Big Ideas Math Algebra 2, 2014
BI
Big Ideas Math Algebra 2, 2014 View details
3. Modeling with Linear Functions
Continue to next subchapter

Exercise 13 Page 27

Graph the data as a scatter plot. Can you draw a trend line?

Is it a linear relationship? Yes.
Example Solution: y=4.27x+1.73
After 15 minutes of walking, you have burned approximately 65.8 calories.

Practice makes perfect

To determine whether the data shows a linear relationship, let's begin by visualizing it.

Observing the Graph

By treating the table as a set of points, we can graph the given data as a scatter plot. Do you see any trends?

It looks like there is some kind of correlation. Let's draw a line of fit, or trend line. To do so, we will draw a line that appears to fit the data closely.

Equation for the Line of Fit

To write an equation for the line of fit we determined above, we first need to use two points on the line to find its slope. Let's use (1,6) and (16,70) in the Slope Formula.
m = y_2-y_1/x_2-x_1
m=70- 6/16- 1
Simplify
m=64/15
m=4.266666...
m≈4.27
Now that we have the slope m≈ 4.27, let's use the point ( 1, 6) in the point-slope form to write and simplify an equation for our line of fit.
y-y_1=m(x-x_1)
y- 6= 4.27(x- 1)
Simplify
y-6=4.27x-4.27
y=4.27x+1.73

Calculating the Projected Value

To calculate the expected output at a given input, we can substitute the value into our equation of fit. In this case, we will substitute 15 for x in our equation and solve for y.
y=4.27x+1.73
y=4.27( 15)+1.73
y=64.05+1.73
y=65.78
y≈65.8
After 15 minutes of walking, you have burned approximately 65.8 calories.