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skad [1K]
3 years ago
9

If an investment of $3000 grows to $4432.37 in eight years with interest compounded annually , what is the interest rate?

Mathematics
1 answer:
melamori03 [73]3 years ago
6 0
\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$4432.37\\
P=\textit{original amount deposited}\to &\$3000\\
r=rate\to r\%\to \frac{r}{100}\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &8
\end{cases}

\bf 4432.37=3000\left(1+\frac{r}{1}\right)^{1\cdot 8}\implies \cfrac{4432.37}{3000}=(1+r)^8
\\\\\\
\sqrt[8]{\cfrac{4432.37}{3000}}=1+r\implies \sqrt[8]{\cfrac{4432.37}{3000}}-1=r
\\\\\\
0.0500001086\approx r\qquad\qquad\cfrac{r\%}{100}\approx 0.0500001086
\\\\\\
r\%\approx 100\cdot 0.0500001086\implies r=\stackrel{\%}{5}
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Answer:

As consequence of the Taylor theorem with integral remainder we have that

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If we ask that f has continuous (n+1)th derivative we can apply the mean value theorem for integrals. Then, there exists c between a and x such that

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}dt = \frac{f^{(n+1)}(c)}{n!} \int^a_x (x-t)^n d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{n+1}}{n+1}\Big|_a^x

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\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{(n+1)}}{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} .

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Step-by-step explanation:

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3 years ago
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Answer:

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Answer:

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Step-by-step explanation:

1. Let's review the information given to us for answering the question correctly:

Amount of milk Amanda use for a batch of pancakes = 1/3 cup

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Let's simplify 11/22, dividing by 11 the numerator and the denominator,

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