A website is offering a savings account that pays 33% compounded continuously. How much interest would a deposit of $3,000 earn over 15 years? (Round your answer to the nearest cent.)
I used f(t)=Pe^rt and got 1000e^-0.045t
I don't know what I did wrong.
Thank you.
I meant to say 3000e^ 0.33(15) and got $62593.57
don' know how you got that final answer.
3000 e^(.33x15)
= 3000 e^4.95
= 3000(141.1749639)
= $423524.89
Quickly, get me that webpage!!!!
To solve this problem, we need to use the formula for compound interest with continuous compounding:
A = P * e^(rt)
Where:
A = the amount of money accumulated after time t
P = the original principal (initial deposit)
r = annual interest rate (as a decimal)
t = time in years
In this case, the initial deposit (P) is $3,000, the interest rate (r) is 33% or 0.33 (as a decimal), and the time period (t) is 15 years. The goal is to determine the interest earned, which is the difference between the final amount accumulated and the initial deposit. So we need to calculate A and subtract the initial deposit P.
Using the formula, we'll have:
A = 3000 * e^(0.33 * 15)
Calculating this expression, we find:
A ≈ 3000 * e^(4.95)
A ≈ 3000 * 142.291
A ≈ 426,873.27
Therefore, the total amount accumulated after 15 years is approximately $426,873.27.
To calculate the interest earned, we subtract the initial deposit of $3,000:
Interest = A - P
Interest ≈ 426,873.27 - 3000
Interest ≈ 423,873.27
Thus, the interest earned on a deposit of $3,000 over 15 years at an interest rate of 33% compounded continuously would be approximately $423,873.27.