at the beginning of the 9th year, namely at the end of the 8th year of it being deposited, so after 8 years then.
[tex] \bf ~~~~~~ \textit{Simple Interest Earned Amount}
\\\\
A=P(1+rt)\qquad
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to& \$4000\\
r=rate\to 6.5\%\to \frac{6.5}{100}\to &0.065\\
t=years\to &8
\end{cases}
\\\\\\
A=4000[1+(0.065)(8)]\implies A=4000(1.52)\implies A=6080 [/tex]
Answer:
D
[tex] \boxed{6080} [/tex]
[tex]\text{A = P(1 + rt)} \Rightarrow A = 4000(1 + (0.065*8)) = 6080 \Rightarrow \boxed {6080}[/tex]
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The formula we use for this problem is:
[tex]A = P(1 + rt)[/tex]
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Now, we would input the numbers into the formula.
A = 4000(1 + (0.065 × 8))
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Hence, that is how the answer would be [tex]\boxed{6080}[/tex]
