[tex]\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill&4370.91\\ P=\textit{original amount deposited}\dotfill \\ r=rate\to 3\%\to \frac{3}{100}\dotfill &0.03\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases}[/tex]
[tex]\bf 4370.91=P\left(1+\frac{0.03}{1}\right)^{1\cdot 3}\implies 4370.91=P(1.03)^3 \\\\\\ 4370.91=1.092727P\implies \cfrac{4370.91}{1.092727}=P\implies 4000.00183\approx P \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
which rounds up to... .yeap, you guessed it.
