Spectrum Math Grade 8 Chapter 6 Lesson 4 Answer Key Creating Equations to Solve Bivariate Problems

Students can use the Spectrum Math Grade 8 Answer Key Chapter 6 Lesson 6.4 Creating Equations to Solve Bivariate Problems as a quick guide to resolve any of their doubts

Spectrum Math Grade 8 Chapter 6 Lesson 6.4 Creating Equations to Solve Bivariate Problems Answers Key

When given a set of bivariate data with a fairly consistent rate of change, a scatter plot with a trend line can be used to create an equation that will approximate the relationship between the two sets of data.

m = $$\frac{522-406}{71.8-65.3}$$ = $$\frac{116}{6.5}$$ = 17.85
y = 17.85x – 759.61

406 = (17.85)(65.3) + b
b = -759.61
Step 1: Use the data set to create a scatter plot with a trend line.
Step 2: Use the 2 points of data closest to the trend line to find the slope of the trend line.
Step 3: Use one point of data with the calculated slope to find the y-intersect of the trend line.
Step 4: Use the calculated slope and y-intersect to state the equation in linear form, y = mx + b.

Use each set of bivariate data to create a scatter plot, trend line, and an equation that approximates the data set.

Question 1.
a.

equation: ______________
Answer:

equation: y = 5.5 x – 4.5

Explanation:
We know that,
When given a set of bivariate data with a fairly consistent rate of change,
a scatter plot with a trend line can be used to create an equation that will approximate the relationship between the two sets of data.
m = $$\frac{12 – 1}{3 – 1}$$ = $$\frac{11}{2}$$ = 5.5
y = 5.5 x + b
12 = 5.5 (3) + b
b = 12 – 16.5 = – 4.5
y = mx + b
y = 5.5 x – 4.5

b.

equation: ______________
Answer:

equation: y = 5.5 x – 4.5

Explanation:
We know that,
When given a set of bivariate data with a fairly consistent rate of change,
a scatter plot with a trend line can be used to create an equation that will approximate the relationship between the two sets of data.

m = $$\frac{1000 – 700}{14 – 12}$$ = $$\frac{300}{2}$$ = 150
y = 150 x + b

500 = 150 (10) + b
b = 500 – 1500 = – 1000
y = mx + b
y = 150 x – 1000

Use each set of bivariate data to create a scatter plot, trend line, and an equation that approximates the data set.

Question 1.
a.

equation: ______________
Answer:

equation: y = 1.75 x – 56.75

Explanation:
We know that,
When given a set of bivariate data with a fairly consistent rate of change,
a scatter plot with a trend line can be used to create an equation that will approximate the relationship between the two sets of data.
m = $$\frac{98 – 84}{30 – 22}$$
= $$\frac{14}{8}$$
= $$\frac{7}{4}$$
= 1.75
y = $$\frac{7}{4}$$ x + b
y = 1.75 x + b

90 = $$\frac{7}{4}$$ (19) + b
90 = 1.75 (19) + b
b = 90 – 33.25 = 56.75
y = mx + b
y = 1.75 x + 56.75

b.

equation: ______________
Answer:

equation:
y = 0.1 x – 10

Explanation:
We know that,
When given a set of bivariate data with a fairly consistent rate of change,
a scatter plot with a trend line can be used to create an equation that will approximate the relationship between the two sets of data.
m = $$\frac{9 – 8}{190 – 180}$$
= $$\frac{1}{10}$$
= 0.1
y = $$\frac{1}{10}$$ x + b
y = 0.1 x + b

7 = $$\frac{1}{10}$$ (170) + b
7 = 0.1 (170) + b
b = 7 – 17
b = -10
y = mx + b
y = 0.1 x – 10

Question 2.
a.

equation: ______________
Answer:

equation:
y = 16.67 x – 783.33

Explanation:
We know that,
When given a set of bivariate data with a fairly consistent rate of change,
a scatter plot with a trend line can be used to create an equation that will approximate the relationship between the two sets of data.
m = $$\frac{175 – 125}{60 – 57}$$
= $$\frac{50}{3}$$
= 16.67
y = $$\frac{50}{3}$$ x + b
y = 16.67x + b

100 = $$\frac{130}{6}$$ (53) + b
100 = 16.67 (53) + b
b = 100 – 883.33
b = -783.33
y = mx + b
y = 16.67 x – 783.33

b.

equation: ______________
Answer:

equation:
y = 2x + 13

Explanation:
We know that,
When given a set of bivariate data with a fairly consistent rate of change,
a scatter plot with a trend line can be used to create an equation that will approximate the relationship between the two sets of data.
m = $$\frac{19 – 15}{4 – 2}$$
= $$\frac{4}{2}$$
= 2
y = 2 x + b
25 = 2(6) + b

b = 13
y = mx + b
y = 2x + 13

Use each set of bivariate data to create a scatter plot, trend line, and an equation that approximates the data set.

Question 1.
a.

equation: ______________
Answer:

equation:
y = 250x – 500

Explanation:
We know that,
When given a set of bivariate data with a fairly consistent rate of change,
a scatter plot with a trend line can be used to create an equation that will approximate the relationship between the two sets of data.
m = $$\frac{1500 – 1000}{7 – 5}$$
= $$\frac{500}{2}$$
= 250
y = 250 x + b
2000 = 250(10) + b

b = 2000 – 2500
b = -500
y = mx + b
y = 250x – 500

b.

equation: ______________
Answer:

equation:
y = x + 11

Explanation:
We know that,
When given a set of bivariate data with a fairly consistent rate of change,
a scatter plot with a trend line can be used to create an equation that will approximate the relationship between the two sets of data.
m = $$\frac{16 – 15}{5 – 4}$$
= $$\frac{1}{1}$$
= 1
y = mx + b
15 = 1 (4) + b
15 = 4 + b

b = 11
y = mx + b
y = (1)x + 11

Question 2.
a.

equation: ______________
Answer:

equation:
y = x + 13

Explanation:
We know that,
When given a set of bivariate data with a fairly consistent rate of change,
a scatter plot with a trend line can be used to create an equation that will approximate the relationship between the two sets of data.
m = $$\frac{26 – 24}{7 – 5}$$
= $$\frac{2}{2}$$
= 1
y = m x + b
15 = 1(2) + b

b = 13
y = mx + b
y = x + 13

b.

equation: ______________
Answer:
equation:
y = 50x – 403

Explanation:
We know that,
When given a set of bivariate data with a fairly consistent rate of change,
a scatter plot with a trend line can be used to create an equation that will approximate the relationship between the two sets of data.
m = $$\frac{64 – 62}{9.7 – 9.3}$$
= $$\frac{2}{0.4}$$
= 50
y = m x + b
62 =50(9.3) + b
62 – 465 = b

b = – 403
y = mx + b
y = 50x – 403

Scroll to Top