Question

ALGEBRA 2: $7,700 is invested in an account earning 7.3% interest (APR), compounded daily.Write a function showing the value of the account after t years, where the annual growth rate can be found from a constant in the function. Round all coefficients inthe function to four decimal places. Also, determine the percentage of growth peryear (APY), to the nearest hundredth of a percent.

Answer 1

Solution:

Given:

Part A:

$\begin{gathered} P=\text{ \ 7,700} \\ r=7.3\text{ \%}=\frac{7.3}{100}=0.073 \\ n=365...............compounded\text{ daily} \end{gathered}$

Using the compound interest formula,

$\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \\ Thus, \\ A=7700(1+\frac{0.073}{365})^{365\times t} \\ A=7700(1+0.0002)^{365t} \\ A=7700(1.0002)^{365t} \end{gathered}$

Therefore, the function showing the value of the account after t-years is:

$A=7700(1.0002)^{365t}$

Part B:

To get the percentage growth per year, we get the amount in the account at the end of two successive years.

At the end of year 1:

$\begin{gathered} A=7700(1.0002)^{365t} \\ \\ when\text{ }t=1 \\ A=7700(1.0002)^{365\times1} \\ A=7700(1.0002)^{365} \\ A=\text{ \ }8283.06 \end{gathered}$

At the end of year 2:

$\begin{gathered} A=7700(1.0002)^{365t} \\ \\ when\text{ }t=2 \\ A=7700(1.0002)^{365\times2} \\ A=7700(1.0002)^{730} \\ A=\text{ \ }8910.28 \end{gathered}$

The percentage growth is gotten using the formula:

$\begin{gathered} PGR=(\frac{Ending\text{ value}}{Beginning\text{ value}}-1)\times100\text{ \%} \\ PGR=(\frac{8910.28}{8283.06}-1)\times100\text{ \%} \\ PGR=(1.0757-1)\times100\text{ \%} \\ PGR=0.0757\times100\text{ \%} \\ PGR=7.57\% \end{gathered}$

Therefore, the percentage of growth per year (APY) is 7.57%