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well, she put 130 bucks, and then she got back 132.6 after one year, so the interest earned is 132.6 - 130 = $2.6, thus
[tex]~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill & \$2.60\\ P=\textit{original amount deposited}\dotfill & \$130\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\dotfill &1 \end{cases} \\\\\\ 2.60 = (130)(\frac{r}{100})(1)\implies 2.6=\cfrac{130r}{100}\implies 260=130r \\\\\\ \cfrac{260}{130}=r\implies 2=r[/tex]