Chapter 4, Problem 89RE

(a)

To determine

To Find: Amount of the investment semiannually.

Answer to Problem 89RE

  $\$16,081.15$

Explanation of Solution

Given: If $\$12000$ is invested at an interest rate of $10\%$ per year and amount to be invested at the end of the $3$ years for the given compounded method.

Given: principal amount = $\$12000$

Rate = $10\%$

Times = $3$

n= $2$

Then

Amount of the investment semiannually is

  $\begin{array}{l}A(t)=P{\left(1+\frac{r}{n}\right)}^{nt}\\ =12000{\left(1+\frac{0.10}{2}\right)}^{2(3)}\\ =\$16,081.15\end{array}$

Hence, The amount of the investment semiannually is $\$16,081.15$

(b)

To determine

To Find: Amount of the investment monthly.

Answer to Problem 89RE

  $\$16,178.18$

Explanation of Solution

Given: If $\$12000$ is invested at an interest rate of $10\%$ per year and amount to be invested at the end of the $3$ years for the given compounded method.

Given: principal amount = $\$12000$

Rate = $10\%$

Times = $3$

n= $12$

Then

Amount of the investment monthly is

  $\begin{array}{l}A(t)=P{\left(1+\frac{r}{n}\right)}^{nt}\\ =12000{\left(1+\frac{0.10}{12}\right)}^{12(3)}\\ =\$16,178.18\end{array}$

Hence, the amount of the investment is $\$16,178.18$

(c)

To determine

To Find: Amount of the investment daily.

Answer to Problem 89RE

  $\$16,197.64$

Explanation of Solution

Given: If $\$12000$ is invested at an interest rate of $10\%$ per year and amount to be invested at the end of the $3$ years for the given compounded method.

Given: principal amount = $\$12000$

Rate = $10\%$

Times = $3$

n= $365$

Then

Amount of the investment daily is

  $\begin{array}{l}A(t)=P{\left(1+\frac{r}{n}\right)}^{nt}\\ =12000{\left(1+\frac{0.10}{365}\right)}^{365(3)}\\ =\$16,197.64\end{array}$

Hence, the amount of the investment daily is $\$16,197.64$

(d)

To determine

To Find: Amount of the investment when compounded continuously.

Answer to Problem 89RE

  $\$16,198.31$

Explanation of Solution

Given: If $\$12000$ is invested at an interest rate of $10\%$ per year and amount to be invested at the end of the $3$ years for the given compounded method.

Given: principal amount = $\$12000$

Rate = $10\%$

Time = $3$

Then

Amount of the investment continuously is

  $\begin{array}{l}A(t)=P{e}^{rt}\\ =12000e^{0.10(3)}\\ =\$16,198.31\end{array}$

Hence, the amount of the investment continuously is $\$16,198.31$