Dependent Variable is SAT-MATH 22-2-20 Degrees of Freedom
Variable
Constant
Per Student
Coefficient
446.990
0.00733
S. E.
37.66
0.0031
t-ratio
11.9
2.37
p-value
< 0.00001
0.0279
The Pennsylvania Department of Education is interested in the relationship between per-student spending and
average SAT Math scores. A scatter plot of the data suggests average SAT Math scores tend to increase as
per-student spending in a school district increases. The Department of Education is interested in whether this
increase is significant. Assuming conditions for inference are satisfied, the p-value for the linear regression
t-test is approximately
A. 0.0001
B. 0.0031
C 0.0140
D. 0.0279
E. 0.2799

okay..i will pick 8!

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.