$24,000 is invested for 3 years with an APR of 2% and daily compounding. Balance in the account after 3. years is $ ___
To calculate the balance in the account after 3 years with daily compounding, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = final amount/balance after time t
P = principal amount (initial investment)
r = annual interest rate (as a decimal)
n = number of times interest is compounded per year
t = number of years
In this case, the principal amount (P) is $24,000, the annual interest rate (r) is 2% (or 0.02 as a decimal), the time (t) is 3 years, and since interest is compounded daily, the number of times interest is compounded per year (n) is 365.
Now, let's plug the values into the formula:
A = 24000(1 + 0.02/365)^(365*3)
To calculate this, we'll follow the order of operations (parentheses, exponentiation, multiplication, and addition/subtraction).
First, divide the annual interest rate by the number of compounding periods per year:
A = 24000(1 + 0.02/365)^(365*3)
= 24000(1 + 0.0000548)^(365*3)
Next, calculate the exponent:
A = 24000(1.0000548)^(365*3)
= 24000(1.0000548)^1095
Finally, multiply the result by the principal amount:
A ≈ 24000 * 1.061844Ш47257
≈ $25,484.27
Therefore, the balance in the account after 3 years is approximately $25,484.27.
P = Po(1+r)^n.
r = (2%/365)/100% = 5.48*10^-5 = Daily %
rate expressed as a decimal.
n = 365Comp/yr * 3yrs = 1095 Compounding
periods.
Plug the above values into the given Eq
and solve for P.
Answer: P = $25,484.19