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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]$\$[/tex] 8,000[tex]$ \\
\hline
1 & $[/tex]\[tex]$ 7,888$[/tex] \\
\hline
2 & [tex]$\$[/tex] 7,775[tex]$ \\
\hline
3 & $[/tex]\[tex]$ 7,661$[/tex] \\
\hline
4 & [tex]$\$[/tex] 7,547[tex]$ \\
\hline
5 & $[/tex]\[tex]$ 7,432$[/tex] \\
\hline
6 & [tex]$\$[/tex] 7,316[tex]$ \\
\hline
7 & $[/tex]\[tex]$ 7,200$[/tex] \\
\hline
8 & [tex]$\$[/tex] 7,083[tex]$ \\
\hline
9 & $[/tex]\[tex]$ 6,966$[/tex] \\
\hline
10 & [tex]$\$[/tex] 6,848[tex]$ \\
\hline
11 & $[/tex]\[tex]$ 6,729$[/tex] \\
\hline
12 & [tex]$\$[/tex] 6,609[tex]$ \\
\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]\[tex]$0$[/tex] in month [tex]\(\square\)[/tex].