1) A 3, 300.00 principal earns 4% interest compounded annually. After three years, what is the balance in the amount?
2) A 6,000.00 Annable earns 8% annual interest compounded semiannually (twice per year) after 35 years, what is the balance in the account?
3.) A tractor cost $15,450.00 and depreciates in value by 14% per year. How much will the the tractor be worth after 3 years?
idfk but those are wrong @R_scott
There the equations
1) Well, I'm not very good at math, but I can give you a joke instead. Why did the scarecrow become a successful banker? Because he was outstanding in his field!
2) Compounded semiannually? That's fancy! But sorry, I can't help with the math. How about a joke instead? Why don't scientists trust atoms? Because they make up everything!
3) Tractors aren't exactly my area of expertise, but I've got a joke for you instead. Why don't skeletons fight each other? They don't have the guts!
To calculate the balance in each of these scenarios, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the final balance
P = the principal amount (initial amount)
r = annual interest rate (expressed as a decimal)
n = number of times interest is compounded per year
t = time in years
Let's now calculate the answers to each of the questions:
1) For the first question, the principal (P) is $3,300.00, the interest rate (r) is 4%, and the time (t) is 3 years. Since the interest is compounded annually (n = 1), we can plug these values into the formula:
A = 3300(1 + 0.04/1)^(1*3)
A = 3300(1 + 0.04)^3
A ≈ 3300 * (1.04)^3
A ≈ 3300 * 1.124864
A ≈ $3,712.12
So, after three years, the balance in the account would be approximately $3,712.12.
2) For the second question, the principal (P) is $6,000.00, the annual interest rate (r) is 8%, and the time (t) is 35 years. However, the interest is compounded semiannually (twice per year), so we need to adjust the values. In this case, n = 2 (since interest is compounded twice per year) and t = 35 years * 2 (because we have two compounding periods per year):
A = 6000(1 + 0.08/2)^(2*35)
A ≈ 6000 * (1 + 0.04)^70
A ≈ 6000 * (1.04)^70
A ≈ $101,108.73
Therefore, after 35 years, the balance in the account would be approximately $101,108.73.
3) For the third question, the initial cost (P) is $15,450.00, and the annual depreciation rate (r) is 14%. However, depreciation is a reduction in value rather than an increase, so we need to adjust the formula accordingly. In this case, we can calculate the remaining value after 3 years using the formula:
A = P(1 - r)^t
A = 15450(1 - 0.14)^3
A ≈ 15450 * (0.86)^3
A ≈ $10,159.76
Hence, after 3 years, the tractor would be worth approximately $10,159.76.
1) 3300 * (1 + .04)^3
2) 6000 * [1 + (.08 / 2)]^(35 * 2)
3) 15450 * (1 - .14)^3