Answer :
Certainly! Let's address each part of the problem step by step.
### Part 1: Investment 1
Given:
- Principal (initial investment): [tex]\( P = 1,817,300 \)[/tex] dollars
- Annual interest rate: [tex]\( r = 12\% \)[/tex] or [tex]\( r = 0.12 \)[/tex]
- Time period: [tex]\( t = 3 \)[/tex] years
- Compounding frequency: twice a year (semi-annually, which means [tex]\( n = 2 \)[/tex])
The formula for compound interest is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Substitute the given values into the formula:
[tex]\[ A = 1,817,300 \left(1 + \frac{0.12}{2}\right)^{2 \times 3} \][/tex]
This simplifies to:
[tex]\[ A = 1,817,300 \left(1 + 0.06\right)^{6} \][/tex]
[tex]\[ A = 1,817,300 \left(1.06\right)^{6} \][/tex]
By calculating the above, we find:
[tex]\[ A \approx 2,577,874.78 \][/tex]
So, the value of the first investment after 3 years is approximately 2,577,874.78 dollars.
### Part 2: Investment 2
Given:
- Principal (initial investment): [tex]\( P = 830,000 \)[/tex] dollars
- Annual Interest rate: [tex]\( r = 10.5\% \)[/tex] or [tex]\( r = 0.105 \)[/tex]
- Time period: [tex]\( t = 2 \)[/tex] years
- Compounding frequency: yearly (annually, which means [tex]\( n = 1 \)[/tex])
The formula for compound interest is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Substitute the given values into the formula:
[tex]\[ A = 830,000 \left(1 + 0.105\right)^{1 \times 2} \][/tex]
This simplifies to:
[tex]\[ A = 830,000 \left(1.105\right)^{2} \][/tex]
By calculating the above, we find:
[tex]\[ A \approx 1,013,450.75 \][/tex]
So, the value of the second investment after 2 years is approximately 1,013,450.75 dollars.
### Summary
1. The value of the first investment after 3 years at 12% annual interest compounded semi-annually is 2,577,874.78 dollars.
2. The value of the second investment after 2 years at 10.5% annual interest compounded annually is 1,013,450.75 dollars.
### Part 1: Investment 1
Given:
- Principal (initial investment): [tex]\( P = 1,817,300 \)[/tex] dollars
- Annual interest rate: [tex]\( r = 12\% \)[/tex] or [tex]\( r = 0.12 \)[/tex]
- Time period: [tex]\( t = 3 \)[/tex] years
- Compounding frequency: twice a year (semi-annually, which means [tex]\( n = 2 \)[/tex])
The formula for compound interest is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Substitute the given values into the formula:
[tex]\[ A = 1,817,300 \left(1 + \frac{0.12}{2}\right)^{2 \times 3} \][/tex]
This simplifies to:
[tex]\[ A = 1,817,300 \left(1 + 0.06\right)^{6} \][/tex]
[tex]\[ A = 1,817,300 \left(1.06\right)^{6} \][/tex]
By calculating the above, we find:
[tex]\[ A \approx 2,577,874.78 \][/tex]
So, the value of the first investment after 3 years is approximately 2,577,874.78 dollars.
### Part 2: Investment 2
Given:
- Principal (initial investment): [tex]\( P = 830,000 \)[/tex] dollars
- Annual Interest rate: [tex]\( r = 10.5\% \)[/tex] or [tex]\( r = 0.105 \)[/tex]
- Time period: [tex]\( t = 2 \)[/tex] years
- Compounding frequency: yearly (annually, which means [tex]\( n = 1 \)[/tex])
The formula for compound interest is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Substitute the given values into the formula:
[tex]\[ A = 830,000 \left(1 + 0.105\right)^{1 \times 2} \][/tex]
This simplifies to:
[tex]\[ A = 830,000 \left(1.105\right)^{2} \][/tex]
By calculating the above, we find:
[tex]\[ A \approx 1,013,450.75 \][/tex]
So, the value of the second investment after 2 years is approximately 1,013,450.75 dollars.
### Summary
1. The value of the first investment after 3 years at 12% annual interest compounded semi-annually is 2,577,874.78 dollars.
2. The value of the second investment after 2 years at 10.5% annual interest compounded annually is 1,013,450.75 dollars.