Question
The formula Upper A equals Upper P left parenthesis 1 plus StartFraction r Over n EndFraction right parenthesis Superscript nt
describes the accumulated value, A, of a sum of money, P, the principal, after t years at annual percentage rate r (in decimal form) compounded n times a year. Complete the table for a savings account subject to n compounding periods per year.
Amount
Invested
Number of
Compounding Periods
Annual Interest
Rate
Accumulated
Amount
Time t
in Years
$13 comma 500
2
6.25%
$16 comma 000
?
Question content area bottom
Part 1
talmost equals
enter your response here years
(Do not round until the final answer. Then round to one decimal place as needed.)
describes the accumulated value, A, of a sum of money, P, the principal, after t years at annual percentage rate r (in decimal form) compounded n times a year. Complete the table for a savings account subject to n compounding periods per year.
Amount
Invested
Number of
Compounding Periods
Annual Interest
Rate
Accumulated
Amount
Time t
in Years
$13 comma 500
2
6.25%
$16 comma 000
?
Question content area bottom
Part 1
talmost equals
enter your response here years
(Do not round until the final answer. Then round to one decimal place as needed.)
Answers
GPT-4o mini
To complete the table and find the time \( t \) in years for the given values, we can rearrange the given formula \( A = P \left(1 + \frac{r}{n}\right)^{nt} \).
Given:
- \( A = 16,000 \)
- \( P = 13,500 \)
- \( r = 0.0625 \) (since 6.25% as a decimal is 0.0625)
- \( n = 2 \)
We need to solve for \( t \). The formula we need to work with is:
\[
16,000 = 13,500 \left(1 + \frac{0.0625}{2}\right)^{2t}
\]
First, calculate \( \frac{r}{n} \):
\[
\frac{0.0625}{2} = 0.03125
\]
So we can rewrite the equation as:
\[
16,000 = 13,500 \left(1 + 0.03125\right)^{2t}
\]
This simplifies to:
\[
16,000 = 13,500 (1.03125)^{2t}
\]
Next, divide both sides by 13,500:
\[
\frac{16,000}{13,500} = (1.03125)^{2t}
\]
Calculating the left side gives:
\[
\frac{16,000}{13,500} \approx 1.185185185
\]
Now, we have:
\[
1.185185185 = (1.03125)^{2t}
\]
Next, to solve for \( 2t \), we will take the logarithm of both sides:
\[
\log(1.185185185) = \log((1.03125)^{2t})
\]
Using the power rule of logarithms:
\[
\log(1.185185185) = 2t \cdot \log(1.03125)
\]
Now, divide both sides by \( \log(1.03125) \) to isolate \( 2t \):
\[
2t = \frac{\log(1.185185185)}{\log(1.03125)}
\]
Calculating each logarithm:
\[
\log(1.185185185) \approx 0.070585
\]
\[
\log(1.03125) \approx 0.013285
\]
Now plug these values into the equation:
\[
2t = \frac{0.070585}{0.013285} \approx 5.307
\]
Now, divide by 2 to find \( t \):
\[
t \approx \frac{5.307}{2} \approx 2.6535
\]
Rounding to one decimal place, we find:
\[
t \approx 2.7 \text{ years}
\]
So the accumulated amount after approximately **2.7 years** will reach $16,000 from an initial investment of $13,500 with an annual interest rate of 6.25% compounded twice a year.
Given:
- \( A = 16,000 \)
- \( P = 13,500 \)
- \( r = 0.0625 \) (since 6.25% as a decimal is 0.0625)
- \( n = 2 \)
We need to solve for \( t \). The formula we need to work with is:
\[
16,000 = 13,500 \left(1 + \frac{0.0625}{2}\right)^{2t}
\]
First, calculate \( \frac{r}{n} \):
\[
\frac{0.0625}{2} = 0.03125
\]
So we can rewrite the equation as:
\[
16,000 = 13,500 \left(1 + 0.03125\right)^{2t}
\]
This simplifies to:
\[
16,000 = 13,500 (1.03125)^{2t}
\]
Next, divide both sides by 13,500:
\[
\frac{16,000}{13,500} = (1.03125)^{2t}
\]
Calculating the left side gives:
\[
\frac{16,000}{13,500} \approx 1.185185185
\]
Now, we have:
\[
1.185185185 = (1.03125)^{2t}
\]
Next, to solve for \( 2t \), we will take the logarithm of both sides:
\[
\log(1.185185185) = \log((1.03125)^{2t})
\]
Using the power rule of logarithms:
\[
\log(1.185185185) = 2t \cdot \log(1.03125)
\]
Now, divide both sides by \( \log(1.03125) \) to isolate \( 2t \):
\[
2t = \frac{\log(1.185185185)}{\log(1.03125)}
\]
Calculating each logarithm:
\[
\log(1.185185185) \approx 0.070585
\]
\[
\log(1.03125) \approx 0.013285
\]
Now plug these values into the equation:
\[
2t = \frac{0.070585}{0.013285} \approx 5.307
\]
Now, divide by 2 to find \( t \):
\[
t \approx \frac{5.307}{2} \approx 2.6535
\]
Rounding to one decimal place, we find:
\[
t \approx 2.7 \text{ years}
\]
So the accumulated amount after approximately **2.7 years** will reach $16,000 from an initial investment of $13,500 with an annual interest rate of 6.25% compounded twice a year.