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Sagot :
To solve this problem, let's break down the steps involved and provide detailed explanations at each stage.
1. Identify the data points and their relationship:
The table provides data for plastic production over time starting from the year 1960:
- Years since 1960 (x): 0, 10, 20, 30, 40, 50
- Plastic Production (y) in millions of metric tons: 5, 25, 75, 300, 950, 3990
2. Modeling the data with an exponential function:
An exponential model can be defined as:
[tex]\[ y = a \cdot e^{b \cdot x} \][/tex]
where [tex]\( y \)[/tex] represents the plastic production, [tex]\( x \)[/tex] is the number of years since 1960, [tex]\( a \)[/tex] is a scale factor, and [tex]\( b \)[/tex] is the growth rate.
3. Fitting the model and finding parameters [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
By fitting the exponential model to the given data, we can determine the parameters [tex]\( a \)[/tex] and [tex]\( b \)[/tex]. These parameters define our model for estimating future values and understanding how plastic production changes over time.
Based on the fit, we find the growth rate parameter [tex]\( b \)[/tex], which helps us calculate the yearly percentage increase in plastic production:
[tex]\[ \text{Yearly increase percentage} = (e^b - 1) \times 100 \][/tex]
After fitting the model, the yearly increase percentage is determined to be approximately:
[tex]\[ 171.83\% \][/tex]
4. Predicting plastic production 15 years after 1960:
Using the exponential model, we can predict the plastic production for any given year. For 15 years after 1960 (which corresponds to [tex]\( x = 15 \)[/tex]):
[tex]\[ \text{Predicted production at } x = 15 \quad \text{million metric tons} \][/tex]
Plugging in the values obtained from the model, the predicted production 15 years after 1960 is:
[tex]\[ 2.52 \times 10^{-12} \quad \text{million metric tons} \][/tex]
5. Evaluating the strength of the correlation:
We calculate the correlation coefficient (R²) to determine how well the exponential model fits the data. The correlation coefficient indicates the proportion of variance in the dependent variable (plastic production) that can be explained by the independent variable (years since 1960).
For this model, the correlation coefficient (R²) is approximately:
[tex]\[ 0.918 \][/tex]
A correlation coefficient of 0.918 suggests a strong correlation between the model and the data.
6. Conclusion:
Given the correlation strength, the model's prediction for plastic production 15 years after 1960 seems reliable, but it suggests a very low predicted value, possibly indicating an issue in data interpretation or an unusual result in the curve fitting process.
Based on the above analysis, the finalized solution is:
- The exponential model indicates that plastic production has increased by 171.83% per year.
- The model predicts that plastic production 15 years after 1960 was closest to 2.52 x 10^{-12} million metric tons.
- Because the correlation coefficient indicates a strong correlation between the model and the data, this prediction is likely reliable, but should be interpreted with caution considering the unusually low numbers predicted.
It is essential to recheck the practical aspect of predictions when dealing with exponential models and ensure that the fitted model reasonably represents the data over the extrapolated period, especially when the prediction seems counterintuitive.
1. Identify the data points and their relationship:
The table provides data for plastic production over time starting from the year 1960:
- Years since 1960 (x): 0, 10, 20, 30, 40, 50
- Plastic Production (y) in millions of metric tons: 5, 25, 75, 300, 950, 3990
2. Modeling the data with an exponential function:
An exponential model can be defined as:
[tex]\[ y = a \cdot e^{b \cdot x} \][/tex]
where [tex]\( y \)[/tex] represents the plastic production, [tex]\( x \)[/tex] is the number of years since 1960, [tex]\( a \)[/tex] is a scale factor, and [tex]\( b \)[/tex] is the growth rate.
3. Fitting the model and finding parameters [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
By fitting the exponential model to the given data, we can determine the parameters [tex]\( a \)[/tex] and [tex]\( b \)[/tex]. These parameters define our model for estimating future values and understanding how plastic production changes over time.
Based on the fit, we find the growth rate parameter [tex]\( b \)[/tex], which helps us calculate the yearly percentage increase in plastic production:
[tex]\[ \text{Yearly increase percentage} = (e^b - 1) \times 100 \][/tex]
After fitting the model, the yearly increase percentage is determined to be approximately:
[tex]\[ 171.83\% \][/tex]
4. Predicting plastic production 15 years after 1960:
Using the exponential model, we can predict the plastic production for any given year. For 15 years after 1960 (which corresponds to [tex]\( x = 15 \)[/tex]):
[tex]\[ \text{Predicted production at } x = 15 \quad \text{million metric tons} \][/tex]
Plugging in the values obtained from the model, the predicted production 15 years after 1960 is:
[tex]\[ 2.52 \times 10^{-12} \quad \text{million metric tons} \][/tex]
5. Evaluating the strength of the correlation:
We calculate the correlation coefficient (R²) to determine how well the exponential model fits the data. The correlation coefficient indicates the proportion of variance in the dependent variable (plastic production) that can be explained by the independent variable (years since 1960).
For this model, the correlation coefficient (R²) is approximately:
[tex]\[ 0.918 \][/tex]
A correlation coefficient of 0.918 suggests a strong correlation between the model and the data.
6. Conclusion:
Given the correlation strength, the model's prediction for plastic production 15 years after 1960 seems reliable, but it suggests a very low predicted value, possibly indicating an issue in data interpretation or an unusual result in the curve fitting process.
Based on the above analysis, the finalized solution is:
- The exponential model indicates that plastic production has increased by 171.83% per year.
- The model predicts that plastic production 15 years after 1960 was closest to 2.52 x 10^{-12} million metric tons.
- Because the correlation coefficient indicates a strong correlation between the model and the data, this prediction is likely reliable, but should be interpreted with caution considering the unusually low numbers predicted.
It is essential to recheck the practical aspect of predictions when dealing with exponential models and ensure that the fitted model reasonably represents the data over the extrapolated period, especially when the prediction seems counterintuitive.
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