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

John estimates the value of his car over time. The equation for the line of best fit is approximated as [tex]$y = -2.9x + 17.7$[/tex], where [tex]$y$[/tex] represents the value in thousands of dollars.

What values complete the residual table?

[tex]$a=$[/tex] [tex]$\square$[/tex] \\
[tex]$b=$[/tex] [tex]$\square$[/tex] \\
[tex]$c=$[/tex] [tex]$\square$[/tex] \\
[tex]$d=$[/tex] [tex]$\square$[/tex]


Sagot :

Sure! Let's work through this step-by-step.

### Given Equation for Line of Best Fit
The equation of the line of best fit is:
[tex]\[ y = -2.9x + 17.7 \][/tex]

Where [tex]\( y \)[/tex] represents the predicted value of the car in thousands of dollars, and [tex]\( x \)[/tex] represents the age of the car in years.

### Step-by-Step Solution

1. Finding [tex]\( a \)[/tex]:
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = -2.9 * 1 + 17.7 = 14.8 \][/tex]
- The residual for [tex]\( x = 1 \)[/tex] is given as [tex]\( 0.2 \)[/tex], thus:
[tex]\[ \text{Given Value} = \text{Predicted Value} + \text{Residual} \][/tex]
[tex]\[ 15 = a + 0.2 \][/tex]
So, [tex]\( a \)[/tex] is:
[tex]\[ a = 14.8 \][/tex]

2. Finding [tex]\( b \)[/tex]:
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ \text{Given Value} = 12 \][/tex]
[tex]\[ \text{Predicted Value} = 11.9 \][/tex]
- Residual is:
[tex]\[ \text{Residual} = \text{Given Value} - \text{Predicted Value} \][/tex]
[tex]\[ b = 12 - 11.9 = 0.1 \][/tex]

3. Finding [tex]\( c \)[/tex]:
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ \text{Given Value} = 9 \][/tex]
[tex]\[ \text{Residual} = 0 \][/tex]
- Since [tex]\(\text{Residual} = 0 \)[/tex]:
[tex]\[ \text{Given Value} = \text{Predicted Value} \][/tex]
Therefore, [tex]\( c = 9 \)[/tex].

4. Finding [tex]\( d \)[/tex]:
- For [tex]\( x = 4 \)[/tex]:
[tex]\[ \text{Given Value} = 5 \][/tex]
[tex]\[ \text{Predicted Value} = 6.1 \][/tex]
- Residual is:
[tex]\[ \text{Residual} = \text{Given Value} - \text{Predicted Value} \][/tex]
[tex]\[ d = 5 - 6.1 = -1.1 \][/tex]

Putting it all together, we get:
[tex]\[ a = 15.0 \][/tex]
[tex]\[ b = 0.1 \][/tex]
[tex]\[ c = 9.0 \][/tex]
[tex]\[ d = -1.1 \][/tex]

The completed residual table is:
\begin{tabular}{|c|c|c|c|}
\hline \begin{tabular}{c}
Age \\
(years)
\end{tabular} & \begin{tabular}{c}
Given \\
Value
\end{tabular} & \begin{tabular}{c}
Predicted \\
Value
\end{tabular} & Residual \\
\hline 1 & 15 & 15.0 & 0.2 \\
\hline 2 & 12 & 11.9 & 0.1 \\
\hline 3 & 9 & 9.0 & 0 \\
\hline 4 & 5 & 6.1 & -1.1 \\
\hline 5 & 4 & 3.2 & 0.8 \\
\hline
\end{tabular}