Answer :
Sure! Let's solve this step-by-step using the compound interest formula.
### Compound Interest Formula
The compound interest formula is given by:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Where:
- [tex]\(A\)[/tex] is the amount of money accumulated after [tex]\(t\)[/tex] years, including interest.
- [tex]\(P\)[/tex] is the principal amount (the initial amount of money).
- [tex]\(r\)[/tex] is the annual interest rate (decimal).
- [tex]\(n\)[/tex] is the number of times that interest is compounded per year.
- [tex]\(t\)[/tex] is the time the money is invested for in years.
### Given Values
- Principal, [tex]\(P = \$40,000\)[/tex]
- Annual interest rate, [tex]\(r = 23.24\% = 0.2324\)[/tex]
- Number of times compounded per year, [tex]\(n = 12\)[/tex]
- Time period, [tex]\(t = 2014 - 1981 = 33\)[/tex] years
### Step-by-Step Calculation
1. Convert the interest rate from a percentage to a decimal:
[tex]\( r = 23.24\% \rightarrow 0.2324 \)[/tex]
2. Determine the number of years the money is invested for:
[tex]\( t = 2014 - 1981 = 33 \)[/tex]
3. Substitute the values into the compound interest formula:
[tex]\[ A = 40000 \left(1 + \frac{0.2324}{12}\right)^{12 \times 33} \][/tex]
4. Calculate the interest rate per period:
[tex]\[ \frac{0.2324}{12} \approx 0.0193667 \][/tex]
5. Calculate the total number of compounding periods:
[tex]\[ 12 \times 33 = 396 \][/tex]
6. Combine the components inside the parentheses:
[tex]\[ 1 + 0.0193667 \approx 1.0193667 \][/tex]
7. Raise the base to the power of the number of compounding periods:
[tex]\[ (1.0193667)^{396} \][/tex]
8. Finally, multiply by the principal amount [tex]\(P\)[/tex]:
[tex]\[ A = 40000 \times (1.0193667)^{396} \][/tex]
### Final Answer
After carrying out the exponentiation and multiplication, the accumulated amount [tex]\(A\)[/tex] is given as:
[tex]\[ A \approx 79,599,344.97 \][/tex]
Thus, if you had invested [tex]$40,000 on August 8, 1981, at an interest rate of 23.24% compounded monthly, by August 8, 2014, you would have approximately \$[/tex]79,599,344.97.
### Compound Interest Formula
The compound interest formula is given by:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Where:
- [tex]\(A\)[/tex] is the amount of money accumulated after [tex]\(t\)[/tex] years, including interest.
- [tex]\(P\)[/tex] is the principal amount (the initial amount of money).
- [tex]\(r\)[/tex] is the annual interest rate (decimal).
- [tex]\(n\)[/tex] is the number of times that interest is compounded per year.
- [tex]\(t\)[/tex] is the time the money is invested for in years.
### Given Values
- Principal, [tex]\(P = \$40,000\)[/tex]
- Annual interest rate, [tex]\(r = 23.24\% = 0.2324\)[/tex]
- Number of times compounded per year, [tex]\(n = 12\)[/tex]
- Time period, [tex]\(t = 2014 - 1981 = 33\)[/tex] years
### Step-by-Step Calculation
1. Convert the interest rate from a percentage to a decimal:
[tex]\( r = 23.24\% \rightarrow 0.2324 \)[/tex]
2. Determine the number of years the money is invested for:
[tex]\( t = 2014 - 1981 = 33 \)[/tex]
3. Substitute the values into the compound interest formula:
[tex]\[ A = 40000 \left(1 + \frac{0.2324}{12}\right)^{12 \times 33} \][/tex]
4. Calculate the interest rate per period:
[tex]\[ \frac{0.2324}{12} \approx 0.0193667 \][/tex]
5. Calculate the total number of compounding periods:
[tex]\[ 12 \times 33 = 396 \][/tex]
6. Combine the components inside the parentheses:
[tex]\[ 1 + 0.0193667 \approx 1.0193667 \][/tex]
7. Raise the base to the power of the number of compounding periods:
[tex]\[ (1.0193667)^{396} \][/tex]
8. Finally, multiply by the principal amount [tex]\(P\)[/tex]:
[tex]\[ A = 40000 \times (1.0193667)^{396} \][/tex]
### Final Answer
After carrying out the exponentiation and multiplication, the accumulated amount [tex]\(A\)[/tex] is given as:
[tex]\[ A \approx 79,599,344.97 \][/tex]
Thus, if you had invested [tex]$40,000 on August 8, 1981, at an interest rate of 23.24% compounded monthly, by August 8, 2014, you would have approximately \$[/tex]79,599,344.97.