Question

Nancy deposits $2500 into an investing account that pays 6.1% annual interest compounded quarterly. What will be the balance after 10 years? Round to the nearest cent.

Answer 1

Step-by-step explanation:

P=principle balance

r= interest rate

n= no. of times interest compounded per year

t= time period.

$A = P {(1 + \frac{r}{n}) }^{nt} \\A = 2500 ({1 + \frac{ \frac{6.1}{100} }{4} )}^{4(10)} \\ A = 2500 {(1 + \frac{0.061}{4} )}^{40} \\ A = 2500 {(1 + 0.01525)}^{40} \\ A = 2500 {(1.01525)}^{40} \\ A = 4579.94$

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Answer 2

Based on the information given the balance after 10 years is $4579.94

Balance after 10 years

Using this formula

Amount=P(1+r/n)^nt

Where:

P=principle balance=2500

r= interest rate=6.1%

n= Number of times interest compounded per year=4

t= time period=10 years

Let plug in the formula

Amount=2500(1+0.061/4)^4×10

Amount=2500(1+0.01525)^40

Amount=2500(1.01525)^40

Amount=$4579.94

Inconclusion  the balance after 10 years is $4579.94

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