Let's go through each part of your question step by step.
A. Completing the Tables
1. Simple Interest Table: The formula for calculating simple interest is: \[ I = P \times r \times t \] Where:
- \( I \) = Interest
- \( P \) = Principal
- \( r \) = Rate (as a decimal)
- \( t \) = Time (in years)
For 1 year:
- \( P = 1000 \)
- \( r = 0.06 \)
- \( t = 1 \)
Calculating interest: \[ I = 1000 \times 0.06 \times 1 = 60 \]
End of Year Balance: \[ \text{Balance} = P + I = 1000 + 60 = 1060 \]
Continuing with the simple interest for subsequent years, simply add \( I = 60 \) for each year:
- 2: Balance = 1060 + 60 = 1120
- 3: Balance = 1120 + 60 = 1180
- 4: Balance = 1180 + 60 = 1240
- 5: Balance = 1240 + 60 = 1300
- 6: Balance = 1300 + 60 = 1360
- 7: Balance = 1360 + 60 = 1420
- 8: Balance = 1420 + 60 = 1480
- 9: Balance = 1480 + 60 = 1540
- 10: Balance = 1540 + 60 = 1600
Simple Interest Table
| t | Principal | Annual Interest | End of year Balance | |----|-----------|----------------|----------------------| | 1 | $1000.00 | $60.00 | $1060.00 | | 2 | $1000.00 | $60.00 | $1120.00 | | 3 | $1000.00 | $60.00 | $1180.00 | | 4 | $1000.00 | $60.00 | $1240.00 | | 5 | $1000.00 | $60.00 | $1300.00 | | 6 | $1000.00 | $60.00 | $1360.00 | | 7 | $1000.00 | $60.00 | $1420.00 | | 8 | $1000.00 | $60.00 | $1480.00 | | 9 | $1000.00 | $60.00 | $1540.00 | | 10 | $1000.00 | $60.00 | $1600.00 |
2. Compound Interest Table: The formula for compound interest is: \[ A = P(1 + r)^t \] Where:
- \( A \) = Amount (or balance after t years)
- \( P \) = Principal
- \( r \) = Rate (as a decimal)
- \( t \) = Time (in years)
Calculating for each year:
- After 1 year: \[ A = 1000(1 + 0.06)^1 = 1000(1.06) = 1060 \]
- After 2 years: \[ A = 1000(1 + 0.06)^2 = 1000(1.06)^2 = 1000(1.1236) = 1123.60 \]
- After 3 years: \[ A = 1000(1 + 0.06)^3 = 1000(1.06)^3 = 1000(1.191016) = 1191.02 \]
- After 4 years: \[ A = 1000(1 + 0.06)^4 = 1000(1.26247696) = 1262.48 \]
- After 5 years: \[ A = 1000(1 + 0.06)^5 = 1000(1.33822594) = 1338.23 \]
- After 6 years: \[ A = 1000(1 + 0.06)^6 = 1000(1.41851917) = 1418.52 \]
- After 7 years: \[ A = 1000(1 + 0.06)^7 = 1000(1.50363033) = 1503.63 \]
- After 8 years: \[ A = 1000(1 + 0.06)^8 = 1000(1.59385336) = 1593.85 \]
- After 9 years: \[ A = 1000(1 + 0.06)^9 = 1000(1.68947818) = 1689.48 \]
- After 10 years: \[ A = 1000(1 + 0.06)^{10} = 1000(1.79084769) = 1790.85 \]
Compound Interest Table
| t | Principal | Annual Interest | End of year Balance | |----|-----------|----------------|----------------------| | 1 | $1000.00 | $60.00 | $1060.00 | | 2 | $1060.00 | $63.60 | $1123.60 | | 3 | $1123.60 | $67.42 | $1191.02 | | 4 | $1191.02 | $71.46 | $1262.48 | | 5 | $1262.48 | $75.75 | $1338.23 | | 6 | $1338.23 | $80.29 | $1418.52 | | 7 | $1418.52 | $86.11 | $1503.63 | | 8 | $1503.63 | $92.22 | $1593.85 | | 9 | $1593.85 | $98.59 | $1689.48 | | 10 | $1689.48 | $106.37 | $1790.85 |
Which type of interest gives the greater balance?
At the end of 10 years:
- Simple Interest Balance: $1600.00
- Compound Interest Balance: $1790.85
So, compound interest gives a greater balance.
B. Graphing the Balances
To graph, you will plot the end-of-year balances for both simple and compound interest on the same graph with the x-axis representing time (years) and the y-axis representing account balance. The simple interest line will be straight, while the compound interest line will be a curve that goes up more sharply as time increases.
C. Calculate Future Balances for $400 with 3.5% Interest Compounded Annually
Using the compound interest formula: \[ A = P(1 + r)^t \]
Where:
- \( P = 400 \)
- \( r = 0.035 \)
-
After 2 years: \[ A = 400(1 + 0.035)^2 = 400(1.035)^2 = 400(1.071225) \approx 428.49 \]
-
After 10 years: \[ A = 400(1 + 0.035)^{10} = 400(1.035)^{10} = 400(1.4107) \approx 564.28 \]
-
After 25 years: \[ A = 400(1 + 0.035)^{25} = 400(1.035)^{25} = 400(2.2082) \approx 883.28 \]
D. Final Balance After 2 Years for $40,000 with 6% Compounded Monthly
Using the formula: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where:
- \( P = 40000 \)
- \( r = 0.06 \)
- \( n = 12 \) (compounded monthly)
- \( t = 2 \)
Calculating: \[ A = 40000 \left(1 + \frac{0.06}{12}\right)^{12 \times 2} = 40000 \left(1 + 0.005\right)^{24} = 40000(1.005)^{24} \]
Calculating \( (1.005)^{24} \): \[ (1.005)^{24} \approx 1.12749 \]
So, \[ A \approx 40000 \times 1.12749 \approx 45099.60 \]
Final Balance β $45,099.60
E. Find the Missing Variables for Compound Interest
a. Find \( I \) \[ P = 2000, , y = 5% = 0.05, , t = 5 \] Using the formula: \[ A = P(1 + r)^t \] \[ A = 2000(1 + 0.05)^5 = 2000(1.27628) \approx 2552.56 \] Now, find \( I \): \[ I = A - P = 2552.56 - 2000 = 552.56 \]
b. Find \( P \) \[ I = 2566.72, , y = 12% = 0.12, , t = 3 \] Using the formula \[ A = P(1 + r)^t \] First, find \( A \): \[ A = I + P \] Assuming \( r = 0.12 \) and \( t = 3 \): \[ 2566.72 = P \cdot (1 + 0.12)^3 \] \[ 2566.72 = P \cdot 1.404928 \] Solving for \( P \): \[ P = \frac{2566.72}{1.404928} \approx 1824.15 \]
c. Find \( R \) \[ I = 842.96, , P = 500, , t = 2 \] Using the rearranged formula: \[ I = P \cdot r \cdot t \] \[ 842.96 = 500 \cdot r \cdot 2 \] \[ r = \frac{842.96}{500 \cdot 2} = \frac{842.96}{1000} \approx 0.84296 \text{ or } 84.296% \]
d. Find \( t \) \[ I = 1738.91, , P = 1500, , y = 3% = 0.03 \] Using the formula: \[ 1738.91 = 1500 \cdot 0.03 \cdot t \] Solving for \( t \): \[ t = \frac{1738.91}{1500 \cdot 0.03} = \frac{1738.91}{45} \approx 38.86 \] So \( t \approx 39 \) years.
Feel free to ask if you need further clarification or assistance!