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Part 1

How much more would you earn in the first investment than in the second​ investment?

​$48 comma 00048,000 

invested for 

3030 

years at 

1212​% 

compounded annually

​$48 comma 00048,000 

invested for 

3030 

years at 

66​% 

compounded monthly

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Part 1

You would earn 

​$enter your response here 

more on the first investment than in the second investment.

​(Round to the nearest dollar as​ needed.) 
In the provided​ formulas, A is the balance in the account after t​ years, P is the principal​ investment, r is the annual interest rate in decimal​ form, n is the number of compounding periods per​ year, and Y is the​ investment's effective annual yield in decimal form.

Upper A equals Upper P left parenthesis 1 plus StartFraction r Over n EndFraction right parenthesis Superscript ntA=P1+rnnt       

Upper P equals StartFraction Upper A Over left parenthesis 1 plus StartFraction r Over n EndFraction right parenthesis Superscript nt EndFractionP=A1+rnnt       

Upper A equals Upper P e Superscript rtA=Pert       

Upper Y equals left parenthesis 1 plus StartFraction r Over n EndFraction right parenthesis Superscript n Baseline minus 1

To find how much more would be earned from the first investment than the second investment, we need to calculate the final amounts for each investment using the compound interest formula:

\[
A = P \left(1 + \frac{r}{n}\right)^{nt}
\]

where:
- $A$ is the amount after $t$ years
- $P$ is the principal (initial investment)
- $r$ is the annual interest rate (decimal form)
- $n$ is the number of times interest is compounded per year
- $t$ is the number of years

### Step 1: Calculate the amount for the first investment
Given:
- $P = 48,000$
- $r = 0.12$ (12%)
- $n = 1$ (compounded annually)
- $t = 30$ years

The formula becomes:
\[
A_1 = 48,000 \left(1 + \frac{0.12}{1}\right)^{1 \times 30}
\]
\[
A_1 = 48,000 \left(1.12\right)^{30}
\]

Calculating $1.12^{30}$:
\[
1.12^{30} \approx 29.96
\]

Thus, 
\[
A_1 \approx 48,000 \times 29.96 = 1,438,080
\]

### Step 2: Calculate the amount for the second investment
Given:
- $P = 48,000$
- $r = 0.06$ (6%)
- $n = 12$ (compounded monthly)
- $t = 30$ years

The formula becomes:
\[
A_2 = 48,000 \left(1 + \frac{0.06}{12}\right)^{12 \times 30}
\]
\[
A_2 = 48,000 \left(1 + 0.005\right)^{360}
\]
\[
A_2 = 48,000 \left(1.005\right)^{360}
\]

Calculating $1.005^{360}$:
\[
1.005^{360} \approx 6.0226
\]

Thus, 
\[
A_2 \approx 48,000 \times 6.0226 = 289,085
\]

### Step 3: Find the difference between the two investments
\[
\text{Difference} = A_1 - A_2
\]
\[
\text{Difference} \approx 1,438,080 - 289,085 = 1,148,995
\]

Therefore, you would earn approximately $1,148,995 more on the first investment than on the second investment.

Do you want more details or have any questions?

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### Related Questions:
1. How would the outcome change if the first investment had a different compounding frequency, such as quarterly?
2. What would happen to the final amounts if both investments were for a shorter duration, like 20 years?
3. How does changing the interest rate affect the difference between the two investments?
4. What is the impact of the initial principal on the overall difference in the earnings between the two options?
5. How does continuous compounding compare to the monthly or annual compounding methods?

### Tip:
Always convert the annual interest rate to decimal form before using it in the compound interest formula for accuracy.

How much more would you earn in the first investment than in the second​ investment? ​$48,000 invested for 30 years at 12% compounded annually versus $48,000 invested for 30 years at 6% compounded monthly.

Learn how to calculate the difference between two investments: $48,000 at 12% compounded annually versus 6% compounded monthly over 30 years. This step-by-step guide covers compound interest formulas and shows you how much more you would earn in the first investment.

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