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2 years ago

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Find the time required for an investment of $3000 to grow to $9000 at an interest rate of 6.5% per year, compounded quarterly. (

Round your answer to two decimal places.)

1 answer:

Vanyuwa [196]2 years ago

8 0

Answer:

$t=17.04\ years$

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

$P=\$3,000\\A=\$9,000\\ r=6.5\%=6.5/100=0.065\\n=4$

substitute in the formula above

$9,000=3,000(1+\frac{0.065}{4})^{4t}$

solve for t

Simplify

$3=(1.01625)^{4t}$

Apply log both sides

$log(3)=log[(1.01625)^{4t}]$

Apply property of logarithms

$log(3)=(4t)log(1.01625)$

$t=log(3)/[4log(1.01625)]$

$t=17.04\ years$

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

The rate of change in the number of hospital outpatient visits, in millions, is given by:

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