A statistics student wants to survey a high school of 910 students concerning support for increasing the number of student parking spots. The student randomly selects 90 students, and finds that 63 support increasing the number of student parking spots. Assuming the conditions for inference have been met, what is the 95% confidence interval for the true proportion of students who would support the increase in the number of parking spots? 0.30 plus-or-minus 1.65 StartRoot StartFraction 0.30 (1 minus 0.30) Over 90 EndFraction EndRoot 0.70 plus-or-minus 1.65 StartRoot StartFraction 0.70 (1 minus 0.70) Over 910 EndFraction EndRoot 0.30 plus-or-minus 1.96 StartRoot StartFraction 0.30 (1 minus 0.30) Over 910 EndFraction EndRoot 0.70 plus-or-minus 1.96 StartRoot StartFraction 0.70 (1 minus 0.70) Over 90 EndFraction EndRoot

i think the answer is fried chicken ; >

step-by-step explanation:

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.