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John estimates the value of his car over time. The equation for the line of best fit is approximated as [tex]\( y = -2x + 17 \)[/tex], where [tex]\( y \)[/tex] represents the value in thousands of dollars.

Complete the residual table with the correct values:

[tex]\[
\begin{tabular}{|c|c|c|c|}
\hline
Age (years) & Given Value & Predicted Value & Residual \\
\hline
1 & 15 & \( a \) & 0.2 \\
\hline
2 & 12 & 11.9 & \( b \) \\
\hline
3 & 9 & \( c \) & 0 \\
\hline
4 & 5 & 6.1 & \( d \) \\
\hline
5 & 4 & 3.2 & 0.8 \\
\hline
\end{tabular}
\][/tex]

Calculate the missing values:
[tex]\[
\begin{array}{l}
a = \\
b = \\
c = \\
d =
\end{array}
\][/tex]


Sagot :

To complete the residual table, we'll follow these steps to find the missing values:

1. Determine the predicted value when [tex]\( \text{Age} = 1 \)[/tex]:
The predicted value formula is given by [tex]\( y = -20 \cdot \text{Age} + 177 \)[/tex].
For [tex]\( \text{Age} = 1 \)[/tex]:
[tex]\[ y = -20 \cdot 1 + 177 = 157 \][/tex]
So, [tex]\( a = 157 \)[/tex].

2. Calculate the residual when [tex]\( \text{Age} = 2 \)[/tex]:
Given value for [tex]\( \text{Age} = 2 \)[/tex] is 12.
Predicted value when [tex]\( \text{Age} = 2 \)[/tex] is 11.9 (from the table).
The residual (difference) is:
[tex]\[ \text{Given value} - \text{Predicted value} = 12 - 11.9 = 0.1 \][/tex]
Hence, [tex]\( b = 0.1 \)[/tex].

3. Determine the predicted value when [tex]\( \text{Age} = 3 \)[/tex]:
For [tex]\( \text{Age} = 3 \)[/tex]:
[tex]\[ y = -20 \cdot 3 + 177 = 117 \][/tex]
Given value for [tex]\( \text{Age} = 3 \)[/tex] is 9.
Residual is given as 0.
Since Residual = [tex]\(\text{Given value} - \text{Predicted value} \)[/tex],
confirming the predicted value, [tex]\( \text{Predicted value} = \text{Given value} - 0 = 9 \)[/tex].

But we need to state the correct predicted value as:
So, [tex]\( c = 117 \)[/tex].

4. Calculate the residual when [tex]\( \text{Age} = 4 \)[/tex]:
Given value for [tex]\( \text{Age} = 4 \)[/tex] is 5.
Predicted value when [tex]\( \text{Age} = 4 \)[/tex] is 6.1 (from the table).
The residual (difference) is:
[tex]\[ \text{Given value} - \text{Predicted value} = 5 - 6.1 = -1.1 \][/tex]
Hence, [tex]\( d = -1.1 \)[/tex].

Therefore, the values that complete the table are as follows:
[tex]\[ \begin{align*} a & = 157 \\ b & = 0.1 \\ c & = 117 \\ d & = -1.1 \\ \end{align*} \][/tex]