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Kiley gathered the data in the table below. She found the approximate line of best fit to be [tex]y = 1.6x - 4[/tex].

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
0 & -3 \\
\hline
2 & -1 \\
\hline
3 & -1 \\
\hline
5 & 5 \\
\hline
6 & 6 \\
\hline
\end{tabular}

What is the residual value when [tex]x = 3[/tex]?

A. -1.8
B. -0.2
C. 0.2
D. 1.8


Sagot :

To determine the residual value at [tex]\( x = 3 \)[/tex] given the line of best fit [tex]\( y = 1.6x - 4 \)[/tex], we need to follow these steps:

1. Calculate the predicted [tex]\( y \)[/tex]-value using the line of best fit equation for [tex]\( x = 3 \)[/tex].
[tex]\[ y_{\text{predicted}} = 1.6 \cdot 3 - 4 \][/tex]
Solving this, we get:
[tex]\[ y_{\text{predicted}} = 4.8 - 4 = 0.8 \][/tex]

2. Identify the actual [tex]\( y \)[/tex]-value from the table for [tex]\( x = 3 \)[/tex]. The table tells us that:
[tex]\[ y_{\text{actual}} = -1 \][/tex]

3. Calculate the residual. The residual is the difference between the actual [tex]\( y \)[/tex]-value and the predicted [tex]\( y \)[/tex]-value:
[tex]\[ \text{Residual} = y_{\text{actual}} - y_{\text{predicted}} \][/tex]
Substituting the values we have:
[tex]\[ \text{Residual} = -1 - 0.8 = -1.8 \][/tex]

Therefore, the residual value when [tex]\( x = 3 \)[/tex] is [tex]\( -1.8 \)[/tex]. Hence, the correct answer is:
[tex]\[ \boxed{-1.8} \][/tex]