Trent opened a savings account and deposited $400.00 as principal. The account earns 10% interest, compounded monthly. What is the balance after 3 years?Use the formula A=P1+rnnt, where A is the balance (final amount), P is the principal (starting amount), r is the interest rate expressed as a decimal, n is the number of times per year that the interest is compounded, and t is the time in years.Round your answer to the nearest cent.
Question
Trent opened a savings account and deposited $400.00 as principal. The account earns 10% interest, compounded monthly. What is the balance after 3 years?Use the formula A=P1+rnnt, where A is the balance (final amount), P is the principal (starting amount), r is the interest rate expressed as a decimal, n is the number of times per year that the interest is compounded, and t is the time in years.Round your answer to the nearest cent.
Solution
To solve this problem, we will use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
- A is the amount of money accumulated after n years, including interest.
- P is the principal amount (the initial amount of money).
- r is the annual interest rate (in decimal).
- n is the number of times that interest is compounded per year.
- t is the time the money is invested for in years.
Given in the problem: P = $400.00 r = 10% = 0.10 (as a decimal) n = 12 (compounded monthly) t = 3 years
Substitute these values into the formula:
A = 400(1 + 0.10/12)^(12*3)
Now, calculate the expression inside the parentheses:
1 + 0.10/12 = 1.008333
Then, raise this result to the power of 36 (which is 12 times 3):
(1.008333)^36 = 1.34392
Finally, multiply this result by the principal amount:
400 * 1.34392 = $537.57
So, the balance in the account after 3 years will be approximately $537.57, rounded to the nearest cent.
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