The J. R. Simplot Company produces frozen French fries that are then sold to customers such as McDonald's. The "prime" line of fries has an average length of 6.00 inches with a standard deviation of 0.50 inch. To make sure that Simplot continues to meet the quality standard for "prime" fries, they plan to select a random sample of n = 100 fries each day. The quality analysts will compute the mean length for the sample. They want to establish limits on either side of the 6.00 inch mean so that the chance of the sample mean falling within the limits is 0.99. What should these limits be?

the answer is (4,3)

since the rule for r0, 90º is (-y,x) and it was already applied to form the image, all you have to do is reverse the answer using this formula.

55.44 inches

step-by-step explanation:

part a. assuming a perfectly linear relationship, we can find the slope from the first two data points.

slope = m = (change in rainfall)/(change in years)

= (61.80 -62.33)/(2005 -2004) = -0.53/1 = -0.53

then the point-slope form of the equation of the line can be written as

y = m(x -h) +k . . . . . m = -0.53, (h, k) = (0, 62.33)

y = -0.53x +62.33 . . . x = years after 2004

part b. in 2017, x = 2017 -2004 = 13. then the predicted rainfall is

y = -0.53·13 +62.33 = 55.44 . . inches

the predicted rainfall in 2017 is 55.44 in.