The least squares regression equation for the data in the table is
= 13.5x + 42.9. the r-value is 0.977.

why might you use a linearized model instead?

A linearized model would be the best in this case, since R \simeq 1 .

Step-by-step explanation:

Since, R-value or the correlation coefficient is 0.977 or very close to 1, so

linearized model will be best suitable as least square regression model.

Since,

when R = 1

the points will accurately fall over a straight line having equation of the form

ax + by + c = 0 where a, b, c are fixed but otherwise arbitrary constants.

answer: a) (0,0), (1,1), (2,8), and (3,27).

step-by-step explanation:

0^3=0

1^3=1

2^3=8

3^3=27

in effect, the exponential equation for this would me y=3^x. each x-coordinate cubed is giving the y coordinate, so a is your answer : )