Respuesta :
bearing in mind "t" in the simple interest equation, is for years, and 9months is just 9 off 12 months in a year
then [tex]\bf \qquad \textit{Simple Interest Earned Amount}\\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\to &\$2700\\ P=\textit{original amount deposited}\\ r=rate\to 11\%\to \frac{11}{100}\to &0.11\\ t=years\to \frac{9}{12}\to &\frac{3}{4} \end{cases} \\\\\\ 2700=P\left( 1+0.11\cdot \frac{3}{4} \right)[/tex]
solve for P
then [tex]\bf \qquad \textit{Simple Interest Earned Amount}\\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\to &\$2700\\ P=\textit{original amount deposited}\\ r=rate\to 11\%\to \frac{11}{100}\to &0.11\\ t=years\to \frac{9}{12}\to &\frac{3}{4} \end{cases} \\\\\\ 2700=P\left( 1+0.11\cdot \frac{3}{4} \right)[/tex]
solve for P
Answer:
$ 2487.13 should be deposited. ( approx )
Step-by-step explanation:
Since, future value formula,
[tex]A=P(1+r)^{n}[/tex]
Where,
P = Principal amount,
r = rate per periods
n = number of periods,
Given,
A = $ 2,700,
t = 9 months,
Annual rate = 11%,
So, the monthly rate, r = [tex]\frac{11}{12}%[/tex] = [tex]\frac{0.11}{12}[/tex]
By substituting the values,
[tex]2700 = P(1+\frac{0.11}{12})^9[/tex]
[tex]\implies P = \frac{2700}{(1+\frac{0.11}{12})^9}=2487.12544296\approx \$ 2487.13[/tex]
