I will do the 2nd part
i = .10/4 = .025
n = 11*4 = 44
amount = 8000(1.025)^44 = 23710.46
do #1, and #3 the same way
#4:
amount = 8000(e^(.1t))
= 8000(e^(.10*11) ) = 24033.33
Find the amount in the bank after 11 years if interest is compounded quaterly ____
Find the amount in the bank after 11 years if interest is compounded monthly ____
Finally, find the amount in the bank after 11 years if interest is compounded continuously
i = .10/4 = .025
n = 11*4 = 44
amount = 8000(1.025)^44 = 23710.46
do #1, and #3 the same way
#4:
amount = 8000(e^(.1t))
= 8000(e^(.10*11) ) = 24033.33
where i is the periodic interest rate as a decimal
and n is the number of interest periods.
in my example, the rate was 10% , compounded quarterly, so I found
i = .10/4 = .025
and in 11 years, for quarterly compounding, there are 11*4 or 44 periods, so n = 44
in #1, i=10, n=11
amount = 8000(1.10)^11 = ...
in #3, i = .10/12 = .00833333
n = 11*12 = 132
A = P(1 + r/n)^(nt)
Where:
A = the final amount
P = the principal amount (initial investment)
r = annual interest rate (expressed as a decimal)
n = number of times interest is compounded per year
t = number of years
For the given question, we have:
P = $8000
r = 10% = 0.10
t = 11 years
1. For annual compounding:
n = 1
Plugging the values into the formula:
A = 8000(1 + 0.10/1)^(1 * 11)
A ≈ $21,391.47
2. For quarterly compounding:
n = 4
Plugging the values into the formula:
A = 8000(1 + 0.10/4)^(4 * 11)
A ≈ $21,462.60
3. For monthly compounding:
n = 12
Plugging the values into the formula:
A = 8000(1 + 0.10/12)^(12 * 11)
A ≈ $21,493.08
4. For continuous compounding:
n → ∞ (as n approaches infinity)
Using the continuous compound interest formula:
A = Pe^(rt)
Plugging the values into the formula:
A = 8000 * e^(0.10 * 11)
A ≈ $22,204.71
Therefore, the amount in the bank after 11 years will be approximately:
- $21,391.47 with annual compounding
- $21,462.60 with quarterly compounding
- $21,493.08 with monthly compounding
- $22,204.71 with continuous compounding