Question

Parker invested $160 in an account paying an interest rate of 9, start fraction, 3, divided by, 8, end fraction9
8
3
​
% compounded continuously. Penelope invested $160 in an account paying an interest rate of 9, one quarter9
4
1
​
% compounded daily. After 6 years, how much more money would Parker have in his account than Penelope, to the nearest dollar?

Answer 1

To calculate the amounts in Parker's and Penelope's accounts after 6 years, we will use the formulas for continuous compounding and daily compounding.

### Parker's Investment:
Parker's investment is compounded continuously, and we use the formula:

\[
A = Pe^{rt}
\]

Where:
- \( A \) = the amount in the account after time \( t \)
- \( P \) = the principal amount (initial investment)
- \( r \) = the annual interest rate (as a decimal)
- \( t \) = the time in years
- \( e \) = Euler's number (approximately 2.71828)

For Parker:
- \( P = 160 \)
- \( r = 9.375\% = \frac{3}{8} = 0.09375 \)
- \( t = 6 \)

Plugging in the values:

\[
A = 160 e^{(0.09375 \times 6)}
\]

Calculating \( 0.09375 \times 6 \):

\[
0.09375 \times 6 = 0.5625
\]

Now substituting back into the formula:

\[
A = 160 e^{0.5625}
\]

Calculating \( e^{0.5625} \):

\[
e^{0.5625} \approx 1.7554
\]

Now calculating \( A \):

\[
A \approx 160 \times 1.7554 \approx 280.88
\]

Parker's total amount after 6 years is approximately **$281.**

### Penelope's Investment:
Penelope's investment is compounded daily, and we use the formula:

\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]

Where:
- \( n \) = the number of compounding periods per year (daily compounding \( n = 365 \))

For Penelope:
- \( P = 160 \)
- \( r = 9.25\% = 0.0925 \)
- \( n = 365 \)
- \( t = 6 \)

Now substituting into the formula:

\[
A = 160 \left(1 + \frac{0.0925}{365}\right)^{365 \times 6}
\]

Calculating \( \frac{0.0925}{365} \):

\[
\frac{0.0925}{365} \approx 0.000253425
\]

Now updating the formula:

\[
A = 160 \left(1 + 0.000253425\right)^{2190}
\]

Calculating \( (1 + 0.000253425)^{2190} \):

\[
(1.000253425)^{2190} \approx 1.230008
\]

Now calculating \( A \):

\[
A \approx 160 \times 1.230008 \approx 196.80
\]

Penelope's total amount after 6 years is approximately **$197.**

### Difference:
Now, we can calculate how much more Parker has than Penelope:

\[
\text{Difference} = A_P - A_{Penelope} = 281 - 197 = 84
\]

Thus, Parker has approximately **$84 more** than Penelope after 6 years.