Four different accounts are described below. Order the accounts according to their values after 20 years, from greatest to least.
1.You deposit 1500$ in an account that earns 5% annual interest compounded quarterly.
2. You deposit 1500$ in an account that earns 6% annual interest compounded monthly
3. You deposit 1400$ in an account that earns 9% annual interest compounded daily
4. You deposit 1000$ in an account that earns 10% annual interest compounded monthly
I'll do the 2nd, you do the others.
2.
i = .06/12 = .005
n = 20*12 = 240
amount = 1000(1 + .005)^240
= 1000(1.005)^240
= 1000(3.310204...)
= $ 3310.20
No incorrect
1. v = 1500 [1 + (.05 / 4)]^(20 * 4)
2. v = 1500 [1 + (.06 / 12)]^(20 * 12)
3. v = 1400 [1 + (.09 / 365)]^(20 * 365)
4. v = 1000 [1 + (.10 / 12)]^(20 * 12)
To determine the values of the four accounts after 20 years, it's helpful to use the compound interest formula:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the account
P = the principal amount (initial deposit)
r = the annual interest rate (in decimal form)
n = the number of times interest is compounded per year
t = the number of years
Let's calculate the future values for each account and then order them accordingly:
1. Account 1:
P = $1500
r = 5% = 0.05
n = 4 (compounded quarterly)
t = 20
Using the formula: A = 1500(1 + 0.05/4)^(4*20) ≈ $4,288.68
2. Account 2:
P = $1500
r = 6% = 0.06
n = 12 (compounded monthly)
t = 20
Using the formula: A = 1500(1 + 0.06/12)^(12*20) ≈ $4,556.18
3. Account 3:
P = $1400
r = 9% = 0.09
n = 365 (compounded daily)
t = 20
Using the formula: A = 1400(1 + 0.09/365)^(365*20) ≈ $5,710.39
4. Account 4:
P = $1000
r = 10% = 0.10
n = 12 (compounded monthly)
t = 20
Using the formula: A = 1000(1 + 0.10/12)^(12*20) ≈ $6,727.50
Now, let's arrange the accounts from greatest to least value after 20 years:
1. Account 4 - $6,727.50
2. Account 3 - $5,710.39
3. Account 2 - $4,556.18
4. Account 1 - $4,288.68
Therefore, the accounts in descending order of their values after 20 years are: Account 4, Account 3, Account 2, Account 1.