Question

hello I seem to be having some difficulty with this problem and need some help thank you

Answer 1

Solution

Given that

The Parker's are saving up to go for a family vacation in 3 years

Number of year, n, is 3 years.

They invested $3100 into an account

Principal, P, is $3100

Annual interest rate of 1.36% compounded monthly

a) To find the amount, A, the Parker's account after 3 years, the formula is

$A=P(1+\frac{r}{n})^{nt}$

Where

$\begin{gathered} r=\frac{1.36}{100}=0.0136 \\ t=12 \end{gathered}$

Substitute the variables into the formula above

$\begin{gathered} A=3100(1+\frac{0.0136}{12})^{3\times12} \\ A=3100(1+\frac{0.0136}{12})^{36}=\text{ \ 3229.02} \\ A=\text{ \ 3229.02 \lparen nearest cent\rparen} \end{gathered}$

Hence, the amount in the Parker's account after 3 years is $3,229.02 (nearest cent)

b) To find the interest, I, earned on the Parker's investment, the formula is

$I=A-P$

Where

$\begin{gathered} A=\text{ \ 3229.02} \\ P=\text{ \ 3100} \end{gathered}$

Substitute the values into the formula to find the interest above

$\begin{gathered} I=A-P \\ I=3229.02-3100=\text{ \ 129.02} \\ I=\text{ \ 129.02} \end{gathered}$

Hence, the interest, I, earned on the Parker's investment is $129.02 (nearest cent)