Need asap.

these are the values in irwin’s data set.

(1, 43), (3, 36), (5,22), (6, 25), (8,14)

irwin determined the equation of a linear regression line, and determined that these are the predicted values.

(1, 44), (3, 35), (5,26), (6, 22), (8,13)

use the point tool to graph the residual plot for the data set.

We know that

the residual value is when you subtract the predicted value from the observed value.
[Residual value] = [Observed value]–[Predicted value]
so
[Observed value]=(1, 43), (3, 36), (5,22), (6, 25), (8,14)
[Predicted value]=(1, 44), (3, 35), (5,26), (6, 22), (8,13)

for x=1
[Residual value] = [43]–[44]-----> -1
point (1,-1)

for x=3
[Residual value] = [36]–[35]-----> 1
point (3,1)

for x=5
[Residual value] = [22]–[26]-----> -4
point (5,-4)

for x=6
[Residual value] = [25]–[22]-----> 3
point (6,3)

for x=8
[Residual value] = [14]–[13]-----> 1
point (8,1)

see the answer in the attached figure

7.3

step-by-step explanation:

81+30=111180 minus 111=691/2×19=9.569÷9.5=7.263157895~7.3