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4 years ago

$1000 invested with compound interest at a rate of 15% per year for 9 years.

Mathematics

1 answer:

ivolga24 [154]4 years ago

5 0

Answer:

The Final Investment Value is $\$3,517.88$

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

$t=9\ years\\ P=\$1,000\\ r=0.15\\n=1$

substitute in the formula above

$A=1,000(1+\frac{0.15}{1})^{1*9}$

$A=\$3,517.88$

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A similar problem is given at brainly.com/question/15165354

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3 years ago

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krek1111 [17]

Answer:

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3 years ago

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If 1=36, 2=38, 3=39, what does 4 equal to? kindly explain

timama [110]

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3 years ago

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Help please?

Alla [95]

$\bf \begin{array}{clclll}
3^{1001}&\qquad &4^{1002}\\
\uparrow &&\uparrow \\
a&&b
\end{array}\\\\
-----------------------------\\\\
(a+b)^2-(a-b)^2\implies (a^2+2ab+b^2)-(a^2-2ab+b^2)
\\\\\\
a^2+2ab+b^2-a^2+2ab-b^2\implies 2ab+2ab\implies 4ab
\\\\\\
now\qquad 4(3^{1001})(4^{1002})=k12^{1001}\qquad 
\begin{cases}
12^{1001}\\
(4\cdot 3)^{1001}\\
4^{1001}\cdot 3^{1001}
\end{cases}$

$\bf \\\\\\
4(3^{1001})(4^{1002})=k(4^{1001}\cdot 3^{1001})\implies \cfrac{4(3^{1001})(4^{1002})}{4^{1001}\cdot 3^{1001}}=k
\\\\\\
4\cdot \cfrac{3^{1001}}{3^{1001}}\cdot \cfrac{4^{1002}}{4^{1001}}=k\implies 4\cdot 4^{1002}4^{-1001}=k\impliedby 
\begin{array}{llll}
\textit{same base}\\
\textit{add the}\\
exponents
\end{array}
\\\\\\
4\cdot 4^{1002-1001}=k\implies 4\cdot 4^1=k\implies 16=k$

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3 years ago