To answer these questions, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = The amount of money in the bank after t years
P = The principal amount (initial investment)
r = The annual interest rate (as a decimal)
n = The number of times interest is compounded per year
t = The number of years
Now, let's calculate the amount in the bank for each case:
a) Compounded Annually
For this case, n = 1 (compounded once per year).
So, A = 8000(1 + 0.10/1)^(1*10)
A = 8000(1.10)^10
A ≈ $21,589.89
b) Compounded Quarterly
For this case, n = 4 (compounded four times per year).
So, A = 8000(1 + 0.10/4)^(4*10)
A = 8000(1.025)^40
A ≈ $22,093.54
c) Compounded Monthly
For this case, n = 12 (compounded twelve times per year).
So, A = 8000(1 + 0.10/12)^(12*10)
A = 8000(1.0083)^120
A ≈ $22,184.82
d) Compounded Continuously
For this case, n approaches infinity, but we can approximate it using the formula:
A = Pe^(rt)
Where e is the base of the natural logarithm (approximately 2.71828).
So, A = 8000e^(0.10*10)
A = 8000e^1.0
A ≈ $22,163.53
Therefore, the amount in the bank after 10 years would be approximately:
a) $21,589.89 (compounded annually)
b) $22,093.54 (compounded quarterly)
c) $22,184.82 (compounded monthly)
d) $22,163.53 (compounded continuously)