Let's work through your questions step by step.
Simple Interest Table Completion
Simple Interest Formula: \[ I = P \cdot r \cdot t \] Where:
- \( I \) is the interest
- \( P \) is the principal
- \( r \) is the interest rate (as a decimal)
- \( t \) is the time in years
Given that the principal \( P = 1000 \) and the interest rate \( r = 0.06 \):
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Year 3:
- Interest: \( 1000 \cdot 0.06 \cdot 3 = 180 \)
- Balance: \( 1000 + 180 = 1180 \)
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Year 4:
- Interest: \( 1000 \cdot 0.06 \cdot 4 = 240 \)
- Balance: \( 1000 + 240 = 1240 \)
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Year 5:
- Interest: \( 1000 \cdot 0.06 \cdot 5 = 300 \)
- Balance: \( 1000 + 300 = 1300 \)
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Year 6:
- Interest: \( 1000 \cdot 0.06 \cdot 6 = 360 \)
- Balance: \( 1000 + 360 = 1360 \)
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Year 7:
- Interest: \( 1000 \cdot 0.06 \cdot 7 = 420 \)
- Balance: \( 1000 + 420 = 1420 \)
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Year 8:
- Interest: \( 1000 \cdot 0.06 \cdot 8 = 480 \)
- Balance: \( 1000 + 480 = 1480 \)
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Year 9:
- Interest: \( 1000 \cdot 0.06 \cdot 9 = 540 \)
- Balance: \( 1000 + 540 = 1540 \)
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Year 10:
- Interest: \( 1000 \cdot 0.06 \cdot 10 = 600 \)
- Balance: \( 1000 + 600 = 1600 \)
Completed Simple Interest Table
| 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 Table Completion
Compound Interest Formula: \[ A = P(1 + \frac{r}{n})^{nt} \] Where:
- \( A \) = amount of money accumulated after n years, including interest.
- \( r \) = annual interest rate (decimal)
- \( n \) = number of times that interest is compounded per year
- \( t \) = time in years
Given \( P = 1000 \), \( r = 0.06 \), and compounding annually (\( n = 1 \)):
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Year 1:
- Balance: \( 1000(1 + 0.06) = 1000(1.06) = 1060.00 \)
- Annual Interest: \( 1060.00 - 1000 = 60.00 \)
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Year 2:
- Balance: \( 1060(1.06) = 1123.60 \)
- Annual Interest: \( 1123.60 - 1060 = 63.60 \)
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Year 3:
- Balance: \( 1123.60(1.06) = 1191.016 \)
- Annual Interest: \( 1191.016 - 1123.60 = 67.416 \)
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Year 4:
- Balance: \( 1191.016(1.06) = 1260.477 \)
- Annual Interest: \( 1260.477 - 1123.60 = 68.461 \)
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Year 5:
- Balance: \( 1260.477(1.06) = 1332.505 \)
- Annual Interest: \( 1332.505 - 1191.016 = 71.529 \)
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Year 6:
- Balance: \( 1332.505(1.06) = 1407.705 \)
- Annual Interest: \( 1407.705 - 1260.477 = 74.552 \)
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Year 7:
- Balance: \( 1407.705(1.06) = 1485.655 \)
- Annual Interest: \( 1485.655 - 1332.505 = 76.982 \)
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Year 8:
- Balance: \( 1485.655(1.06) = 1566.19 \)
- Annual Interest: \( 1566.19 - 1407.705 = 79.025 \)
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Year 9:
- Balance: \( 1566.19(1.06) = 1650.114 \)
- Annual Interest: \( 1650.114 - 1485.655 = 84.459 \)
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Year 10:
- Balance: \( 1650.114(1.06) = 1740.122 \)
- Annual Interest: \( 1740.122 - 1566.19 = 91.932 \)
Completed Compound Interest Table
| 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 | $71.53 | $1,260.48 | | 5 | $1,260.48 | $74.55 | $1,332.51 | | 6 | $1,332.51 | $78.04 | $1,407.71 | | 7 | $1,407.71 | $81.56 | $1,485.66 | | 8 | $1,485.66 | $84.32 | $1,566.19 | | 9 | $1,566.19 | $88.56 | $1,650.11 | |10 | $1,650.11 | $91.93 | $1,740.12 |
Which Type of Interest Gives the Greater Balance?
At the end of 10 years:
- Simple Interest Balance: $1,600.00
- Compound Interest Balance: $1,740.12
The compound interest gives a greater balance.
Graphing the Results
Use a graphing tool to plot the following points:
- For Simple Interest: (1, 1060), (2, 1120), (3, 1180), ..., (10, 1600)
- For Compound Interest: (1, 1060), (2, 1123.60), (3, 1191.02), ..., (10, 1740.12)
Future Values for $400 at 3.5% Interest
Using the compound interest formula with \( P = 400 \), \( r = 0.035 \), and \( n = 1 \):
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After 2 Years: \[ A = 400(1 + 0.035)^2 = 400(1.035)^2 \approx 400(1.071225) \approx 428.49 \]
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After 10 Years: \[ A = 400(1 + 0.035)^{10} = 400(1.035)^{10} \approx 400(1.4105) \approx 564.21 \]
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After 25 Years: \[ A = 400(1 + 0.035)^{25} \approx 400(2.308) \approx 923.20 \]
Future Value of $40,000 at 6% Compounded Monthly
Using \( P = 40000 \), \( r = 0.06 \), \( n = 12 \), and \( t = 2 \):
\[ A = P(1 + \frac{r}{n})^{nt} = 40000(1 + \frac{0.06}{12})^{12 \times 2} \] \[ A = 40000(1 + 0.005)^{24} = 40000(1.005)^{24} \] \[ A \approx 40000(1.12749) \approx 45099.56 \]
Simple Interest Calculations
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Finding I with \( P = 2000 \), \( R = 0.05 \), \( t = 5 \): \[ I = P \cdot R \cdot t = 2000 \cdot 0.05 \cdot 5 = 500 \]
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Finding P with \( I = 842.96 \), \( R = 0.12 \), \( t = 3 \): \[ P = \frac{I}{R \cdot t} = \frac{842.96}{0.12 \cdot 3} = \frac{842.96}{0.36} \approx 2340.44 \]
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Finding R with \( I = 1738.91 \), \( P = 1500 \), \( t = 3 \): \[ R = \frac{I}{P \cdot t} = \frac{1738.91}{1500 \cdot 3} \approx \frac{1738.91}{4500} \approx 0.3863 \approx 38.63% \]
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Finding t with \( I = 1738.91 \), \( P = 1500 \), \( R = 0.03 \): \[ t = \frac{I}{P \cdot R} = \frac{1738.91}{1500 \cdot 0.03} = \frac{1738.91}{45} \approx 38.85 \text{ years} \]
If further clarification or information is needed, feel free to ask!