How much money would you need to deposit today at 9% annual interestcompounded monthly to have $12000 in the account after 6 years?

To solve this problem, we need to use the formula for compound interest, which is:

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

Where: A = the amount of money accumulated after n years, including interest. P = the principal amount (the initial amount of money) r = annual interest rate (in decimal) n = number of times that interest is compounded per year t = time the money is invested for in years

We need to rearrange the formula to solve for P, the initial deposit:

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

We know that A = $12000, r = 9% or 0.09 (in decimal), n = 12 (since the interest is compounded monthly), and t = 6 years.

Substituting these values into the formula gives:

P = 12000 / (1 + 0.09/12)^(12*6)

Now, calculate the value inside the brackets:

= 12000 / (1 + 0.0075)^(72)

= 12000 / (1.0075)^(72)

Now, calculate the exponent:

= 12000 / 1.7138

Finally, divide 12000 by 1.7138 to find P:

P = $7000.37

So, you would need to deposit approximately $7000.37 today to have$12000 in the account after 6 years with an annual interest rate of 9% compounded monthly.