Question

Find the value of $ 15,000 at the end of one year if it is invested in an account that has an interest rate of 4.95 % and is compounded in accordance with the rules below. a. compounded monthly b. compounded daily​ (assuming a​ 365-day year) c. compounded quarterly

Answer 1

A)

$\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\ 15000\\ r=rate\to 4.95\%\to \frac{4.95}{100}\to &0.0495\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{twelve months, thus} \end{array}\to &12\\ t=years\to &1 \end{cases} \\\\\\ A=15000\left(1+\frac{0.0495}{12}\right)^{12\cdot 1}$

b)

$\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\ 15000\\ r=rate\to 4.95\%\to \frac{4.95}{100}\to &0.0495\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{365 days, thus} \end{array}\to &365\\ t=years\to &1 \end{cases} \\\\\\ A=15000\left(1+\frac{0.0495}{365}\right)^{365\cdot 1}$

c)

$\bf \qquad \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\to &\ 15000\\ r=rate\to 4.95\%\to \frac{4.95}{100}\to &0.0495\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{four quarters, thus} \end{array}\to &4\\ t=years\to &1 \end{cases} \\\\\\ A=15000\left(1+\frac{0.0495}{4}\right)^{4\cdot 1}$