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.