Let's solve this step by step:
### Part (A)
We need to determine the following values from the problem:
- [tex]\( P \)[/tex]: the initial deposit.
- [tex]\( r \)[/tex]: the annual percentage rate written as a decimal.
- [tex]\( n \)[/tex]: the number of times the interest is compounded per year.
Given in the problem:
- Raquel deposits [tex]\( \$ 10,000 \)[/tex], so [tex]\( P = 10000 \)[/tex].
- The annual percentage rate is [tex]\( 5\% \)[/tex]. As a decimal, [tex]\( r = 0.05 \)[/tex].
- The interest is compounded quarterly, which means 4 times a year. So, [tex]\( n = 4 \)[/tex].
So, we have:
- [tex]\( P = 10000 \)[/tex]
- [tex]\( r = 0.05 \)[/tex]
- [tex]\( n = 4 \)[/tex]
### Part (B)
We need to calculate how much money Raquel will have in the account in 8 years using the formula [tex]\( A = P \left( 1 + \frac{r}{n} \right)^{nt} \)[/tex].
Substituting the known values:
- [tex]\( P = 10000 \)[/tex]
- [tex]\( r = 0.05 \)[/tex]
- [tex]\( n = 4 \)[/tex]
- [tex]\( t = 8 \)[/tex] (since we are calculating for 8 years)
Applying these into the formula:
[tex]\[ A = 10000 \left( 1 + \frac{0.05}{4} \right)^{4 \cdot 8} \][/tex]
Now calculating the values step-by-step:
1. [tex]\( \frac{r}{n} = \frac{0.05}{4} = 0.0125 \)[/tex]
2. [tex]\( 1 + \frac{r}{n} = 1 + 0.0125 = 1.0125 \)[/tex]
3. [tex]\( n \cdot t = 4 \cdot 8 = 32 \)[/tex]
4. [tex]\( (1.0125)^{32} \)[/tex]
After computing:
[tex]\[ A = 10000 \times (1.0125)^{32} \approx 14881.305085948254 \][/tex]
Rounding to the nearest penny:
[tex]\[ A \approx 14881.31 \][/tex]
So, the amount of money Raquel will have in the account in 8 years is:
[tex]\[ \boxed{14881.31} \][/tex]
### Summary
(A) The values are:
- [tex]\( P = 10000 \)[/tex]
- [tex]\( r = 0.05 \)[/tex]
- [tex]\( n = 4 \)[/tex]
(B) The amount Raquel will have in the account in 8 years is:
[tex]\[ \boxed{14881.31} \][/tex]
