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Question

An illustration of a scatterplot graph shows x-axis ranging from 0 to 10 in increments of 1 and y-axis ranging from 0 to 50 in increments of 10. Multiple points are plotted around a line that points downward to the right with an arrowhead on the bottom.
The line passes approximately through left parenthesis 0 comma 45 right parenthesis, left parenthesis 2 comma 25 right parenthesis, left parenthesis 3 comma 15 right parenthesis, and left parenthesis 4.5 comma 0 right parenthesis. 13 points are plotted close to the line.

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

(1 point)
y=

Answer 1

To determine the equation of the trend line in slope-intercept form (y = mx + b), we need to find the slope (m) and the y-intercept (b) from the points you provided.

From your description, the line passes approximately through the following points:

• (0, 45)
• (2, 25)
• (3, 15)
• (4.5, 0)

Let's start by calculating the slope (m) using the points (0, 45) and (4.5, 0):

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

Using the points (0, 45) and (4.5, 0):

• \( (x_1, y_1) = (0, 45) \)
• \( (x_2, y_2) = (4.5, 0) \)

Now, plug in the values: \[ m = \frac{0 - 45}{4.5 - 0} = \frac{-45}{4.5} = -10 \]

Now that we have the slope (m = -10), we can use the y-intercept (b) from the point (0, 45) where the line crosses the y-axis. Thus, \( b = 45 \).

Now we can write the equation in slope-intercept form: \[ y = -10x + 45 \]

Thus, the equation of the trend line is: \[ y = -10x + 45 \]