Respuesta :
[tex]~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill &\$\stackrel{12000~~ + ~~1200}{13200}\\ P=\textit{original amount deposited}\dotfill &\$12000\\ r=rate\to r\%\to \frac{r}{100}\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1.8 \end{cases}[/tex]
[tex]13200=12000\left(1+\frac{~\frac{r}{100}~}{1}\right)^{1\cdot 1.8}\implies \cfrac{13200}{12000}=\left( 1+\cfrac{r}{100} \right)^{1.8} \\\\\\ \cfrac{11}{10}=\left(\cfrac{100 + r}{100} \right)^{1.8}\implies \cfrac{11}{10}=\left(\cfrac{100 + r}{100} \right)^{\frac{9}{5}} \\\\\\ \left( \cfrac{11}{10} \right)^{\frac{5}{9}}=\left[ \left(\cfrac{100 + r}{100} \right)^{\frac{9}{5}} \right]^{\frac{5}{9}}\implies \left( \cfrac{11}{10} \right)^{\frac{5}{9}}=\cfrac{100+r}{100}[/tex]
[tex]\sqrt[9]{\left( \cfrac{11}{10} \right)^5}=\cfrac{100+r}{100}\implies 100\sqrt[9]{\left( \cfrac{11}{10} \right)^5}=100+r \\\\\\ 100\sqrt[9]{\left( \cfrac{11}{10} \right)^5}-100 = r\implies 5.44\approx r[/tex]
