If $7,800 is deposited into an account paying 6% interest compounded annually (at the end of each year), how much money is in the account after 2 years?
$7,800 * 0.06 = $468
$7,800 + 468 = $8,268
$8,268 * 0.06 = $496.08
$8,268 + 496.08 = ?
X = 7800 (1 + .06)(1 +.06)
X = future value = Current value times (1 + interest rate) to the power of how many times the interest is calculated.
This should help you calculate similar problems too.
I hope this helps. Thanks for asking.
To find the amount of money in the account after 2 years with compounded interest, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the amount of money in the account after the specified time period
P = the principal amount (initial deposit)
r = annual interest rate (as a decimal)
n = number of times interest is compounded per year
t = number of years
Let's plug in the given values into the formula:
P = $7,800
r = 6% = 0.06
n = 1 (compounded annually)
t = 2
A = 7800(1 + 0.06/1)^(1*2)
A = 7800(1 + 0.06)^2
A = 7800(1.06)^2
To calculate this we do:
A = 7800 * (1.06)^2.
The compounding formula raised "1.06" to the power of "2", representing the two years.
Calculating this would give you the final amount of money in the account after 2 years with compounded interest.