Respuesta :
[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 &\$4000\\ r=rate\to 5.5\%\to \frac{5.5}{100}\dotfill &0.055\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases} \\\\\\ A=4000\left(1+\frac{0.055}{1}\right)^{1\cdot 4}\implies A=4000(1.055)^4\implies A\approx 4955.2986[/tex]
Answer:
$ 4955.30 ( approx )
Step-by-step explanation:
The formula for compound interest is,
[tex] A(t)=P(1+i)^t[/tex]
Where, P is the principal amount,
i is the rate per period,
t is the number of periods,
Here, P = $ 4000,
i = 5.5% = 0.055
t = 4 years,
By substituting the values,
The amount in the account after 4 years would be,
[tex]A=4000(1+0.055)^4=4000(1.055)^4=4955.2986025\approx \$4955.30[/tex]
