How much money will you need to invest initially to have $10,500.00 in two and one-quarter years if the money is compounded monthly at an annual rate of 3.25%?

1.). Compound Interest Formula: A = P(1 + r/n)^(nt)

2.). Balance: ===> $10,500.00

3.). Time in Years: ===> 2 1 / 4 ===> 9 / 4

4.). Continuous Compound Monthly / Annual Rate: ===> 3.25 %

n is the number of times compounded yearly: A = P ( 1 + r/n )^nt.

r is interest rate as a decimal: ==> r = 3 1/4 %, ==> (12 + 1)/4 %, ==> 13/4 %, ==> 13/400 ===> 0.0325

t is the time in years: ==> t = two and one - quarter years ==> 2 1/4 years ==>

[ {2(4) + 1}/4 ] years ==> [ (8 + 1) / 4 ] years ===> 9/4 years

Solution: Final Amount: $9,760.55, would be the initial investment.

n compounding per year : A = P ( 1 + r/n )^nt.

10,500.00 ===> P ( 1 + 0.0325/12 ) (12) (9/4)

10,500.00 ===> P ( 1.00270833 )^27

10,500.00 ===> P ( 1.075758673 )

(10,500.00) / (1.075758673) ==> P

P ==> 9,760.55 dollars.

Therefore, The initial investment ==> principle ==> $9,760.55 dollars; is your final answer.

Hope that helps : )