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Mackenzie is going to invest in an account paying an interest rate of 6.5% compounded monthly. How much would Mackenzie need to invest, to the nearest ten dollars, for the value of the account to reach $194,000 in 11 years?

User Donkim
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2 Answers

3 votes

Answer:

95090

Step-by-step explanation:

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

A= 194000

T= 11

R=0.065

N=12

User RoMEoMusTDiE
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3 votes

Final answer:

Mackenzie would need to invest approximately $97,560, to the nearest ten dollars, in an account with a 6.5% interest rate compounded monthly to have $194,000 in 11 years.

Step-by-step explanation:

To determine how much Mackenzie would need to invest to have $194,000 in her account in 11 years at an interest rate of 6.5% compounded monthly, we need to use the formula for compound interest:

P = A / (1 + r/n)(nt)

Where:

P is the principal amount (the initial amount of money)

A is the amount of money accumulated after n years, including interest.

r is the annual interest rate (decimal)

n is the number of times that interest is compounded per year

t is the time the money is invested for, in years

Let's plug in the given values:

A = $194,000

r = 0.065 (6.5% as a decimal)

n = 12 (since interest is compounded monthly)

t = 11 (the number of years the money will be invested)

We can now solve for P:

P = 194,000 / (1 + 0.065/12)(12*11)

Calculating the above expression with a calculator, we get:

P ≈ $97,560.29

To the nearest ten dollars, Mackenzie would need to invest $97,560.

User Christian Goltz
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