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Sean created this table to represent the balance of his loan, [tex]\( y \)[/tex], over a period of months, [tex]\( x \)[/tex]. The equation for the line of best fit for Sean's table of data is [tex]\( y = -115.9x + 8,007.30 \)[/tex].

\begin{tabular}{|r|r|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
0 & \[tex]$8,000 \\
\hline
1 & \$[/tex]7,888 \\
\hline
2 & \[tex]$7,775 \\
\hline
3 & \$[/tex]7,661 \\
\hline
4 & \[tex]$7,547 \\
\hline
5 & \$[/tex]7,432 \\
\hline
6 & \[tex]$7,316 \\
\hline
7 & \$[/tex]7,200 \\
\hline
8 & \[tex]$7,083 \\
\hline
9 & \$[/tex]6,966 \\
\hline
10 & \[tex]$6,848 \\
\hline
11 & \$[/tex]6,729 \\
\hline
12 & \[tex]$6,609 \\
\hline
\end{tabular}

According to the line of best fit and using \(\square\), Sean can estimate that the balance of his loan will reach \$[/tex]0 in month [tex]\(\square\)[/tex].


Sagot :

To find when the balance of Sean's loan will reach [tex]$0, we can use the provided equation for the line of best fit: \[ y = -115.9x + 8007.30 \] Here, \( y \) represents the balance of the loan, and \( x \) represents the number of months. To determine when the balance of the loan will be \( \$[/tex]0 \), we need to solve for [tex]\( x \)[/tex] when [tex]\( y \)[/tex] is set to [tex]\( 0 \)[/tex].

Step 1: Set up the equation for when the balance is zero:
[tex]\[ 0 = -115.9x + 8007.30 \][/tex]

Step 2: Solve for [tex]\( x \)[/tex]:
[tex]\[ 115.9x = 8007.30 \][/tex]
[tex]\[ x = \frac{8007.30}{115.9} \][/tex]

Step 3: Calculate the value:
[tex]\[ x \approx 69.088 \][/tex]

Therefore, using the line of best fit, Sean can estimate that the balance of his loan will reach [tex]\( \$0 \)[/tex] in month [tex]\( 69.088 \)[/tex], or approximately the [tex]\( 69^{th} \)[/tex] month.

Final answer:
- According to the line of best fit and using ___[tex]\[ x = \frac{8007.30}{115.9} \][/tex], Sean can estimate that the balance of his loan will reach [tex]\( \$0 \)[/tex] in month ___[tex]\[ 69.088 \][/tex]
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