Question

Chester hopes to earn $1200 in interest in 2.6 years time from $48,000 that he has available to invest. To decide if it's feasible to do this by investing in an account that

compounds annually, he needs to determine the annual interest rate such an account would have to offer for him to meet his goal. What would the annual rate of
Interest have to be? Round to two decimal places.

Answer 1

`~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\dotfill & \stackrel{ 48000+1200 }{\$ 49200}\\ P=\textit{original amount deposited}\dotfill &\$48000\\ r=rate\to r\%\to (r)/(100)\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2.6 \end{cases}`

`49200 = 48000\left(1+( ~~ (r)/(100) ~~ )/(1)\right)^(1\cdot 2.6) \implies \cfrac{49200}{48000}=\left( 1+\cfrac{r}{100} \right)^(2.6) \\\\\\ \cfrac{41}{40}=\left( \cfrac{100+r}{100} \right)^(2.6)\implies \sqrt[2.6]{\cfrac{41}{40}}=\cfrac{100+r}{100} \\\\\\ 100\sqrt[2.6]{\cfrac{41}{40}}=100+r\implies 100\sqrt[2.6]{\cfrac{41}{40}}-100=r\implies \stackrel{ \% }{0.95}\approx r`