To find the slope of the line defined by the points \( (60, 0) \) and \( (85, 30) \), we can use the slope formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Plugging in our points:
- \( (x_1, y_1) = (60, 0) \)
- \( (x_2, y_2) = (85, 30) \)
We get:
\[ m = \frac{30 - 0}{85 - 60} = \frac{30}{25} = \frac{6}{5} = 1.2 \]
So the slope of the trend line is \( 1.2 \).
Now, to find the expected value of \( y \) when \( x = 90 \) using the trend line, we first need to find the y-intercept of the line. Since we have a slope \( m = 1.2 \) and one point \( (60, 0) \), we can use the point-slope form of the line:
\[ y - y_1 = m(x - x_1) \]
Using point \( (60, 0) \):
\[ y - 0 = 1.2(x - 60) \]
Simplifying:
\[ y = 1.2x - 1.2 \times 60 \] \[ y = 1.2x - 72 \]
Now we can use this equation to find the expected value of \( y \) when \( x = 90 \):
\[ y = 1.2(90) - 72 \] \[ y = 108 - 72 \] \[ y = 36 \]
Thus, the expected value of \( y \) when \( x = 90 \) is:
\[ \text{The expected value of } y \text{ is } 36. \]