Question

Suppose that $6000 is placed in an account that pays 19% interest compounded each year. Assume that no withdrawals are made from the account.

Answer 1

We are going to use the formula for the compound interest, which is

`A=P\cdot(1+(r)/(n))^(nt)`

A = the future value of the investment

P = the principal investment amount (the initial deposit or loan amount)

r = the annual interest rate (decimal)

n = the number of times that interest is compounded per unit t

t = the time the money is invested or borrowed for

Replacing the values in the first question we have:

`\begin{gathered} A=P\cdot(1+(r)/(n))^(nt) \\ A=6000,r=0.19,n=1,t=1 \\ A=6000\cdot(1+(0.19)/(1))^1=7140 \end{gathered}`

Answer for the first question is : $7140

Then, replacing the values in the second question we have:

`\begin{gathered} A=P\cdot(1+(r)/(n))^(nt) \\ A=6000,r=0.19,n=1,t=2 \\ A=6000\cdot(1+(0.19)/(1))^2=8497 \end{gathered}`

Answer for the second question is : $8497