Question
A TV manufacturer produces 500 TVs in its first year and plans to increase production by 8 percent each year for the next 10 years. How many total TVs will have been produced at the end of the tenth year?(1 point)
Responses
5,687
5,687
7,243
7,243
8,949
8,949
6,856
Responses
5,687
5,687
7,243
7,243
8,949
8,949
6,856
Answers
GPT-4o mini
To find the total number of TVs produced over 10 years, we can determine the number of TVs produced each year and then sum them up.
1. **Year 1**: 500 TVs
2. **Year 2**: 500 * (1 + 0.08) = 500 * 1.08 = 540 TVs
3. **Year 3**: 540 * (1 + 0.08) = 540 * 1.08 = 583.2 TVs
4. **Year 4**: 583.2 * (1 + 0.08) = 583.2 * 1.08 = 629.856 TVs
5. **Year 5**: 629.856 * (1 + 0.08) = 629.856 * 1.08 ≈ 679.45 TVs
6. **Year 6**: 679.45 * (1 + 0.08) = 679.45 * 1.08 ≈ 733.29 TVs
7. **Year 7**: 733.29 * (1 + 0.08) = 733.29 * 1.08 ≈ 790.56 TVs
8. **Year 8**: 790.56 * (1 + 0.08) = 790.56 * 1.08 ≈ 852.70 TVs
9. **Year 9**: 852.70 * (1 + 0.08) = 852.70 * 1.08 ≈ 919.21 TVs
10. **Year 10**: 919.21 * (1 + 0.08) = 919.21 * 1.08 ≈ 992.69 TVs
Next, we sum up the production over the 10 years.
Here's a simplified calculation using geometric series:
The production for each year can be calculated as:
- Year 1: \( 500 \)
- Year 2: \( 500 \times (1.08)^1 \)
- Year 3: \( 500 \times (1.08)^2 \)
- Year 4: \( 500 \times (1.08)^3 \)
- Year 5: \( 500 \times (1.08)^4 \)
- Year 6: \( 500 \times (1.08)^5 \)
- Year 7: \( 500 \times (1.08)^6 \)
- Year 8: \( 500 \times (1.08)^7 \)
- Year 9: \( 500 \times (1.08)^8 \)
- Year 10: \( 500 \times (1.08)^9 \)
Thus, the total TVs produced over the 10 years can be computed using the formula for the sum of a geometric series:
\[
S_n = a \frac{(1 - r^n)}{(1 - r)}
\]
Where:
- \( a = 500 \) (the number of TVs produced in the first year)
- \( r = 1.08 \) (the growth rate)
- \( n = 10 \) (the number of years)
Calculating:
\[
S_{10} = 500 \frac{(1 - (1.08)^{10})}{(1 - 1.08)} = 500 \frac{(1 - 2.158925)}{-0.08}
\]
\[
S_{10} = 500 \times \frac{-1.158925}{-0.08} \approx 500 \times 14.48656 \approx 7243
\]
So, the total number of TVs produced at the end of the tenth year is approximately **7243**.
Thus, the correct answer is **7,243**.
1. **Year 1**: 500 TVs
2. **Year 2**: 500 * (1 + 0.08) = 500 * 1.08 = 540 TVs
3. **Year 3**: 540 * (1 + 0.08) = 540 * 1.08 = 583.2 TVs
4. **Year 4**: 583.2 * (1 + 0.08) = 583.2 * 1.08 = 629.856 TVs
5. **Year 5**: 629.856 * (1 + 0.08) = 629.856 * 1.08 ≈ 679.45 TVs
6. **Year 6**: 679.45 * (1 + 0.08) = 679.45 * 1.08 ≈ 733.29 TVs
7. **Year 7**: 733.29 * (1 + 0.08) = 733.29 * 1.08 ≈ 790.56 TVs
8. **Year 8**: 790.56 * (1 + 0.08) = 790.56 * 1.08 ≈ 852.70 TVs
9. **Year 9**: 852.70 * (1 + 0.08) = 852.70 * 1.08 ≈ 919.21 TVs
10. **Year 10**: 919.21 * (1 + 0.08) = 919.21 * 1.08 ≈ 992.69 TVs
Next, we sum up the production over the 10 years.
Here's a simplified calculation using geometric series:
The production for each year can be calculated as:
- Year 1: \( 500 \)
- Year 2: \( 500 \times (1.08)^1 \)
- Year 3: \( 500 \times (1.08)^2 \)
- Year 4: \( 500 \times (1.08)^3 \)
- Year 5: \( 500 \times (1.08)^4 \)
- Year 6: \( 500 \times (1.08)^5 \)
- Year 7: \( 500 \times (1.08)^6 \)
- Year 8: \( 500 \times (1.08)^7 \)
- Year 9: \( 500 \times (1.08)^8 \)
- Year 10: \( 500 \times (1.08)^9 \)
Thus, the total TVs produced over the 10 years can be computed using the formula for the sum of a geometric series:
\[
S_n = a \frac{(1 - r^n)}{(1 - r)}
\]
Where:
- \( a = 500 \) (the number of TVs produced in the first year)
- \( r = 1.08 \) (the growth rate)
- \( n = 10 \) (the number of years)
Calculating:
\[
S_{10} = 500 \frac{(1 - (1.08)^{10})}{(1 - 1.08)} = 500 \frac{(1 - 2.158925)}{-0.08}
\]
\[
S_{10} = 500 \times \frac{-1.158925}{-0.08} \approx 500 \times 14.48656 \approx 7243
\]
So, the total number of TVs produced at the end of the tenth year is approximately **7243**.
Thus, the correct answer is **7,243**.