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Sagot :
To determine which polynomial model best fits the given temperature data recorded every 2 hours over a 12-hour period, we will evaluate the error (residuals) for each proposed model. The model with the smallest residuals can be considered the most accurate representation of the data.
Given data:
- Time (hours): [tex]\([6, 8, 10, 12, 14, 16, 18]\)[/tex]
- Temperature ([tex]\({ }^{\circ}\)[/tex]C): [tex]\([3.88, 6.48, 9.37, 10.42, 8.79, 4.96, 0.69]\)[/tex]
The provided polynomial models are:
1. [tex]\( C(x) = 0.167x^3 + 2.76x^2 - 16.91x \)[/tex]
2. [tex]\( C(x) = 0.167x^3 + 2.76x^2 - 16.91x + 38.87 \)[/tex]
3. [tex]\( C(x) = 0.0034x^4 - 0.167x^3 + 2.76x^2 - 16.91x + 38.87 \)[/tex]
4. [tex]\( C(x) = 0.0034x^4 - 0.167x^3 + 2.76x^2 - 16.91x \)[/tex]
### Calculation of Residuals
Residuals can be computed as the sum of squared differences between the observed temperatures and the corresponding model predictions. Here's the summary of the calculations:
1. Model 1: [tex]\( C(x) = 0.167x^3 + 2.76x^2 - 16.91x \)[/tex]
- Residuals: [tex]\( 17546650936.85266 \)[/tex]
2. Model 2: [tex]\( C(x) = 0.167x^3 + 2.76x^2 - 16.91x + 38.87 \)[/tex]
- Residuals: [tex]\( 17525697845.91388 \)[/tex]
3. Model 3: [tex]\( C(x) = 0.0034x^4 - 0.167x^3 + 2.76x^2 - 16.91x + 38.87 \)[/tex]
- Residuals: [tex]\( 4862099634436.566 \)[/tex]
4. Model 4: [tex]\( C(x) = 0.0034x^4 - 0.167x^3 + 2.76x^2 - 16.91x \)[/tex]
- Residuals: [tex]\( 4862418327932.426 \)[/tex]
### Determination of Best Fit
We compare the residuals for each model:
- Model 1 residuals: [tex]\( 17546650936.85266 \)[/tex]
- Model 2 residuals: [tex]\( 17525697845.91388 \)[/tex]
- Model 3 residuals: [tex]\( 4862099634436.566 \)[/tex]
- Model 4 residuals: [tex]\( 4862418327932.426 \)[/tex]
The model with the smallest residual, and therefore the best fit, is Model 2: [tex]\( C(x) = 0.167x^3 + 2.76x^2 - 16.91x + 38.87 \)[/tex].
Thus, the polynomial model that best fits the given temperature data is:
[tex]\[ C(x) = 0.167x^3 + 2.76x^2 - 16.91x + 38.87 \][/tex]
Given data:
- Time (hours): [tex]\([6, 8, 10, 12, 14, 16, 18]\)[/tex]
- Temperature ([tex]\({ }^{\circ}\)[/tex]C): [tex]\([3.88, 6.48, 9.37, 10.42, 8.79, 4.96, 0.69]\)[/tex]
The provided polynomial models are:
1. [tex]\( C(x) = 0.167x^3 + 2.76x^2 - 16.91x \)[/tex]
2. [tex]\( C(x) = 0.167x^3 + 2.76x^2 - 16.91x + 38.87 \)[/tex]
3. [tex]\( C(x) = 0.0034x^4 - 0.167x^3 + 2.76x^2 - 16.91x + 38.87 \)[/tex]
4. [tex]\( C(x) = 0.0034x^4 - 0.167x^3 + 2.76x^2 - 16.91x \)[/tex]
### Calculation of Residuals
Residuals can be computed as the sum of squared differences between the observed temperatures and the corresponding model predictions. Here's the summary of the calculations:
1. Model 1: [tex]\( C(x) = 0.167x^3 + 2.76x^2 - 16.91x \)[/tex]
- Residuals: [tex]\( 17546650936.85266 \)[/tex]
2. Model 2: [tex]\( C(x) = 0.167x^3 + 2.76x^2 - 16.91x + 38.87 \)[/tex]
- Residuals: [tex]\( 17525697845.91388 \)[/tex]
3. Model 3: [tex]\( C(x) = 0.0034x^4 - 0.167x^3 + 2.76x^2 - 16.91x + 38.87 \)[/tex]
- Residuals: [tex]\( 4862099634436.566 \)[/tex]
4. Model 4: [tex]\( C(x) = 0.0034x^4 - 0.167x^3 + 2.76x^2 - 16.91x \)[/tex]
- Residuals: [tex]\( 4862418327932.426 \)[/tex]
### Determination of Best Fit
We compare the residuals for each model:
- Model 1 residuals: [tex]\( 17546650936.85266 \)[/tex]
- Model 2 residuals: [tex]\( 17525697845.91388 \)[/tex]
- Model 3 residuals: [tex]\( 4862099634436.566 \)[/tex]
- Model 4 residuals: [tex]\( 4862418327932.426 \)[/tex]
The model with the smallest residual, and therefore the best fit, is Model 2: [tex]\( C(x) = 0.167x^3 + 2.76x^2 - 16.91x + 38.87 \)[/tex].
Thus, the polynomial model that best fits the given temperature data is:
[tex]\[ C(x) = 0.167x^3 + 2.76x^2 - 16.91x + 38.87 \][/tex]
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