Discover new information and insights with the help of IDNLearn.com. Ask your questions and receive accurate, in-depth answers from our knowledgeable community members.
Sagot :
Alright, let's solve each part step-by-step.
### Part (a)
First, we derive the formula for the investment growth. The problem states that the initial investment is [tex]\(\$ 88,000\)[/tex] with an annual interest rate of [tex]\(4.5\%\)[/tex], compounded quarterly.
1. Quarterly interest rate:
The annual interest rate is [tex]\(4.5\%\)[/tex] which is equivalent to [tex]\(0.045\)[/tex] as a decimal. Since the interest is compounded quarterly, we divide this annual rate by 4 (the number of quarters in a year) to get the quarterly interest rate:
[tex]\[ \text{Quarterly interest rate} = \frac{0.045}{4} = 0.01125 \][/tex]
2. Growth factor per quarter:
The growth factor per quarter is then:
[tex]\[ 1 + 0.01125 = 1.01125 \][/tex]
3. Number of compounding periods in [tex]\(t\)[/tex] years:
If [tex]\(t\)[/tex] is the number of years, then the total number of compounding periods in [tex]\(t\)[/tex] years is [tex]\(4 \cdot t\)[/tex], because interest is compounded quarterly.
4. Formulating the investment function:
Using the compound interest formula [tex]\(A(t) = P \left(1 + \frac{r}{n}\right)^{nt}\)[/tex], we substitute [tex]\(P = 88,000\)[/tex], the quarterly growth factor [tex]\(1.01125\)[/tex], and number of periods [tex]\(4t\)[/tex]:
[tex]\[ A(t) = 88,000 \times (1.01125)^{4t} \][/tex]
Thus, the function for the amount to which the investment grows after [tex]\(t\)[/tex] years is:
[tex]\[ A(t) = 88,000 \times (1.01125)^{4t} \][/tex]
### Part (b)
Next, let's find the amount of money in the account at specific points in time: [tex]\(t = 0\)[/tex], [tex]\(t = 4\)[/tex], [tex]\(t = 5\)[/tex], and [tex]\(t = 10\)[/tex] years. We will use the function derived above and evaluate it at these values of [tex]\(t\)[/tex].
1. At [tex]\(t = 0\)[/tex]:
[tex]\[ A(0) = 88,000 \times (1.01125)^{4 \times 0} = 88,000 \times (1.01125)^0 = 88,000 \times 1 = 88,000 \][/tex]
The amount of money in the account at [tex]\(t = 0\)[/tex] years is [tex]\(\$88,000\)[/tex].
2. At [tex]\(t = 4\)[/tex]:
[tex]\[ A(4) = 88,000 \times (1.01125)^{4 \times 4} = 88,000 \times (1.01125)^{16} \][/tex]
This evaluates to approximately [tex]\(105,249\)[/tex]. The amount of money in the account at [tex]\(t = 4\)[/tex] years is [tex]\(\$105,249\)[/tex].
3. At [tex]\(t = 5\)[/tex]:
[tex]\[ A(5) = 88,000 \times (1.01125)^{4 \times 5} = 88,000 \times (1.01125)^{20} \][/tex]
This evaluates to approximately [tex]\(110,066\)[/tex]. The amount of money in the account at [tex]\(t = 5\)[/tex] years is [tex]\(\$110,066\)[/tex].
4. At [tex]\(t = 10\)[/tex]:
[tex]\[ A(10) = 88,000 \times (1.01125)^{4 \times 10} = 88,000 \times (1.01125)^{40} \][/tex]
This evaluates to approximately [tex]\(137,665\)[/tex]. The amount of money in the account at [tex]\(t = 10\)[/tex] years is [tex]\(\$137,665\)[/tex].
### Summary
Part (a): The function for the amount to which the investment grows after [tex]\(t\)[/tex] years is:
[tex]\[ A(t) = 88,000 \times (1.01125)^{4t} \][/tex]
Part (b): The amounts of money in the account at various times are:
- At [tex]\(t = 0\)[/tex] years: [tex]\(\$88,000\)[/tex]
- At [tex]\(t = 4\)[/tex] years: [tex]\(\$105,249\)[/tex]
- At [tex]\(t = 5\)[/tex] years: [tex]\(\$110,066\)[/tex]
- At [tex]\(t = 10\)[/tex] years: [tex]\(\$137,665\)[/tex]
### Part (a)
First, we derive the formula for the investment growth. The problem states that the initial investment is [tex]\(\$ 88,000\)[/tex] with an annual interest rate of [tex]\(4.5\%\)[/tex], compounded quarterly.
1. Quarterly interest rate:
The annual interest rate is [tex]\(4.5\%\)[/tex] which is equivalent to [tex]\(0.045\)[/tex] as a decimal. Since the interest is compounded quarterly, we divide this annual rate by 4 (the number of quarters in a year) to get the quarterly interest rate:
[tex]\[ \text{Quarterly interest rate} = \frac{0.045}{4} = 0.01125 \][/tex]
2. Growth factor per quarter:
The growth factor per quarter is then:
[tex]\[ 1 + 0.01125 = 1.01125 \][/tex]
3. Number of compounding periods in [tex]\(t\)[/tex] years:
If [tex]\(t\)[/tex] is the number of years, then the total number of compounding periods in [tex]\(t\)[/tex] years is [tex]\(4 \cdot t\)[/tex], because interest is compounded quarterly.
4. Formulating the investment function:
Using the compound interest formula [tex]\(A(t) = P \left(1 + \frac{r}{n}\right)^{nt}\)[/tex], we substitute [tex]\(P = 88,000\)[/tex], the quarterly growth factor [tex]\(1.01125\)[/tex], and number of periods [tex]\(4t\)[/tex]:
[tex]\[ A(t) = 88,000 \times (1.01125)^{4t} \][/tex]
Thus, the function for the amount to which the investment grows after [tex]\(t\)[/tex] years is:
[tex]\[ A(t) = 88,000 \times (1.01125)^{4t} \][/tex]
### Part (b)
Next, let's find the amount of money in the account at specific points in time: [tex]\(t = 0\)[/tex], [tex]\(t = 4\)[/tex], [tex]\(t = 5\)[/tex], and [tex]\(t = 10\)[/tex] years. We will use the function derived above and evaluate it at these values of [tex]\(t\)[/tex].
1. At [tex]\(t = 0\)[/tex]:
[tex]\[ A(0) = 88,000 \times (1.01125)^{4 \times 0} = 88,000 \times (1.01125)^0 = 88,000 \times 1 = 88,000 \][/tex]
The amount of money in the account at [tex]\(t = 0\)[/tex] years is [tex]\(\$88,000\)[/tex].
2. At [tex]\(t = 4\)[/tex]:
[tex]\[ A(4) = 88,000 \times (1.01125)^{4 \times 4} = 88,000 \times (1.01125)^{16} \][/tex]
This evaluates to approximately [tex]\(105,249\)[/tex]. The amount of money in the account at [tex]\(t = 4\)[/tex] years is [tex]\(\$105,249\)[/tex].
3. At [tex]\(t = 5\)[/tex]:
[tex]\[ A(5) = 88,000 \times (1.01125)^{4 \times 5} = 88,000 \times (1.01125)^{20} \][/tex]
This evaluates to approximately [tex]\(110,066\)[/tex]. The amount of money in the account at [tex]\(t = 5\)[/tex] years is [tex]\(\$110,066\)[/tex].
4. At [tex]\(t = 10\)[/tex]:
[tex]\[ A(10) = 88,000 \times (1.01125)^{4 \times 10} = 88,000 \times (1.01125)^{40} \][/tex]
This evaluates to approximately [tex]\(137,665\)[/tex]. The amount of money in the account at [tex]\(t = 10\)[/tex] years is [tex]\(\$137,665\)[/tex].
### Summary
Part (a): The function for the amount to which the investment grows after [tex]\(t\)[/tex] years is:
[tex]\[ A(t) = 88,000 \times (1.01125)^{4t} \][/tex]
Part (b): The amounts of money in the account at various times are:
- At [tex]\(t = 0\)[/tex] years: [tex]\(\$88,000\)[/tex]
- At [tex]\(t = 4\)[/tex] years: [tex]\(\$105,249\)[/tex]
- At [tex]\(t = 5\)[/tex] years: [tex]\(\$110,066\)[/tex]
- At [tex]\(t = 10\)[/tex] years: [tex]\(\$137,665\)[/tex]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Your questions deserve accurate answers. Thank you for visiting IDNLearn.com, and see you again for more solutions.