How do you do both questions?

1. We have all the information we need here to set up our 'least-squares regression equation.'

Given: r = 0.68, x = 260, y = 70, s₁ = 33, s₂ = 6, remember that the general equation for our least-squares line is in the form y = a + bx. The slope of the regression equation is the linear correlation coefficient multiplied by the standard deviation for x:

b = r(s₂ / s₁) = 0.68(6 / 33) = 0.12363...

Remember that the constant of the regression equation is the mean of y decreased by the product of the slope and the mean of x:

a = y - bx = 70 -  0.12363(260) = 37.8562

The equation of the least-squares regression line would then be y = 37.8562 + 0.12363x. We know that the Jane's previous score was 279, so if we plug that into our equation as 'x' we will receive Jane's final score. The set up will be y = 37.8562 + 0.12363(279) = 72.34897. Your solution for the first part is absolutely correct!

2. Actually the values of 'y' could have been much larger. r^2 = 0.4624, thus the proportion of y explained by x is only about 46%. It would negatively affect her final course grade.

41*67/2=1373.5

step-by-step explanation:

i just used a calculator