To determine the correct quadratic function based on the provided data and to find the profit for the year 2010, we need to use the given data points (x, y):
x: 0, 2, 3, 5, 8, 11
y: 5.7, 4.9, 4.8, 5.2, 7.3, 11.2
Steps to Find the Quadratic Function that Best Models the Data:
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Using the Data Points:
- We can use regression analysis to find the quadratic function \(f(x) = ax^2 + bx + c\) that fits the points.
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Calculating the Coefficients:
- Let's assume that you already performed this regression and obtained a quadratic function \(f(x)\).
Given that we have potential functions and results from calculations, we need to evaluate them:
- \(f(x) = 0.14x^2 - 0.94x + 5.93\)
- \(f(x) = 0.1x^2 - 0.6x + 5.7\)
The year 2010 corresponds to \(x = 20\) (since x represents the years since 1990).
Finding Profit for 2010:
- Using the proposed functions to find the values at \( x = 20 \):
For the first function:
\[ f(20) = 0.14(20^2) - 0.94(20) + 5.93 \] \[ = 0.14(400) - 18.8 + 5.93 \] \[ = 56 - 18.8 + 5.93 \] \[ = 56 - 18.8 + 5.93 = 43.13 \text{ (in millions of dollars)} \]
For the second function:
\[ f(20) = 0.1(20^2) - 0.6(20) + 5.7 \] \[ = 0.1(400) - 12 + 5.7 \] \[ = 40 - 12 + 5.7 \] \[ = 40 - 12 + 5.7 = 33.7 \text{ (in millions of dollars)} \]
Profit in Dollars:
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From \(f(20) = 43.13\) million, the profit would be: \[ 43.13 \text{ million} \rightarrow 43,130,000 \]
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From \(f(20) = 33.7\) million, the profit would be: \[ 33.7 \text{ million} \rightarrow 33,700,000 \]
Conclusion:
Based on the evaluations, the profit for the year 2010 is:
- $43,130,000 from the function \(f(x) = 0.14x^2 - 0.94x + 5.93\).
- $33,700,000 from the function \(f(x) = 0.1x^2 - 0.6x + 5.7\).
Thus, it appears you have calculated the profit from both functions correctly.
If you have to choose based on the provided functions,
- The profit of $43,130,000 (from \(f(x) = 0.14x^2 - 0.94x + 5.93\)) seems to be the correct result for the profit in 2010.
2 answers