Question

If 44700 dollars is invested at an interest rate of 5 percent per year, find the value of the investment at the end of 5 years for the following compounding methods, to the nearest cent.

(a) Annual: $
(b) Semiannual: $
(c) Monthly: $
(d) Daily: $

Answer 1

Annual


`\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$44700\\ r=rate\to 5\%\to (5)/(100)\to &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to &1\\ t=years\to &5 \end{cases} \\\\\\ A=44700\left(1+(0.05)/(1)\right)^(1\cdot 5)\implies A=44700(1.05)^5`

Semi-Annual


`\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$44700\\ r=rate\to 5\%\to (5)/(100)\to &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{semi-annual, thus twice} \end{array}\to &2\\ t=years\to &5 \end{cases} \\\\\\ A=44700\left(1+(0.05)/(2)\right)^(2\cdot 5)\implies A=44700(1.025)^(10)`

Monthly


`\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$44700\\ r=rate\to 5\%\to (5)/(100)\to &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\to &12\\ t=years\to &5 \end{cases} \\\\\\ A=44700\left(1+(0.05)/(12)\right)^(12\cdot 5)\implies A=44700(\stackrel{\approx}{1.00417})^(60)`

Daily, assuming 365 days per year


`\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+(r)/(n)\right)^(nt) \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\$44700\\ r=rate\to 5\%\to (5)/(100)\to &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{daily, thus 365} \end{array}\to &365\\ t=years\to &5 \end{cases} \\\\\\ A=44700\left(1+(0.05)/(365)\right)^(365\cdot 5)\implies A=44700(\stackrel{\approx}{1.000137})^(1825)`