1. A 3,500.00 principal earns 3% interest, compounded semi annually. After 20 years, what is the balance in the account
A. 7,700.00
B. 4,713.99
C. 5,600.00
D. 6,349.06
2. A boat costs 92,000.00 and depreciates in value by 15% per year. How much will the boat be worth after 10 years?
A. 18,112.45
B. 78,200.00
C. 18,941.98
D. 69,000.00
If anyone could help that would be great.Thank you so much!
To solve these questions, we need to use the compound interest formula for the first question and the depreciation formula for the second question. Let's break down the steps for each question:
1. Calculating the balance in the account after 20 years with compound interest:
Formula: A = P(1 + r/n)^(nt)
Where:
A is the final amount after time t,
P is the principal (initial amount),
r is the annual interest rate (as a decimal),
n is the number of times interest is compounded per year, and
t is the number of years.
Plug in the given values:
P = $3,500.00
r = 3% (or 0.03 as a decimal)
n = 2 (compounded semiannually)
t = 20 years
Calculate:
A = $3,500.00 * (1 + 0.03/2)^(2*20)
The resulting balance in the account after 20 years is approximately $4,713.99. Therefore, the answer is B. 4,713.99.
2. Finding the boat's worth after 10 years with annual depreciation:
Formula: A = P(1 - r)^t
Where:
A is the final amount after time t,
P is the initial amount,
r is the annual depreciation rate (as a decimal),
and t is the number of years.
Plug in the given values:
P = $92,000.00
r = 15% (or 0.15 as a decimal)
t = 10 years
Calculate:
A = $92,000.00 * (1 - 0.15)^10
The resulting worth of the boat after 10 years is approximately $18,112.45. Therefore, the answer is A. 18,112.45.
Hope this helps!
Thank you so much! I really appreciate it!
18112.44519935 and that for the second one?
tia, you are correct. ( sry ik this is late )
thank you
#1: 3500(1+.035/2)^(2*20)
#2: 92000*0.85^10