The line of best fit to model the data in the table is [tex]y = 5.2x - 0.4[/tex].

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
1 & 8 \\
\hline
2 & 13 \\
\hline
3 & 18 \\
\hline
4 & 23 \\
\hline
5 & 24 \\
\hline
\end{tabular}

What is the residual for [tex]$x = 5$[/tex]?

A. [tex]$-1.6$[/tex]
B. [tex]$-0.6$[/tex]
C. [tex]$0.6$[/tex]
D. [tex]$1.6$[/tex]



Answer :

To solve the problem of finding the residual for [tex]\( x = 5 \)[/tex] using the line of best fit [tex]\( y = 5.2x - 0.4 \)[/tex], we can follow these steps:

1. Identify the actual [tex]\( y \)[/tex] value for [tex]\( x = 5 \)[/tex]:
From the given table, when [tex]\( x = 5 \)[/tex], the actual [tex]\( y \)[/tex] value is [tex]\( y_{\text{actual}} = 24 \)[/tex].

2. Calculate the predicted [tex]\( y \)[/tex] value using the line of best fit:
Plug [tex]\( x = 5 \)[/tex] into the equation of the line of best fit, [tex]\( y = 5.2x - 0.4 \)[/tex]:
[tex]\[ y_{\text{predicted}} = 5.2 \cdot 5 - 0.4 = 26 - 0.4 = 25.6 \][/tex]

3. Calculate the residual:
The residual is the difference between the actual [tex]\( y \)[/tex] value and the predicted [tex]\( y \)[/tex] value. This can be expressed as:
[tex]\[ \text{Residual} = y_{\text{actual}} - y_{\text{predicted}} \][/tex]
Substituting the values, we get:
[tex]\[ \text{Residual} = 24 - 25.6 = -1.6 \][/tex]

The residual for [tex]\( x = 5 \)[/tex] is [tex]\(-1.6\)[/tex]. Hence, the correct answer is:
[tex]\(-1.6\)[/tex].