Question

now lets compare interest compounded continuously to simple interest. suppose that \\$75,000 is invested in two accounts: one earns 3\\% simple interest, the other earns 3\\% compounded continuously. fill in the following tables with the future value of each account after each term. use the formula \\(a = pe^{rt}\\) for interest compounded continuously. round your answers to the nearest cent.

\

$$\begin{tabular}{|c|c|c|c|} \\hline \\textbf{years} & 5 & 10 & 15 \\\\ \\hline \\textbf{3\\% simple interest} & \\$ & \\$ & \\$ \\\\ \\hline \\textbf{3\\% compounded continuously} & \\$ & \\$ & \\$ \\\\ \\hline \\end{tabular}$$

\

$$\begin{tabular}{|c|c|c|c|} \\hline \\textbf{years} & 20 & 25 & 30 \\\\ \\hline \\textbf{3\\% simple interest} & \\$ & \\$ & \\$ \\\\ \\hline \\textbf{3\\% compounded continuously} & \\$ & \\$ & \\$ \\\\ \\hline \\end{tabular}$$

Explanation:

<pre_analysis>
{
"quality": "clear",
"question_count": 1,
"skills_matched": [
"step_cot",
"explain_then_answer",
"tutor"
],
"needs_drawing": false,
"concepts_used": [
"Compound Interest",
"Future Value"
],
"new_concepts": [
"Continuous Compounding",
"Simple Interest"
],
"current_concepts": [
"Simple Interest",
"Continuous Compounding",
"Compound Interest",
"Future Value"
]
}
</pre_analysis>

<reasoning>

Identify given parameters

We have principal \(P = 75,000\) and annual interest rate \(r = 0.03\).
We need to calculate the Future Value for both accounts at \(t = 5, 10, 15, 20, 25, 30\) years.

Calculate simple interest values

Using the Future Value formula for simple interest:
\[A = P(1 + rt)\]

• For \(t = 5\): \(A = 75,000(1 + 0.03 \times 5) = 75,000(1.15) = 86,250.00\)
• For \(t = 10\): \(A = 75,000(1 + 0.03 \times 10) = 75,000(1.30) = 97,500.00\)
• For \(t = 15\): \(A = 75,000(1 + 0.03 \times 15) = 75,000(1.45) = 108,750.00\)
• For \(t = 20\): \(A = 75,000(1 + 0.03 \times 20) = 75,000(1.60) = 120,000.00\)
• For \(t = 25\): \(A = 75,000(1 + 0.03 \times 25) = 75,000(1.75) = 131,250.00\)
• For \(t = 30\): \(A = 75,000(1 + 0.03 \times 30) = 75,000(1.90) = 142,500.00\)

Calculate continuous compounding values

Using the Compound Interest formula for continuous compounding:
\[A = P e^{rt}\]

• For \(t = 5\): \(A = 75,000 e^{0.03 \times 5} = 75,000 e^{0.15} \approx 75,000 \times 1.16183424 = 87,137.57\)
• For \(t = 10\): \(A = 75,000 e^{0.03 \times 10} = 75,000 e^{0.30} \approx 75,000 \times 1.34985881 = 101,239.41\)
• For \(t = 15\): \(A = 75,000 e^{0.03 \times 15} = 75,000 e^{0.45} \approx 75,000 \times 1.56831219 = 117,623.41\)
• For \(t = 20\): \(A = 75,000 e^{0.03 \times 20} = 75,000 e^{0.60} \approx 75,000 \times 1.82211880 = 136,658.91\)
• For \(t = 25\): \(A = 75,000 e^{0.03 \times 25} = 75,000 e^{0.75} \approx 75,000 \times 2.11700002 = 158,775.00\)
• For \(t = 30\): \(A = 75,000 e^{0.03 \times 30} = 75,000 e^{0.90} \approx 75,000 \times 2.45960311 = 184,470.23\)

</reasoning>

<answer>

Years	5	10	15
3% compounded continuously	\$87,137.57	\$101,239.41	\$117,623.41

Years	20	25	30
3% compounded continuously	\$136,658.91	\$158,775.00	\$184,470.23

</answer>

<post_analysis>
{
"subject": "Mathematics",
"question_type": "Fill-in-the-blank",
"knowledge_point": [
"Mathematics",
"Algebra",
"Continuous Compounding"
]
}
</post_analysis>

Answer:

<pre_analysis>
{
"quality": "clear",
"question_count": 1,
"skills_matched": [
"step_cot",
"explain_then_answer",
"tutor"
],
"needs_drawing": false,
"concepts_used": [
"Compound Interest",
"Future Value"
],
"new_concepts": [
"Continuous Compounding",
"Simple Interest"
],
"current_concepts": [
"Simple Interest",
"Continuous Compounding",
"Compound Interest",
"Future Value"
]
}
</pre_analysis>

<reasoning>

Identify given parameters

We have principal \(P = 75,000\) and annual interest rate \(r = 0.03\).
We need to calculate the Future Value for both accounts at \(t = 5, 10, 15, 20, 25, 30\) years.

Calculate simple interest values

Using the Future Value formula for simple interest:
\[A = P(1 + rt)\]

• For \(t = 5\): \(A = 75,000(1 + 0.03 \times 5) = 75,000(1.15) = 86,250.00\)
• For \(t = 10\): \(A = 75,000(1 + 0.03 \times 10) = 75,000(1.30) = 97,500.00\)
• For \(t = 15\): \(A = 75,000(1 + 0.03 \times 15) = 75,000(1.45) = 108,750.00\)
• For \(t = 20\): \(A = 75,000(1 + 0.03 \times 20) = 75,000(1.60) = 120,000.00\)
• For \(t = 25\): \(A = 75,000(1 + 0.03 \times 25) = 75,000(1.75) = 131,250.00\)
• For \(t = 30\): \(A = 75,000(1 + 0.03 \times 30) = 75,000(1.90) = 142,500.00\)

Calculate continuous compounding values

Using the Compound Interest formula for continuous compounding:
\[A = P e^{rt}\]

• For \(t = 5\): \(A = 75,000 e^{0.03 \times 5} = 75,000 e^{0.15} \approx 75,000 \times 1.16183424 = 87,137.57\)
• For \(t = 10\): \(A = 75,000 e^{0.03 \times 10} = 75,000 e^{0.30} \approx 75,000 \times 1.34985881 = 101,239.41\)
• For \(t = 15\): \(A = 75,000 e^{0.03 \times 15} = 75,000 e^{0.45} \approx 75,000 \times 1.56831219 = 117,623.41\)
• For \(t = 20\): \(A = 75,000 e^{0.03 \times 20} = 75,000 e^{0.60} \approx 75,000 \times 1.82211880 = 136,658.91\)
• For \(t = 25\): \(A = 75,000 e^{0.03 \times 25} = 75,000 e^{0.75} \approx 75,000 \times 2.11700002 = 158,775.00\)
• For \(t = 30\): \(A = 75,000 e^{0.03 \times 30} = 75,000 e^{0.90} \approx 75,000 \times 2.45960311 = 184,470.23\)

</reasoning>

<answer>

Years	5	10	15
3% compounded continuously	\$87,137.57	\$101,239.41	\$117,623.41

Years	20	25	30
3% compounded continuously	\$136,658.91	\$158,775.00	\$184,470.23

</answer>

<post_analysis>
{
"subject": "Mathematics",
"question_type": "Fill-in-the-blank",
"knowledge_point": [
"Mathematics",
"Algebra",
"Continuous Compounding"
]
}
</post_analysis>