Question

The savings account offering which of these APRs and compounding periods offers the best APY?

Answer 1

$\bf \qquad \qquad \textit{Annual Yield Formula} \\\\ ~~~~~~~~~~~~\textit{4.0784\% compounded monthly}\\\\ ~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1 \\\\ \begin{cases} r=rate\to 4.0784\%\to \frac{4.0784}{100}\to &0.040784\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\to &12 \end{cases} \\\\\\ \left(1+\frac{0.040784}{12}\right)^{12}-1\\\\ -------------------------------\\\\ ~~~~~~~~~~~~\textit{4.0798\% compounded semiannually}$

$\bf ~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1 \\\\ \begin{cases} r=rate\to 4.0798\%\to \frac{4.0798}{100}\to &0.040798\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{semi-annually, thus twice} \end{array}\to &2 \end{cases} \\\\\\ \left(1+\frac{0.040798}{2}\right)^{2}-1\\\\ -------------------------------$

$\bf ~~~~~~~~~~~~\textit{4.0730\% compounded daily}\\\\ ~~~~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1 \\\\ \begin{cases} r=rate\to 4.0730\%\to \frac{4.0730}{100}\to &0.040730\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{daily, thus 365} \end{array}\to &365 \end{cases} \\\\\\ \left(1+\frac{0.040730}{365}\right)^{365}-1$

Answer 2

Answer:

Option C is correct.

Step-by-step explanation:

The formula is = $(1+\frac{r}{n})^{n}-1$

r = rate of interest

n = number of times its compounded

1. 4.0784% compounded monthly

here n = 12

$(1+\frac{0.040784}{12})^{12} -1$ = 1.0403-1 = 0.0403

2. 4.0798% compounded semiannually

here n = 2

$(1+\frac{0.040798}{12})^{2} -1$ = 1.0066-1 = 0.0066

3. 4.0730% compounded daily

here n = 365

$(1+\frac{0.040730}{12})^{365}$ = 3.328-1 = 2.328