Respuesta :
[tex]\bf \qquad \textit{Continuously Compounding Interest Earned Amount}\\\\
A=Pe^{rt}\qquad
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to& \$2000\\
r=rate\to 3.1\%\to \frac{3.1}{100}\to &0.031\\
t=years\to &1,5,20
\end{cases}
\\\\\\
\stackrel{\textit{for 1 year}}{A=2000e^{0.031\cdot 1}}\qquad \stackrel{\textit{for 5 years}}{A=2000e^{0.031\cdot 5}}\qquad \stackrel{\textit{for 20 years}}{A=2000e^{0.031\cdot 20}}\\\\
-------------------------------\\\\[/tex]
[tex]\bf \qquad \qquad \textit{Annual Yield Formula} \\\\ ~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1 \\\\ \begin{cases} r=rate\to 3.1\%\to \frac{3.1}{100}\to &0.031\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{continuously, 365 days then} \end{array}\to &365 \end{cases} \\\\\\ \left(1+\frac{0.031}{365}\right)^{365}-1 \approx 0.03148414 \\\\\\ 0.03148414\cdot 100\implies 3.148\% \approx 3.15\%[/tex]
[tex]\bf \qquad \qquad \textit{Annual Yield Formula} \\\\ ~~~~~~~~~\left(1+\frac{r}{n}\right)^{n}-1 \\\\ \begin{cases} r=rate\to 3.1\%\to \frac{3.1}{100}\to &0.031\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{continuously, 365 days then} \end{array}\to &365 \end{cases} \\\\\\ \left(1+\frac{0.031}{365}\right)^{365}-1 \approx 0.03148414 \\\\\\ 0.03148414\cdot 100\implies 3.148\% \approx 3.15\%[/tex]
Answer:
$ 2062.97 ( approx )
Step-by-step explanation:
Since, the amount formula for continuous compounding,
[tex]A=Pe^{rt}[/tex]
Where, P is the principal amount,
r is the rate per period,
t is the number of years,
e is euclid number,
Here,
P = $ 2000,
r = 3.1% = 0.031,
t = 1 year,
Hence, the balance would be,
[tex]A=2000 e^{0.031\times 1}[/tex]
[tex] = 2000 e^{0.031}[/tex]
[tex]=\$ 2062.97100777[/tex]
[tex]\approx \$ 2062.97[/tex]
