The data show Mark’s commute times, in minutes, based on how many minutes after 7:00 A.M.

Question

The data show Mark’s commute times, in minutes, based on how many minutes after 7:00 A.M. he leaves his home. Use the points (1, 24) and (15, 17) to write an equation for the line of fit in slope-intercept form, where x is the number of minutes after 7:00 A.M. and y is the commute time, in minutes. If the trend continues, what would be the approximate commute time if Mark leaves at 7:18 A.M.?

Minutes after 7:00 A.M.

1

2

4

4

6

8

10

12

14

15

Commute

Time (min)

24

22

24

20

22

19

22

19

15

17

The line of best fit is y = 

If the trend continues, the approximate commute time if Mark leaves at 7:18 A.M. is [blank]   minutes.

Answer 1

y=-.4875x+24.11

use linear regression

if you find it tedious to calculate use an online regression calculator. it will do all the work for you, to get the above then plus in x=18 to calculate commute time at 7:18

this assumes the 10 data points were significant to he problem.

but you try to make 8 of them extraneous, focusing only on two points (1, 24) and (15,17) as if the rest were a meaningless distraction

with just those two points you get a different answer. m = change in y over change in x =(17-24)/(15-1) = -7/14 = -1/2

y =-x/2 +b, plug in either point to solve for b=y intercept

24 =-1/2 +b

b = 24.5

y =--x/2 +24.5 which is fairly close to the regression model

y= -0,4875 x +24.11

then plug in 18 to get

y =-18/2+24.5 = -9 +24.5 = 15.5

y = 15 1/2

Answer 2

Start with two-point form

(y - 24) / (x - 1) = (17 - 24) / 15 - 1)