To find the amount of money in Nico's savings account each year for 8 years, we can use the formula for compound interest:
\[ A = P(1 + r)^t \]
where:
- \( A \) is the amount of money in the account after \( t \) years,
- \( P \) is the initial principal balance ($1000 in this case),
- \( r \) is the annual interest rate (8%, or 0.08), and
- \( t \) is the number of years the money is invested or borrowed.
Now let's calculate the amount for each year from 1 to 8:
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Year 0 (initial deposit): \[ A = 1000(1 + 0.08)^0 = 1000 \]
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Year 1: \[ A = 1000(1 + 0.08)^1 = 1000(1.08) = 1080 \]
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Year 2: \[ A = 1000(1 + 0.08)^2 = 1000(1.08^2) = 1000(1.1664) \approx 1166.40 \]
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Year 3: \[ A = 1000(1 + 0.08)^3 = 1000(1.08^3) = 1000(1.259712) \approx 1259.71 \]
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Year 4: \[ A = 1000(1 + 0.08)^4 = 1000(1.08^4) = 1000(1.36048896) \approx 1360.49 \]
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Year 5: \[ A = 1000(1 + 0.08)^5 = 1000(1.08^5) = 1000(1.469328) \approx 1469.33 \]
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Year 6: \[ A = 1000(1 + 0.08)^6 = 1000(1.08^6) = 1000(1.586874) \approx 1586.87 \]
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Year 7: \[ A = 1000(1 + 0.08)^7 = 1000(1.08^7) = 1000(1.7138) \approx 1717.99 \]
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Year 8: \[ A = 1000(1 + 0.08)^8 = 1000(1.08^8) = 1000(1.85093) \approx 1850.93 \]
Now, summarizing the amounts after each year for 8 years:
- Year 0: $1,000.00
- Year 1: $1,080.00
- Year 2: $1,166.40
- Year 3: $1,259.71
- Year 4: $1,360.49
- Year 5: $1,469.33
- Year 6: $1,586.87
- Year 7: $1,717.99
- Year 8: $1,850.93
Next, let's find the amount of money in Nico's savings account after 18 years using the same formula:
\[ A = 1000(1 + 0.08)^{18} \]
Calculating: \( 1.08^{18} \approx 4.015 \).
So: \[ A = 1000 \times 4.015 = 4015.00 \]
After 18 years, Nico will have $4,015.00 in his savings account.
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