Respuesta :
Answer:
A=Pe^rt
P= princible (1300)
e= (2.71828)- function on a graphing calculator
r = interest rate (.05 or 5%)
t = time (10 years)
A = 1300e^.05(10)
A = 1300e^.5
A = 2143.337652
A = 2143
[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 &\$1300\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &10 \end{cases}[/tex]
[tex]\bf A=1300\left(1+\frac{0.05}{1}\right)^{1\cdot 10}\implies A=1300(1.05)^{10}\implies \stackrel{\textit{rounded up}}{A=2118}[/tex]