Jose invests 3250 dollars at 6% interest compounded annually. What will be the balance in the account after 3.5 years?
A. 3932.50 dollars
B. 3985.23 dollars
C. 4752 dollars
D. 5200 dollars
P = Po(1+r)^n
Po = $3250.
r = 6%/100% = 0.06
n = 1Comp./yr * 3.5yr. = 3.5 Compounding
periods.
Plug the above values into the given Eq
and solve for P.
I disagree with @Henry because thats not was the answer to the question that she gave give her more imformation bout the question u gave her something different and u suppose to step by step with her so she get the answer
To find the balance in the account after 3.5 years, we can use the formula for compound interest:
\[A = P \times (1 + r/n)^(n \times t)\]
Where:
A = final amount (balance) in the account
P = principal amount (initial investment) = $3250
r = annual interest rate (as a decimal) = 6% = 0.06
n = number of times interest is compounded per year = 1 (annually)
t = number of years = 3.5
Plugging in the values in the formula, we get:
\[A = 3250 \times (1 + 0.06/1)^(1 \times 3.5)\]
Simplifying this equation, we have:
\[A = 3250 \times (1.06)^3.5\]
Using a calculator or a computer program, we can compute that:
\[A \approx 3985.23\]
Therefore, the balance in the account after 3.5 years will be approximately $3985.23.
The correct answer is B. 3985.23 dollars.