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Answer :
Sure! Let's take a look at each part of the problem and solve it step by step, as if we're calculating the values ourselves.
1. Expression 1: [tex]\( A(30) = 10000 \times e^{0.08 \times 30} \)[/tex]
In this expression, you're calculating a future amount using continuous compounding interest. We use the formula:
[tex]\[
A(30) = 10000 \times e^{0.08 \times 30}
\][/tex]
After computing this, we find that:
[tex]\[
A(30) \approx 110231.76
\][/tex]
2. Expression 2: [tex]\( 10000 = P \times e^{0.08 \times 30} \)[/tex]
Here, you're finding the present value [tex]\( P \)[/tex] needed to reach 10000 after continuous compounding. You rearrange the formula to solve for [tex]\( P \)[/tex]:
[tex]\[
P = \frac{10000}{e^{0.08 \times 30}}
\][/tex]
Once calculated:
[tex]\[
P \approx 907.18
\][/tex]
3. Expression 3: [tex]\( A(30) = 10000 \times \left(1 + \frac{0.08}{4}\right)^{4 \times 30} \)[/tex]
This expression is calculating the amount using quarterly compounding. The formula used here is:
[tex]\[
A(30) = 10000 \times \left(1 + \frac{0.08}{4}\right)^{120}
\][/tex]
After calculating, the result is approximately:
[tex]\[
A(30) \approx 107651.63
\][/tex]
4. Expression 4: [tex]\( A(30) = 10000 \times 0.08 \times 30 \)[/tex]
This expression is a straightforward calculation of simple interest earned over a certain time period:
[tex]\[
A(30) = 10000 \times 0.08 \times 30
\][/tex]
This computes to:
[tex]\[
A(30) = 24000.00
\][/tex]
These detailed calculations provide the results for each part as shown.
1. Expression 1: [tex]\( A(30) = 10000 \times e^{0.08 \times 30} \)[/tex]
In this expression, you're calculating a future amount using continuous compounding interest. We use the formula:
[tex]\[
A(30) = 10000 \times e^{0.08 \times 30}
\][/tex]
After computing this, we find that:
[tex]\[
A(30) \approx 110231.76
\][/tex]
2. Expression 2: [tex]\( 10000 = P \times e^{0.08 \times 30} \)[/tex]
Here, you're finding the present value [tex]\( P \)[/tex] needed to reach 10000 after continuous compounding. You rearrange the formula to solve for [tex]\( P \)[/tex]:
[tex]\[
P = \frac{10000}{e^{0.08 \times 30}}
\][/tex]
Once calculated:
[tex]\[
P \approx 907.18
\][/tex]
3. Expression 3: [tex]\( A(30) = 10000 \times \left(1 + \frac{0.08}{4}\right)^{4 \times 30} \)[/tex]
This expression is calculating the amount using quarterly compounding. The formula used here is:
[tex]\[
A(30) = 10000 \times \left(1 + \frac{0.08}{4}\right)^{120}
\][/tex]
After calculating, the result is approximately:
[tex]\[
A(30) \approx 107651.63
\][/tex]
4. Expression 4: [tex]\( A(30) = 10000 \times 0.08 \times 30 \)[/tex]
This expression is a straightforward calculation of simple interest earned over a certain time period:
[tex]\[
A(30) = 10000 \times 0.08 \times 30
\][/tex]
This computes to:
[tex]\[
A(30) = 24000.00
\][/tex]
These detailed calculations provide the results for each part as shown.
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