If Ron put $1000 of his money into a bank account that earns a 2.3% annual interest rate, how much will he earn after 10 years if the interest is compounded
A. yearly, B monthly? C. continuously?
If it is compounded n times per year, then after t years, the interest earned is
1000[(1+.023/n)^(n*t) - 1]
continuous interest is
1000(e^(.023t) - 1)
To calculate the amount that Ron will earn after 10 years, we need to use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
- A is the final amount
- P is the principal amount (initial deposit)
- r is the annual interest rate (as a decimal)
- n is the number of times the interest is compounded per year
- t is the number of years
Let's calculate the amount Ron will earn using this formula for the given scenarios:
A. Yearly Compounding:
In this case, the interest is compounded once a year, so n = 1.
Let's substitute the given values into the formula:
A = 1000(1 + 0.023/1)^(1*10)
A = 1000(1 + 0.023)^10
A = 1000(1.023)^10
A ≈ $1258.71
So, if the interest is compounded yearly, Ron will earn approximately $1258.71 after 10 years.
B. Monthly Compounding:
In this case, the interest is compounded 12 times a year, so n = 12.
Let's substitute the given values into the formula:
A = 1000(1 + 0.023/12)^(12*10)
A = 1000(1 + 0.0019)^120
A ≈ $1259.71
So, if the interest is compounded monthly, Ron will earn approximately $1259.71 after 10 years.
C. Continuously Compounding:
In this case, the interest is compounded continuously, which means n approaches infinity.
The formula for continuous compound interest is:
A = Pe^(rt)
Let's substitute the given values into the formula:
A = 1000e^(0.023*10)
A ≈ $1259.71
So, if the interest is compounded continuously, Ron will also earn approximately $1259.71 after 10 years.