Sure! Let's start by defining the functions for each account's future value after [tex]\( t \)[/tex] years using the given parameters.
### Abby's Account (Compounded Quarterly)
Abby's account pays an annual interest rate of [tex]\( 4.2\% \)[/tex] compounded quarterly. The formula for compound interest is given by:
[tex]\[ A(t) = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
where:
- [tex]\( P \)[/tex] is the principal amount (\[tex]$8000),
- \( r \) is the annual interest rate as a decimal (0.042),
- \( n \) is the number of times the interest is compounded per year (4, since it's quarterly),
- \( t \) is the time in years.
So, the function \( A(t) \) for Abby's account is:
\[ A(t) = 8000 \left(1 + \frac{0.042}{4}\right)^{4t} \]
### Brett's Account (Compounded Continuously)
Brett's account pays an annual interest rate of \( 3.9\% \) compounded continuously. The formula for continuous compound interest is given by:
\[ B(t) = P e^{rt} \]
where:
- \( P \) is the principal amount (\$[/tex]8000),
- [tex]\( r \)[/tex] is the annual interest rate as a decimal (0.039),
- [tex]\( t \)[/tex] is the time in years,
- [tex]\( e \)[/tex] is the base of the natural logarithm (approximately 2.71828).
So, the function [tex]\( B(t) \)[/tex] for Brett's account is:
[tex]\[ B(t) = 8000 e^{0.039t} \][/tex]
### Example Calculation for [tex]\( t = 1 \)[/tex] Year
Let's calculate the value of each account after [tex]\( 1 \)[/tex] year.
For Abby's Account:
[tex]\[ A(1) = 8000 \left(1 + \frac{0.042}{4}\right)^{4 \times 1} \][/tex]
[tex]\[ A(1) \approx 8000 \left(1 + 0.0105\right)^4 \][/tex]
[tex]\[ A(1) \approx 8000 \left(1.0105\right)^4 \][/tex]
[tex]\[ A(1) \approx 8341.33 \][/tex] (rounded to two decimal places)
For Brett's Account:
[tex]\[ B(1) = 8000 e^{0.039 \times 1} \][/tex]
[tex]\[ B(1) \approx 8000 e^{0.039} \][/tex]
[tex]\[ B(1) \approx 8000 \times 1.0398 \][/tex]
[tex]\[ B(1) \approx 8318.16 \][/tex] (rounded to two decimal places)
Therefore, after 1 year, Abby's account will have approximately \[tex]$8341.33 and Brett's account will have approximately \$[/tex]8318.16.
