Question

using the trend line ( y = 8x + 5 ), for the data shown, complete the residual table.

distance (mi)	2	2.5	3	3.5	4

distance (mi)	2	2.5	3	3.5	4

Explanation:

First, we need to find the equation of the trend line \( y = 8x + 5 \). Then, for each given distance \( x \), we calculate the predicted time \( \hat{y} = 8x + 5 \), and then find the residual which is \( \text{Residual} = \text{Actual Time} - \text{Predicted Time} \).

Step 1: Calculate predicted time for \( x = 2 \)

For \( x = 2 \), the predicted time \( \hat{y} = 8(2) + 5 = 16 + 5 = 21 \). The actual time is 23. So the residual is \( 23 - 21 = 2 \).

Step 2: Calculate predicted time for \( x = 2.5 \)

For \( x = 2.5 \), \( \hat{y} = 8(2.5) + 5 = 20 + 5 = 25 \). Actual time is 29. Residual: \( 29 - 25 = 4 \).

Step 3: Calculate predicted time for \( x = 3 \)

For \( x = 3 \), \( \hat{y} = 8(3) + 5 = 24 + 5 = 29 \). Actual time is 34. Residual: \( 34 - 29 = 5 \).

Step 4: Calculate predicted time for \( x = 3.5 \)

For \( x = 3.5 \), \( \hat{y} = 8(3.5) + 5 = 28 + 5 = 33 \). Actual time is 34. Residual: \( 34 - 33 = 1 \).

Step 5: Calculate predicted time for \( x = 4 \)

For \( x = 4 \), \( \hat{y} = 8(4) + 5 = 32 + 5 = 37 \). Actual time is 40. Residual: \( 40 - 37 = 3 \).

Answer:

For distance \( 2 \) m, residual is \( 2 \); for \( 2.5 \) m, residual is \( 4 \); for \( 3 \) m, residual is \( 5 \); for \( 3.5 \) m, residual is \( 1 \); for \( 4 \) m, residual is \( 3 \). So the residual table (Time column for residuals) is: \( 2, 4, 5, 1, 3 \)