The study mentioned in Exercise 5 also included measurement of the outside diameters of the 16 bushings. Two of the students measured each of the bushings, with the results given here. Bushing 1 2 3 4 Student A .3690 .3690 .3690 .3700 Student B .3690 .3695 .3695 .3695 Bushing 5 6 7 8Student A .3695 .3700 .3695 .3690 Student B .3695 .3700 .3700 .3690Brushing 9 10 11 12 Student A .3690 .3695 .3690 .3690 Student B .3700 .3690 .3695 .3695 Bushing 13 14 15 16 Student A .3695 .3700 .3690 .3690Student B .3690 .3695 .3690 .3690(a) If you want to compare the two students average measurements, the methods of formulas (6.35), (6.36) and (6.38) are not appropriate. Why? (b) Make a 95% two-sided confidence interval for the mean difference in outside diameter measurements for the two students.

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hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.