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: no

step-by-step explanation:

you are only given the cost not the amount of pounds.

see attached

step-by-step explanation:

given the function as;

write the equation replacing f(x) with y

form   and fill a table containing values of y with corresponding values of x

you can take a range for   values of x as -3≤x≤3

solving for x=-3

for x=-2

repeat the same procedure to get the values of y for the remaining ,x=-1 to x=3

plot the values of y against values   of x and obtain the following graph