SCATTER PLOTS & LINES OF FIT
As the x-values increase,
12
m=
8
4
m=
1000
2000
3000
16 - 2
4000 - 500
14
= 0.004
3500
As the x-values increase, the y-values
(20, 14) (120, 2)
16
12
m=
8
4
0
.
y = mx + b
2 = 0.004(500) + b
2=2+b
0=b
(500, 2) (4000, 16)
16
0
increase
the y-values
m=
20 40 60 80 100 120 140
2 - 14
120 - 20
-12
100
= -0.12
y = 0.004x
decrease
.
y = mx + b
14 = -0.12(20) + b
14 = -2.4 + b
16.4 = b
y = -0.12x + 16.4
As the x-values increase, the y-values increase and decrease
.
(18, 58) (39, 58)
80
60
m=
58 - 58
m=
0
21
40
20
20
40
60
80
39 - 18
=0
y = mx + b
58 = 0(18) + b
58 = 0 + b
58 = b
y = 0x + 58
y = 58
Created by @iteachalgebra
SCATTER PLOTS & LINES OF FIT
Given the table of values, create a scatter plot from the data.
Then calculate the line of best fit and use the equation to make
inferences.
Test
Grade
10
65
40
90
30
80
25
70
50
85
15
50
40
95
25
75
60
95
30
70
100
80
60
grade
Minutes
Studying
40
20
(10, 65) (60, 95)
m=
m=
95 - 65
60 - 10
30
50
= 0.6
Based on the data, what grade
would be likely if a student
studied for 35 minutes?
y = 0.6x + 59
y = 0.6(35) + 59
y = 80
a test score of 80
20
40
60
80
minutes studying
100
y = mx + b
65 = 0.6(10) + b
65 = 6 + b
y = 0.6x + 59
59 = b
Based on the data, how long
would a student have to study to
get 100?
y
100
41
68.3
=
=
=
=
0.6x + 59
0.6x + 59
0.6x
about 68.3 mins
x
Created by @iteachalgebra