# Linear Models Worksheet

Linear Models Worksheet
• Page 1
1.
Which is a method of estimating the co-ordinates of a point that lie between two given data points?
 a. Linear extrapolation b. Linear interpolation

#### Solution:

Linear interpolation is a method of estimating the co-ordinates of a point that lie between two given data points.

2.
Which is a method of estimating the co-ordinates of a point that lie to the right or left of all of the given data points?
 a. Linear extrapolation b. Linear interpolation

#### Solution:

Linear extrapolation is a method of estimating the co-ordinates of a point that lie to the right or left of all of the given data points.

3.
The number of women competing in a beauty contest from 1995 to 1999 is shown in the table. Is it reasonable to represent the data by a linear model? Which method can be used to estimate the number of women competing in the year 2000?

 a. yes, linear extrapolation b. yes, linear interpolation c. no, linear extrapolation d. no, linear interpolation

#### Solution:

Draw a scatter plot of the data to decide whether the data can be represented by a linear model.

From the scatter plot, you can see that the data fall almost exactly on a line.

So, a linear model is appropriate.

Linear extrapolation is a method of estimating the co-ordinates of a point that lie to the right or left of all of the given data points.

So, linear extrapolation is used to estimate the number of women competing in the year 2000.

4.
The cost $x$ on advertising in millions of dollars of a company can be modeled by the equation $x$ = 11256 + 7538$t$, where $t$ represents the number of years since 1990. Use the model to predict the total cost of advertising in 2003.
 a. $177,092,000,000 b.$162,016,000,000 c. $109,250,000,000 d.$33,870,000,000

#### Solution:

x = 11256 + 7538t
[Given linear model.]

To find the total cost of advertising in 2003, substitute 13 for t.
[t = number of years since 1990 to 2003, 2003 - 1990 = 13.]

x = 11256 + 7538(13) = 109,250
[Replace t with 13.]

So, the total cost of advertising in 2003 is \$109,250,000,000.

5.
The scatter plot shows the change in gravity (rounded to the nearest tenth) as the altitude changes. State whether it is reasonable for the graph to be represented by a linear model. Which method can be used to find the change in gravity at an altitude of 300 km?

 a. no, linear interpolation b. no, linear extrapolation c. yes, linear extrapolation d. yes, linear interpolation

#### Solution:

From the scatter plot, you can see that the data fall almost exactly on a line.

So, a linear model is appropriate.

Linear interpolation is a method of estimating the co-ordinates of a point that lie in between the given data points.

So, linear interpolation is used to find the change in gravity at an altitude of 300 km.

6.
The number of viewers $x$ (in millions) of a sports channel can be modeled by the equation, $x$ = 7$t$ +12, where $t$ represents the number of years counted from year 1980, ($t$ = 0 for 1980, 1 for 1981, etc.). Use the linear model to predict the number of viewers in 2007.
 a. 189 millions b. 46 millions c. 201 millions d. 194 millions

#### Solution:

x = 7t + 12
[Given linear model.]

To find the number of viewers in 2007, substitute t as 27 since t is the number of years since 1980.
[2007 - 1980 = 27.]

x = 7(27) + 12
[Substitute t = 27.]

x = 201
[Solve for x.]

So, the number of viewers in 2007 would be 201 millions.

7.
A data set can be represented by a linear model if the data points on the scatter plot _____.
 a. lie in negative quadrants only b. lie close to a straight line c. form a smooth curve d. lie in positive quadrants only

#### Solution:

A data set can be represented by a linear model if the data points on the scatter plot lie close to a straight line.

8.
The US postage rates have been increasing by 1 cent per year since 1970. The postage rates in 1980 were 40 cents per ounce. Predict the postage rates in the year 2010.
 a. 72 cents b. 75 cents c. 70 cents d. 80 cents

#### Solution:

Let r be the variable representing the postage rates and t be the variable representing the number of years.

Slope m = 1.
[The postal rates per ounce have been increasing by 1 cent per year.]

t = 0 represents year 1970. So, t = 10 represents year 1980.

(t1, r1) = (10, 40) is a point on the line whose slope is 1.

The equation of the line passing through the point (t1, r1) with slope m in point-slope form is r - r1 = m(t - t1).

r - 40 = 1(t - 10)
[Substitute t1 = 10, r1 = 40 and m = 1 in the equation.]

r - 40 = t - 10

r = t + 30

The linear model for the postage rates per ounce since 1970 is r = t + 30.

t = 40 represents year 2010.

r = (40) + 30
[Substitute t = 40 in the linear model.]

r = 70
[Simplify.]

The postage rates in 2010 could be 70 cents.

9.
Use the scatter plot and the trend line to find the approximate value of the altitude that corresponds to the gravity of 9.25 m/s2.

 a. 250 km b. 240 km c. 130 km d. 186 km

#### Solution:

Draw a horizontal line starting from 9.25 m/s2 in the scatter plot. This line intersects the trend line at a point.

From the intersection, draw a vertical line that intersects the x-axis. This intersection point on the x-axis gives the required altitude as the x-axis represents the altitude.

So, when the gravity is 9.25 m/s2, the altitude is approximately equal to 186 km.

10.
The scatter plot shows the population of Alabama and California from 1998 to 2002. Describe the trend in the scatter plot.

 a. no trend b. positive trend for both c. positive trend for California and negative trend for Alabama d. negative trend for both

#### Solution:

The yellow and the blue points in the plot represent the population of Alabama and California respectively, in the corresponding year.

As the year increases, the population in the two states also tend to increase. So, the scatter plot shows a positive trend.