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