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Find out how long it takes a $3,500 investment to double if it is invested at 7%
compounded quarterly Round to the nearest tenth of a year. Use the formula
A = P(1+ )
T
iit

Respuesta :

adioabiola

Answer:

9.99 years

Step-by-step explanation:

P=$3,500

r=7%=0.07

n=4(quarterly)

A= double of $3,500=

$3,500×2=$7,000

t=?

A=p(1+r/n)^nt

$7,000=$3,500(1+0.07/4)^4t

$7,000=$3,500(1+0.0175)^4t

$7,000=$3,500(1.0175)^4t

Divide both sides by $3,500

2=(1.0175)^4t

Take the log to base 10 of both sides

log2=4t × log1.0175

0.30103=4t × 0.00753

0.30103=4(0.00753)t

0.30103=0.03012t

t=0.30103/0.03012

t=9.99435

Approximately

9.99 years

raphealnwobi

It would take 10 years for the investment to be doubled.

Compound interest is given by:

[tex]A=P(1+\frac{r}{n} )^{nt}[/tex]

Where A is the final amount, P is the principal, r is the rate, n is the number of times compounded in the time (t) period

Given that:

r = 7% = 0.07, P = 3500, A = 2 * 3500 = 7000, n = 4 quarterly. Hence:

[tex]7000=3500(1+\frac{0.07}{4})^{4t} \\\\2=(1.0175)^{4t} \\\\ln(2)=4t* ln(1.0175)\\\\t=10\ years[/tex]

It would take 10 years for the investment to be doubled.

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