Answer is: A. 0.50 [x + (x + 0.25x)], B. 0.50 (x + 1.25x), E.0.50 (2.25x)
x y x^2 xy
2 100 4 200
4 83 16 332
6 50 36 300
8 35 64 280
10 23 100 230
∑x = 30; ∑y = 291; ∑x^2 = 220; ∑xy = 1,342; n = 5
![b= \frac{n\Sigma(xy)-(\Sigma x)(\Sigma y)}{n(\Sigma x^2) - (\Sigma x)^2} = \frac{5(1,342)-(30)(291)}{5(220)-(30)^2} \\ = \frac{6,710-8,730}{1,100-900} = \frac{-2,020}{200} =-10.1 \\ a = \frac{\Sigma y-b\Sigma x}{n} = \frac{291-(-10.1)(30)}{5} = \frac{291+303}{5}= \frac{594}{5} =118.8](https://tex.z-dn.net/?f=b%3D%20%5Cfrac%7Bn%5CSigma%28xy%29-%28%5CSigma%20x%29%28%5CSigma%20y%29%7D%7Bn%28%5CSigma%20x%5E2%29%20-%20%28%5CSigma%20x%29%5E2%7D%20%3D%20%5Cfrac%7B5%281%2C342%29-%2830%29%28291%29%7D%7B5%28220%29-%2830%29%5E2%7D%20%20%5C%5C%20%3D%20%5Cfrac%7B6%2C710-8%2C730%7D%7B1%2C100-900%7D%20%3D%20%5Cfrac%7B-2%2C020%7D%7B200%7D%20%3D-10.1%20%5C%5C%0Aa%20%3D%20%20%5Cfrac%7B%5CSigma%20y-b%5CSigma%20x%7D%7Bn%7D%20%3D%20%5Cfrac%7B291-%28-10.1%29%2830%29%7D%7B5%7D%20%3D%20%5Cfrac%7B291%2B303%7D%7B5%7D%3D%20%5Cfrac%7B594%7D%7B5%7D%20%3D118.8)
The <span>least squares regression line equation is given by y = bx + a = -10.1x + 118.9.</span>
The answer is less then six