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To find the principal investment [tex]\( P \)[/tex] from the given information, we need to rearrange the compound interest formula and solve for [tex]\( P \)[/tex]. The given formula is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
where:
- [tex]\( A \)[/tex] is the final amount, which is $14,326.68.
- [tex]\( r \)[/tex] is the annual interest rate, which is 0.039 (3.9%).
- [tex]\( n \)[/tex] is the number of compounding periods per year, which is 12.
- [tex]\( t \)[/tex] is the number of years, which is 20.
First, let's determine the interest rate per compounding period and the number of compounding periods over the entire timeframe:
1. The interest rate per period:
[tex]\[ \frac{r}{n} = \frac{0.039}{12} \][/tex]
2. The number of compounding periods:
[tex]\[ nt = 12 \times 20 = 240 \][/tex]
We can now replace these values in the formula to isolate [tex]\( P \)[/tex]:
[tex]\[ P = \frac{A}{\left(1 + \frac{0.039}{12}\right)^{240}} \][/tex]
Next, we calculate the expression in the denominator:
1. Calculate [tex]\( \frac{0.039}{12} \)[/tex]:
[tex]\[ \frac{0.039}{12} = 0.00325 \][/tex]
2. Add 1 to the result:
[tex]\[ 1 + 0.00325 = 1.00325 \][/tex]
3. Raise the sum to the power of 240:
[tex]\[ 1.00325^{240} \][/tex]
Finally, divide [tex]\( A \)[/tex] by this result to find the principal [tex]\( P \)[/tex]:
[tex]\[ P = \frac{14,326.68}{1.00325^{240}} \][/tex]
When you perform the above calculation, you find that:
[tex]\[ P \approx 6,575.75 \][/tex]
Therefore, the value of the principal investment, rounded to the nearest hundredths place, is [tex]\( \boxed{6,575.75} \)[/tex]. This matches the first answer option given.
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
where:
- [tex]\( A \)[/tex] is the final amount, which is $14,326.68.
- [tex]\( r \)[/tex] is the annual interest rate, which is 0.039 (3.9%).
- [tex]\( n \)[/tex] is the number of compounding periods per year, which is 12.
- [tex]\( t \)[/tex] is the number of years, which is 20.
First, let's determine the interest rate per compounding period and the number of compounding periods over the entire timeframe:
1. The interest rate per period:
[tex]\[ \frac{r}{n} = \frac{0.039}{12} \][/tex]
2. The number of compounding periods:
[tex]\[ nt = 12 \times 20 = 240 \][/tex]
We can now replace these values in the formula to isolate [tex]\( P \)[/tex]:
[tex]\[ P = \frac{A}{\left(1 + \frac{0.039}{12}\right)^{240}} \][/tex]
Next, we calculate the expression in the denominator:
1. Calculate [tex]\( \frac{0.039}{12} \)[/tex]:
[tex]\[ \frac{0.039}{12} = 0.00325 \][/tex]
2. Add 1 to the result:
[tex]\[ 1 + 0.00325 = 1.00325 \][/tex]
3. Raise the sum to the power of 240:
[tex]\[ 1.00325^{240} \][/tex]
Finally, divide [tex]\( A \)[/tex] by this result to find the principal [tex]\( P \)[/tex]:
[tex]\[ P = \frac{14,326.68}{1.00325^{240}} \][/tex]
When you perform the above calculation, you find that:
[tex]\[ P \approx 6,575.75 \][/tex]
Therefore, the value of the principal investment, rounded to the nearest hundredths place, is [tex]\( \boxed{6,575.75} \)[/tex]. This matches the first answer option given.
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