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

We are going to use the formula for the compound interest, which is
[tex]A=P\cdot(1+\frac{r}{n})^{nt}[/tex]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:
[tex]\begin{gathered} A=P\cdot(1+\frac{r}{n})^{nt} \\ A=6000,r=0.19,n=1,t=1 \\ A=6000\cdot(1+\frac{0.19}{1})^1=7140 \end{gathered}[/tex]Answer for the first question is : $7140
Then, replacing the values in the second question we have:
[tex]\begin{gathered} A=P\cdot(1+\frac{r}{n})^{nt} \\ A=6000,r=0.19,n=1,t=2 \\ A=6000\cdot(1+\frac{0.19}{1})^2=8497 \end{gathered}[/tex]Answer for the second question is : $8497