Respuesta :
we'll be assuming this is with simple interest.
now, a year has 12 months, thus in 18 months there are 18/12 years.
[tex]\bf ~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\to &\$1050\\ P=\textit{original amount deposited}\to& \$1020\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\to \frac{18}{12}\to &\frac{3}{2} \end{cases}[/tex]
[tex]\bf 1050=1020\left[ 1+r\left( \frac{3}{2} \right) \right]\implies \cfrac{1050}{1020}=1+r\left( \frac{3}{2} \right) \implies \cfrac{35}{34}=1+r\left( \frac{3}{2} \right) \\\\\\ \cfrac{35}{34}-1=\cfrac{3}{2}r\implies \cfrac{1}{34}=\cfrac{3}{2}r\implies \cfrac{2\cdot 1}{3\cdot 34}=r\implies \cfrac{1}{51}=r \\\\\\ r\%=100\cdot \cfrac{1}{51}\implies r\approx\stackrel{\%}{1.96078431372549}[/tex]
now, a year has 12 months, thus in 18 months there are 18/12 years.
[tex]\bf ~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\to &\$1050\\ P=\textit{original amount deposited}\to& \$1020\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\to \frac{18}{12}\to &\frac{3}{2} \end{cases}[/tex]
[tex]\bf 1050=1020\left[ 1+r\left( \frac{3}{2} \right) \right]\implies \cfrac{1050}{1020}=1+r\left( \frac{3}{2} \right) \implies \cfrac{35}{34}=1+r\left( \frac{3}{2} \right) \\\\\\ \cfrac{35}{34}-1=\cfrac{3}{2}r\implies \cfrac{1}{34}=\cfrac{3}{2}r\implies \cfrac{2\cdot 1}{3\cdot 34}=r\implies \cfrac{1}{51}=r \\\\\\ r\%=100\cdot \cfrac{1}{51}\implies r\approx\stackrel{\%}{1.96078431372549}[/tex]
