[tex]\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$50\\ r=rate\to 10\%\to \frac{10}{100}\dotfill &0.10\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases}[/tex]
[tex]\bf A=50\left(1+\frac{0.10}{1}\right)^{1\cdot 2}\implies A=50(1.1)^2\implies A=60.5 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{earned interest}}{60.5 - 50\implies 10.5}~\hfill[/tex]
Answer:
$10.50
Step-by-step explanation:
(see attached for reference)
The formula for compound interest is:
A = P [1 + (r/n) ]^(nt)
where
A = Final amount
P = Principal = $50
r = Annual interest rate = 10% = 0.1
n = 1 (because compounded annually)
t = time = 2 years
substituting this into the equation:
A = P [1 + (r/n) ]^(nt)
= 50 [ 1 + (0.1/1) ] ^ [(1)(2)]
= 50 [ 1 + 0.1] ^ 2
= 50 (1.1)²
= $60.50
Interest = Final amount - Principal
=$60.50 - $50.00
= $10.50
