Answer:
A) $3,570
B) $4,248.30
Explanation:
We were given that:
Principal, P = $3,000
Interest Rate, r = 19% = 19/100 = 0.19
Compounding, n = yearly = 1
Time, t = ?
We will proceed to solve for the amount in the account as shown below:
A) At the end of one year:
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ Time,t=1year \\ A=3000(1+\frac{0.19}{1})^{1*1} \\ A=3000(1+0.19)^1 \\ A=3000(1.19) \\ A=\text{\$}3,570 \end{gathered}[/tex]B) At the end of two years:
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ Time,t=2years \\ A=3000(1+\frac{0.19}{1})^{1*2} \\ A=3000(1+0.19)^2 \\ A=3000(1.19)^2 \\ A=\text{\$}4,248.30 \end{gathered}[/tex]