Bivariate Data Analysis & Probability 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.

(1 point)
y=
x +

1 answer

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

From the points you provided:

  • Point 1: (0, 25)
  • Point 2: (1, 30)
  • Point 3: (2, 35)
  • Point 4: (7, 60)
  • Point 5: (10, 75)

Step 1: Calculate the slope (m)
The slope \(m\) can be calculated using any two points. Let's use points (1, 30) and (2, 35):

\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{35 - 30}{2 - 1} = \frac{5}{1} = 5 \]

Step 2: Use one point to find the y-intercept (b)
We can use point (0, 25) to find \(b\). Using the slope-intercept form:

\[ y = mx + b \ 25 = 5(0) + b \ b = 25 \]

Final Equation
Now that we have \(m\) and \(b\), we can write the equation of the trend line:

\[ y = 5x + 25 \]

So, in slope-intercept form, the equation is:

\[ y = 5x + 25 \]