Chapter 6.3, Problem 12Q

a.

To determine

To write: a function that represents the balance after t years.

Answer to Problem 12Q

  $f(t)=500{(1.03)}^t.$

Explanation of Solution

Given information: Deposit $\$500$ in a savings account that earn $3\%$ annual interest

compounded monthly.

Formula Used:

The formula for compound amount is:

  $f(t)=P{\left(1+\frac{r}{n}\right)}^{nt}$ .

Calculation:

  $\begin{array}{c}P=500\\ r=3\%=0.03\\ n=12\\ f(t)=P{\left(1+\frac{r}{n}\right)}^{nt}\\ f(t)=500{\left(1+\frac{0.03}{12}\right)}^{12t}\\ $ f(t)=500{(1.0025)}^{12t}$$ f(t)=500{\left[{(1.0025)}^{12}\right]}^t$$ f(t)=500{(1.03)}^t$\end{array}$

b.

To determine

To find: what is the balance of the account after $36$ months, $10$ years.

Answer to Problem 12Q

  $f(3)\approx \$\mathrm{546.35.}$

  $f(10)\approx \$671.96$ .

Explanation of Solution

Given information: Deposit $\$500$ in a savings account that earn $3\%$ annual interest

compounded monthly.

Calculation:

Have to compute the balance after 36 months:

  $\begin{array}{l}36\text{months}=3\text{years}\\ t=3\\ $ f(3)=500{(1.03)}^3\approx \mathrm{546.35.}$\end{array}$

Have to compute the balance after:

  $\begin{array}{l}t=10\\ $ f(10)=500{(1.03)}^{10}\approx 671.96$\end{array}$