Let's start by completing the tables for simple interest and compound interest:
Simple Interest
To calculate simple interest using the formula
\[ I = P \cdot r \cdot t \]
where:
- \( I \) = interest earned
- \( P \) = principal amount (initial deposit)
- \( r \) = interest rate (as a decimal)
- \( t \) = time (in years)
Using your values:
- \( P = 1000 \)
- \( r = 0.06 \)
We can fill out the table as follows:
| t | Principal | Annual Interest | End of Year Balance | |---|---------------|------------------|---------------------| | 1 | $1,000.00 | $60.00 | $1,060.00 | | 2 | $1,000.00 | $60.00 | $1,120.00 | | 3 | $1,000.00 | $60.00 | $1,180.00 | | 4 | $1,000.00 | $60.00 | $1,240.00 | | 5 | $1,000.00 | $60.00 | $1,300.00 | | 6 | $1,000.00 | $60.00 | $1,360.00 | | 7 | $1,000.00 | $60.00 | $1,420.00 | | 8 | $1,000.00 | $60.00 | $1,480.00 | | 9 | $1,000.00 | $60.00 | $1,540.00 | | 10| $1,000.00 | $60.00 | $1,600.00 |
Compound Interest
To calculate compound interest using the formula
\[ A = P(1 + r)^t \]
where:
- \( A \) = the amount of money accumulated after n years, including interest
- \( P = 1,000 \) (initial deposit)
- \( r = 0.06 \) (interest rate)
- \( t \) = time (in years)
We can fill out the compound interest table as follows:
| t | Principal | Annual Interest | End of Year Balance | |---|----------------|-------------------|---------------------| | 1 | $1,000.00 | $60.00 | $1,060.00 | | 2 | $1,060.00 | $63.60 | $1,123.60 | | 3 | $1,123.60 | $67.42 | $1,191.02 | | 4 | $1,191.02 | $71.46 | $1,262.48 | | 5 | $1,262.48 | $75.75 | $1,338.23 | | 6 | $1,338.23 | $80.29 | $1,418.52 | | 7 | $1,418.52 | $85.11 | $1,503.63 | | 8 | $1,503.63 | $90.22 | $1,593.85 | | 9 | $1,593.85 | $95.63 | $1,689.48 | | 10| $1,689.48 | $101.37 | $1,790.85 |
Comparison
From the tables, we see that the balances at the end of 10 years are:
- Simple Interest: $1,600.00
- Compound Interest: $1,790.85
Conclusion: Compound interest gives a greater balance than simple interest.
Graph the Balances
When graphing, the X-axis can represent the number of years (0 to 10) and the Y-axis represents the account balance. You will plot two lines, one for simple interest that increases linearly and one for compound interest that curves upward.
Additional Calculations
-
For the account that earns 3.5% interest compounded annually:
- \( P = 400 \)
- \( r = 0.035 \)
After 2 years: \[ A = 400(1 + 0.035)^2 = 400(1.071225) \approx 428.49 \]
After 10 years: \[ A = 400(1 + 0.035)^{10} = 400(1.410802) \approx 564.32 \]
After 25 years: \[ A = 400(1 + 0.035)^{25} = 400(2.314783) \approx 925.91 \]
-
For the firefighter's investment:
- $40,000 invested at 6% for 2 years compounded monthly.
- The formula is: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where \( n = 12 \) (monthly compounding). \[ A = 40000 \left(1 + \frac{0.06}{12}\right)^{12 \times 2} = 40000 \left(1.005)^{24} \approx 40000(1.12749) \approx 45099.60 \]
-
Calculating Interest (I):
- For \( P = 2000, R = 5% (0.05), t = 5 \): \[ I = P \cdot r \cdot t = 2000 \cdot 0.05 \cdot 5 = 500 \]
-
Finding Principal (P):
- \( I = 842.96, R = 12% (0.12), t = 3 \): \[ P = \frac{I}{r \cdot t} = \frac{842.96}{0.12 \cdot 3} \approx 2,341.67 \]
-
Finding Rate (R):
- \( I = 842.96, P = 500, t = 2 \): \[ R = \frac{I}{P \cdot t} = \frac{842.96}{500 \cdot 2} \approx 0.16859 \approx 16.86% \]
-
Finding Time (t):
- \( I = 1738.91, P = 1500, R = 3% (0.03) \): \[ t = \frac{I}{P \cdot R} = \frac{1738.91}{1500 \cdot 0.03} \approx 38.63 \text{ years} \]
This should cover all the calculations and questions you posed! Let me know if you need any further assistance.