5.12: Problem Solving with Linear Models
What if you've plotted some data points, with the \begin{align*}x-\end{align*}coordinates of the points representing the number of years a teacher has been teaching at a school and the \begin{align*}y-\end{align*}coordinates representing his salary? Suppose that you've found the line of best fit to be \begin{align*}y=1500x+28,000\end{align*}. If the teacher has been teaching at the school for 8 years, could you use the line of best fit to predict how much his salary will be after he's taught for 12 years? How would you do it? In this Concept, you'll learn how to answer real-world questions like these by using a linear model.
Guidance
The previous lessons have focused on writing equations and determining lines of best fit. When we fit a line to data using interpolation, extrapolation, or linear regression, it is called linear modeling.
A model is an equation that best describes the data graphed in the scatter plot.
Example A
Dana heard something very interesting at school. Her teacher told her that if you divide the circumference of a circle by its diameter you always get the same number. She tested this statement by measuring the circumference and diameter of several circular objects. The following table shows her results.
From this data, estimate the circumference of a circle whose diameter is 12 inches.
Solution:
Begin by creating a scatter plot and drawing the line of best fit.
Object | Diameter (inches) | Circumference (inches) |
---|---|---|
Table | 53 | 170 |
Soda can | 2.25 | 7.1 |
Cocoa tin | 4.2 | 12.6 |
Plate | 8 | 25.5 |
Straw | 0.25 | 1.2 |
Propane tank | 13.3 | 39.6 |
Hula hoop | 34.25 | 115 |
Find the equation of the line of best fit using points (0.25, 1.2) and (8, 25.5).
\begin{align*}&\text{Slope} && m = \frac{25.5 - 12}{8 - 0.25} = \frac{24.3}{7.75} = 3.14\\ &&&y = 3.14x + b\\ &&&1.2 = 3.14(0.25) + b \Rightarrow b = 0.42\\ &\text{Equation} && y = 3.14x + 0.42\end{align*}
\begin{align*}&\text{Diameter} = 12 \ \text{inches} \Rightarrow y = 3.14 (12) + 0.42 = \underline{38.1 \ \text{inches}}\end{align*}
In this problem, the \begin{align*}\text{slope} = 3.14\end{align*}. This number should be very familiar to you—it is the number pi rounded to the hundredths place. Theoretically, the circumference of a circle divided by its diameter is always the same and it equals 3.14 or \begin{align*}\pi\end{align*}.
Example B
Using Dana's data from Example A, estimate the circumference of a circle whose diameter is 25 inches.
Solution:
The equation \begin{align*}y=3.14x+0.42\end{align*} of the relationship between diameter and circumference from Example A applies here.
\begin{align*}&\text{Diameter} = 25 \ \text{inches} \Rightarrow y = 3.14 (25) + 0.42 = \underline{78.92 \ \text{inches}}\end{align*}
A circle with a diameter of 25 inches will have a circumference that is approximately 78.92 inches.
Example C
Using Dana's data from Example A, estimate the circumference of a circle whose diameter is 60 inches.
Solution:
The equation \begin{align*}y=3.14x+0.42\end{align*} of the relationship between diameter and circumference from Example A applies here.
\begin{align*}&\text{Diameter} = 60 \ \text{inches} \Rightarrow y = 3.14 (60) + 0.42 = \underline{188.82 \ \text{inches}}\end{align*}
A circle with a diameter of 60 inches will have a circumference that is approximately 188.82 inches.
Guided Practice
A cylinder is filled with water to a height of 73 centimeters. The water is drained through a hole in the bottom of the cylinder and measurements are taken at two-second intervals. The table below shows the height of the water level in the cylinder at different times.
Time (seconds) | Water level (cm) |
---|---|
0.0 | 73 |
2.0 | 63.9 |
4.0 | 55.5 |
6.0 | 47.2 |
8.0 | 40.0 |
10.0 | 33.4 |
12.0 | 27.4 |
14.0 | 21.9 |
16.0 | 17.1 |
18.0 | 12.9 |
20.0 | 9.4 |
22.0 | 6.3 |
24.0 | 3.9 |
26.0 | 2.0 |
28.0 | 0.7 |
30.0 | 0.1 |
Find the water level at 15 seconds.
Solution:
Begin by graphing the scatter plot. As you can see below, a straight line does not fit the majority of this data. Therefore, there is no line of best fit. Instead, use interpolation.
To find the value at 15 seconds, connect points (14, 21.9) and (16, 17.1) and find the equation of the straight line.
\begin{align*}m=\frac{17.1-21.9}{16-14}=\frac{-4.8}{2}=-2.4\end{align*}
\begin{align*}y=-2.4x+b \Rightarrow 21.9 = -2.4(14)+b \Rightarrow b =55.5\end{align*}
Equation \begin{align*}y=-2.4x+55.5\end{align*}
Substitute \begin{align*}x=15\end{align*} and obtain \begin{align*}y=-2.4(15)+55.5=19.5 \ cm\end{align*}.
Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Using a Linear Model (12:14)
- What is a mathematical model?
- What is linear modeling? What are the options to determine a linear model?
- Using the Water Level data, use interpolation to determine the height of the water at 17 seconds.
Use the Life Expectancy table below to answer the questions.
- Make a scatter plot of the data.
- Use a line of best fit to estimate the life expectancy of a person born in 1955.
- Use linear interpolation to estimate the life expectancy of a person born in 1955.
- Use a line of best fit to estimate the life expectancy of a person born in 1976.
- Use linear interpolation to estimate the life expectancy of a person born in 1976.
- Use a line of best fit to estimate the life expectancy of a person born in 2012.
- Use linear extrapolation to estimate the life expectancy of a person born in 2012.
- Which method gives better estimates for this data set? Why?
Birth Year | Life expectancy in years |
---|---|
1930 | 59.7 |
1940 | 62.9 |
1950 | 68.2 |
1960 | 69.7 |
1970 | 70.8 |
1980 | 73.7 |
1990 | 75.4 |
2000 | 77 |
The table below lists the high temperature for the first day of each month in 2006 in San Diego, California (Weather Underground). Use this table to answer the questions.
- Draw a scatter plot of the data.
- Use a line of best fit to estimate the temperature in the middle of the \begin{align*}4^{th}\end{align*} month (month 4.5).
- Use linear interpolation to estimate the temperature in the middle of the \begin{align*}4^{th}\end{align*} month (month 4.5).
- Use a line of best fit to estimate the temperature for month 13 (January 2007).
- Use linear extrapolation to estimate the temperature for month 13 (January 2007).
- Which method gives better estimates for this data set? Why?
Month number | Temperature \begin{align*}(F)\end{align*} |
---|---|
1 | 63 |
2 | 66 |
3 | 61 |
4 | 64 |
5 | 71 |
6 | 78 |
7 | 88 |
8 | 78 |
9 | 81 |
10 | 75 |
11 | 68 |
12 | 69 |
Mixed Review
- Simplify \begin{align*}6t(-2+7t)-t(4+3t)\end{align*}.
- Solve for \begin{align*}y: \frac{119}{8}=\frac{-10}{3}\left (y+\frac{7}{5}\right)-\frac{5}{2}\end{align*}.
- Determine the final cost. Original cost of jacket: $45.00; 15% markup; and 8% sales tax.
- Write as a fraction: 0.096.
- Is this function an example of direct variation? \begin{align*}g(t)=-35t+15\end{align*}. Explain your answer.
- The following data shows the number of youth-aged homicides at school during various years (source: http://nces.ed.gov/programs/crimeindicators/crimeindicators2009/tables/table_01_1.asp). \begin{align*}&\text{Year} && 1993 && 1995 && 1997 && 1999 && 2001 && 2003 && 2005 && 2007\\
&\# && 34 && 28 && 28 && 33 && 14 && 18 && 22 && 30\end{align*}
- Graph this data and connect the data points.
- What conclusions can you make regarding this data?
- There seems to be a large drop in school homicides between 1999 and 2001. What could have happened to cause such a large gap?
- Make a prediction about 2009 using this data.
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Here you'll use some of the linear modeling tools learned in previous Concepts to solve real-world problems.