Luke deposits 3,500 dollars into each of two savings accounts account 1 earns 3%annual simple interest account 2 earns 3% interest compounded annually luke does not make any additional deposits or withdrawals what is the sum of the balance of account 1 and 2 at the end of 4 years
To find the balance of account 1 after 4 years, we use the simple interest formula:
A = P(1 + rt)
Where:
A = Final balance
P = Principal (initial deposit)
r = Annual interest rate
t = Time (in years)
For account 1:
P1 = $3,500
r1 = 3% = 0.03
t1 = 4 years
A1 = 3500(1 + 0.03(4))
A1 = 3500(1 + 0.12)
A1 = 3500(1.12)
A1 = $3,920
For account 2, the interest is compounded annually, so we use the compound interest formula:
A = P(1 + r/n)^(nt)
Where:
A = Final balance
P = Principal (initial deposit)
r = Annual interest rate
n = Number of times the interest is compounded per year
t = Time (in years)
For account 2:
P2 = $3,500
r2 = 3% = 0.03
n2 = 1 (compounded annually)
t2 = 4 years
A2 = 3500(1 + 0.03/1)^(1*4)
A2 = 3500(1 + 0.03)^(4)
A2 = 3500(1.03)^(4)
A2 = 3500(1.125508754) [rounded to 9 decimal places]
A2 = $3,939.78
The sum of the balances of account 1 and 2 after 4 years is:
$3,920 + $3,939.78 = $7,859.78
Try again the answers are either 7,859.28$, 3,920.00$, 3,939.28$ 4,359.28$
Apologies for the previous response. Let's recalculate the balances using the correct formulas.
For account 1:
P1 = $3,500
r1 = 3% = 0.03
t1 = 4 years
A1 = P1(1 + rt1)
A1 = 3500(1 + 0.03*4)
A1 = 3500(1 + 0.12)
A1 = 3500(1.12)
A1 = $3,920
For account 2:
P2 = $3,500
r2 = 3% = 0.03
n2 = 1 (compounded annually)
t2 = 4 years
A2 = P2(1 + r2/n2)^(n2*t2)
A2 = 3500(1 + 0.03/1)^(1*4)
A2 = 3500(1 + 0.03)^(4)
A2 = 3500(1.03)^(4)
A2 ≈ 3500(1.1255) [rounded to 4 decimal places]
A2 ≈ $3,920.25
The sum of the balances of account 1 and 2 after 4 years is:
$3,920 + $3,920.25 = $7,840.25
Therefore, the correct answer is $7,840.25.