Scatter Plots & Data Analysis Unit Test
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Question
Use the image to answer the question.
An illustration of a scatterplot graph shows x-axis ranging from 0 to 15 in increments of 1 and y-axis ranging from 0 to 75 in increments of 5. Multiple points are plotted and a line points upward to the right with an arrowhead on the top.
The line passes approximately through left parenthesis 0 comma 25 right parenthesis, left parenthesis 1 comma 30 right parenthesis, left parenthesis 2 comma 35 right parenthesis, left parenthesis 7 comma 60 right parenthesis, and left parenthesis 10 comma 75 right parenthesis. 5 points are plotted to the line.
Write an equation in slope-intercept form of the trend line.
To find the equation of the trend line in slope-intercept form, we need to determine the slope and y-intercept.
We can calculate the slope using the formula:
slope = (change in y)/(change in x)
From the given points, the change in x is (10 - 0) = 10 and the change in y is (75 - 25) = 50.
So, the slope is:
slope = 50/10 = 5
Next, we can use the point-slope form of a linear equation to find the y-intercept.
y - y1 = m(x - x1)
We can choose any point on the line to use in the equation. Let's use the point (0, 25).
Plugging in the values, we get:
y - 25 = 5(x - 0)
y - 25 = 5x
y = 5x + 25
Therefore, the equation of the trend line in slope-intercept form is:
y = 5x + 25