Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
To solve this problem, we will approach it step-by-step:
### Part (a): Graph the Function
1. Identify the equation:
[tex]\[ y = 0.401x^2 + 20.836x + 171.4 \][/tex]
Here, \( y \) represents the spending in millions of dollars, and \( x \) represents the number of years since 1988.
2. Determine the range for \( x \):
You are asked to graph the function for \( x \) in the interval [0, 10].
3. Calculate \( y \) values:
Substitute various values of \( x \) from 0 to 10 into the equation to find the corresponding \( y \) values. Here are some calculations for selected points:
- For \( x = 0 \): \( y = 0.401(0)^2 + 20.836(0) + 171.4 = 171.4 \)
- For \( x = 2 \): \( y = 0.401(2)^2 + 20.836(2) + 171.4 = 218.408 \)
- For \( x = 4 \): \( y = 0.401(4)^2 + 20.836(4) + 171.4 = 279.944 \)
- For \( x = 6 \): \( y = 0.401(6)^2 + 20.836(6) + 171.4 = 355.008 \)
- For \( x = 8 \): \( y = 0.401(8)^2 + 20.836(8) + 171.4 = 443.6 \)
- For \( x = 10 \): \( y = 0.401(10)^2 + 20.836(10) + 171.4 = 545.72 \)
4. Plot the function:
You can now plot these points and draw a smooth curve through them. On the x-axis, mark the years from 0 (1988) to 10 (1998), and on the y-axis, mark the spending in millions of dollars.
### Part (b): Find the Spending in 2001
1. Determine \( x \) for the year 2001:
Since \( x = 0 \) corresponds to 1988, \( x \) in 2001 would be:
[tex]\[ x = 2001 - 1988 = 13 \][/tex]
2. Substitute \( x = 13 \) into the equation to find \( y \):
[tex]\[ y = 0.401(13)^2 + 20.836(13) + 171.4 \][/tex]
3. Solve for \( y \):
[tex]\[ y = 0.401(169) + 20.836(13) + 171.4 \][/tex]
[tex]\[ y = 67.769 + 271.568 + 171.4 \][/tex]
[tex]\[ y = 510.737 \][/tex]
So, the spending by this model in 2001 is approximately \( 510.737 \) million dollars.
### Part (c): Determine if the Value is an Interpolation or an Extrapolation
1. Define interpolation and extrapolation:
- Interpolation is the process of estimating unknown values that fall within the range of known values.
- Extrapolation is the process of estimating values outside the range of known values.
2. Check the range of known values:
The model is provided for \( x \) from 0 (1988) to 10 (1998).
3. Determine the nature of the estimate:
Since the year 2001 (where \( x = 13 \)) falls outside the given range of 1988 to 1998 (where \( x \) ranges from 0 to 10), the value calculated in part (b) is an extrapolation.
In summary:
- Graph: Plot the equation over the interval [0, 10].
- Spending in 2001: Approximately \$510.737 million.
- Nature of Estimate: The value is an extrapolation.
### Part (a): Graph the Function
1. Identify the equation:
[tex]\[ y = 0.401x^2 + 20.836x + 171.4 \][/tex]
Here, \( y \) represents the spending in millions of dollars, and \( x \) represents the number of years since 1988.
2. Determine the range for \( x \):
You are asked to graph the function for \( x \) in the interval [0, 10].
3. Calculate \( y \) values:
Substitute various values of \( x \) from 0 to 10 into the equation to find the corresponding \( y \) values. Here are some calculations for selected points:
- For \( x = 0 \): \( y = 0.401(0)^2 + 20.836(0) + 171.4 = 171.4 \)
- For \( x = 2 \): \( y = 0.401(2)^2 + 20.836(2) + 171.4 = 218.408 \)
- For \( x = 4 \): \( y = 0.401(4)^2 + 20.836(4) + 171.4 = 279.944 \)
- For \( x = 6 \): \( y = 0.401(6)^2 + 20.836(6) + 171.4 = 355.008 \)
- For \( x = 8 \): \( y = 0.401(8)^2 + 20.836(8) + 171.4 = 443.6 \)
- For \( x = 10 \): \( y = 0.401(10)^2 + 20.836(10) + 171.4 = 545.72 \)
4. Plot the function:
You can now plot these points and draw a smooth curve through them. On the x-axis, mark the years from 0 (1988) to 10 (1998), and on the y-axis, mark the spending in millions of dollars.
### Part (b): Find the Spending in 2001
1. Determine \( x \) for the year 2001:
Since \( x = 0 \) corresponds to 1988, \( x \) in 2001 would be:
[tex]\[ x = 2001 - 1988 = 13 \][/tex]
2. Substitute \( x = 13 \) into the equation to find \( y \):
[tex]\[ y = 0.401(13)^2 + 20.836(13) + 171.4 \][/tex]
3. Solve for \( y \):
[tex]\[ y = 0.401(169) + 20.836(13) + 171.4 \][/tex]
[tex]\[ y = 67.769 + 271.568 + 171.4 \][/tex]
[tex]\[ y = 510.737 \][/tex]
So, the spending by this model in 2001 is approximately \( 510.737 \) million dollars.
### Part (c): Determine if the Value is an Interpolation or an Extrapolation
1. Define interpolation and extrapolation:
- Interpolation is the process of estimating unknown values that fall within the range of known values.
- Extrapolation is the process of estimating values outside the range of known values.
2. Check the range of known values:
The model is provided for \( x \) from 0 (1988) to 10 (1998).
3. Determine the nature of the estimate:
Since the year 2001 (where \( x = 13 \)) falls outside the given range of 1988 to 1998 (where \( x \) ranges from 0 to 10), the value calculated in part (b) is an extrapolation.
In summary:
- Graph: Plot the equation over the interval [0, 10].
- Spending in 2001: Approximately \$510.737 million.
- Nature of Estimate: The value is an extrapolation.
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
