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Answer :
To solve this problem and predict the company's profits in 2010 using a power function, follow these steps:
1. Understand the Data Rescaling:
- Start by considering the given year data (2003 to 2008) and profits.
- Rescale the year data so that the year 2003 corresponds to [tex]\(x = 1\)[/tex]. This is done by subtracting 2002 from each year value. Thus, the years become [tex]\(x = 1\)[/tex] for 2003, [tex]\(x = 2\)[/tex] for 2004, and so on up to [tex]\(x = 6\)[/tex] for 2008.
2. Formulate the Problem:
- We have the following pairs: [tex]\((1, 31.3)\)[/tex], [tex]\((2, 32.7)\)[/tex], [tex]\((3, 31.8)\)[/tex], [tex]\((4, 33.7)\)[/tex], [tex]\((5, 35.9)\)[/tex], [tex]\((6, 36.1)\)[/tex].
- We are tasked with fitting a power function of the form [tex]\(P(x) = a \cdot x^b\)[/tex] to this data, where [tex]\(P(x)\)[/tex] represents the profit in millions of dollars and [tex]\(x\)[/tex] is the rescaled year value.
3. Calculate Parameters for the Power Function:
- Fit the power function to the data points to determine the parameters [tex]\(a\)[/tex] and [tex]\(b\)[/tex].
- Suppose we find coefficients [tex]\(a ≈ 30.7515\)[/tex] and [tex]\(b ≈ 0.0789\)[/tex].
4. Predict the Profit for 2010:
- In the rescaled year data, 2010 corresponds to [tex]\(x = 2010 - 2002 = 8\)[/tex].
- Substitute [tex]\(x = 8\)[/tex] into the power function: [tex]\(P(8) = a \cdot (8^b)\)[/tex].
5. Calculate the Predicted Profit:
- Using the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] we have determined, calculate [tex]\(P(8)\)[/tex].
- This gives us a predicted profit of approximately [tex]\(36.24\)[/tex] million dollars for the year 2010.
Given the predicted value does not exactly match any of the provided options but is closest to $36.2 million, which is slightly off from the calculated prediction, leading us to choose "None of the above" as the best fitting answer from the given options.
1. Understand the Data Rescaling:
- Start by considering the given year data (2003 to 2008) and profits.
- Rescale the year data so that the year 2003 corresponds to [tex]\(x = 1\)[/tex]. This is done by subtracting 2002 from each year value. Thus, the years become [tex]\(x = 1\)[/tex] for 2003, [tex]\(x = 2\)[/tex] for 2004, and so on up to [tex]\(x = 6\)[/tex] for 2008.
2. Formulate the Problem:
- We have the following pairs: [tex]\((1, 31.3)\)[/tex], [tex]\((2, 32.7)\)[/tex], [tex]\((3, 31.8)\)[/tex], [tex]\((4, 33.7)\)[/tex], [tex]\((5, 35.9)\)[/tex], [tex]\((6, 36.1)\)[/tex].
- We are tasked with fitting a power function of the form [tex]\(P(x) = a \cdot x^b\)[/tex] to this data, where [tex]\(P(x)\)[/tex] represents the profit in millions of dollars and [tex]\(x\)[/tex] is the rescaled year value.
3. Calculate Parameters for the Power Function:
- Fit the power function to the data points to determine the parameters [tex]\(a\)[/tex] and [tex]\(b\)[/tex].
- Suppose we find coefficients [tex]\(a ≈ 30.7515\)[/tex] and [tex]\(b ≈ 0.0789\)[/tex].
4. Predict the Profit for 2010:
- In the rescaled year data, 2010 corresponds to [tex]\(x = 2010 - 2002 = 8\)[/tex].
- Substitute [tex]\(x = 8\)[/tex] into the power function: [tex]\(P(8) = a \cdot (8^b)\)[/tex].
5. Calculate the Predicted Profit:
- Using the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] we have determined, calculate [tex]\(P(8)\)[/tex].
- This gives us a predicted profit of approximately [tex]\(36.24\)[/tex] million dollars for the year 2010.
Given the predicted value does not exactly match any of the provided options but is closest to $36.2 million, which is slightly off from the calculated prediction, leading us to choose "None of the above" as the best fitting answer from the given options.
Thank you for reading the article The following table of values gives a company s annual profits in millions of dollars Rescale the data so that the year 2003 corresponds to. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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