How much should be invested now at 7.55% compounded annually to have $46,000 in 12 years?
P = Po(1+r)^n = 46,000.
r = 7.55%/100% = 0.0755.
n = 1comp./yr. * 12yrs. = 12 Compounding periods.
Po = ?.
Po = 46,000/(1.0755)^12 = $19,205.82
To determine the amount that should be invested now at 7.55% compounded annually to accrue $46,000 in 12 years, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future value (desired amount in this case), which is $46,000,
P = the principal amount (initial investment),
r = the annual interest rate (in decimal form), which is 7.55% or 0.0755,
n = the number of times interest is compounded per year, which is once annually in this case, and
t = the number of years the money is invested for, which is 12 years.
Now, we can rearrange the formula to solve for P:
P = A / (1 + r/n)^(nt)
Plugging in the given values:
P = $46,000 / (1 + 0.0755/1)^(1*12)
Simplifying further:
P = $46,000 / (1.0755)^12
Using a calculator or a computational tool, we can raise (1.0755) to the power of 12:
P ≈ $46,000 / 1.9397
P ≈ $23,723.06
Therefore, approximately $23,723.06 should be invested now at 7.55% compounded annually to have $46,000 in 12 years.