Question

Use the image to answer the question.

A scatterplot graph shows x-axis ranging from 0 to 100 and y-axis ranging from 0 to 50. Both axes are drawn in increments of 5 but labeled in increments of 10. Multiple points are plotted around a line that points upward to the right.
The line passes through left parenthesis 0 comma 30 right parenthesis, left parenthesis 15 comma 40 right parenthesis, and left parenthesis 30 comma 50 right parenthesis. 14 points are plotted close to or on the line. The line has an arrowhead on the top of it.

Write an equation in slope-intercept form of the trend line.

Answer 1

To write the equation of the trend line in slope-intercept form (y = mx + b), we need to determine the slope (m) and the y-intercept (b) based on the points the line passes through.

1. **Finding the slope (m):**
The slope can be calculated using two points on the line. Let's use the points (0, 30) and (15, 40).

The formula for the slope (m) is:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

Substituting the points (0, 30) and (15, 40):
\[
m = \frac{40 - 30}{15 - 0} = \frac{10}{15} = \frac{2}{3}
\]

2. **Finding the y-intercept (b):**
The y-intercept is the y-value when x = 0. From the point (0, 30), we see that the y-intercept (b) is 30.

3. **Writing the equation:**
Now that we have the slope (m = 2/3) and the y-intercept (b = 30), we can write the equation of the line in slope-intercept form:
\[
y = \frac{2}{3}x + 30
\]

Thus, the equation of the trend line is:
\[
y = \frac{2}{3}x + 30
\]