed. determine the 99% confidence interval for the mean repair cost for the refrigerators. assume the population is approximately normal. step 1 of 2 : find the critical value that should be used in constructing the confidence interval. round your answer to three decimal places.

Critical value z = 2.574

Confidence interval = ( $71.596 to $93.264)

Step-by-step explanation:

Mean x = $82.43

Standard deviation r = $13.96

Number of sample n = 11

Confidence interval of 99%

Critical value = z = t(a/2)

a = 1 - 0.99 = 0.01

a/2 = 0.01/2 = 0.005

z= t(0.005) = 2.57

z = 2.574

Confidence interval = x+/-(z×r/√n)

= 82.43 +/- (2.574 × 13.96/√11)

= 82.43 +/-10.834

Confidence interval = ( 71.596 to 93.264)

answer: 1,380

step-by-step explanation:

1: you turn %15 into a decimal by moving the decimal point two places to the left.

2: and then you multiply .15 by 1,200 , which will get you 180.00

3: from there you will add 180 to 1,200

4: which will get you 1,380

hope it *im kind of slow so idk if my answer is right*

hi there!

step-by-step explanation:

a roof pitch is basically for every inch rise there is 12 inches run.

the height of the roof is also known as the rafter.

formula: rafter^2 = rise^2 + run^2

rafter^2 = 5^2 + 12^2

rafter^2 = 25 + 144

rafter ^2 = 169"^2

the height of the roof from the lowest point to the highest point is 169"^2.

hope this : )