Question
he table gives the wind-chill temperature when the outside temperature is 20°F. Complete parts (a) through (c) below.
Wind
(mph)
Wind
Chill (degreesF)
Wind
(mph)
Wind
Chill (degreesF)
5
13
35
0
10
10
40
negative 1
15
7
45
negative 2
20
5
50
negative 3
25
3
55
negative 3
30
1
60
negative 4
Question content area bottom
Part 1
a. Use x as the wind speed and create a quadratic model for the data.
yequals(
enter your response here)xsquaredplus(
enter your response here)xplus(
enter your response here)
(Type integers or decimals rounded to three decimal places as needed.)
Wind
(mph)
Wind
Chill (degreesF)
Wind
(mph)
Wind
Chill (degreesF)
5
13
35
0
10
10
40
negative 1
15
7
45
negative 2
20
5
50
negative 3
25
3
55
negative 3
30
1
60
negative 4
Question content area bottom
Part 1
a. Use x as the wind speed and create a quadratic model for the data.
yequals(
enter your response here)xsquaredplus(
enter your response here)xplus(
enter your response here)
(Type integers or decimals rounded to three decimal places as needed.)
Answers
GPT-4o mini
To create a quadratic model from the given data, we can fit a quadratic equation of the form:
\[
y = ax^2 + bx + c
\]
where \(y\) is the wind chill temperature and \(x\) is the wind speed (in mph).
We can use the data points provided to create a system of equations that we will then solve for the coefficients \(a\), \(b\), and \(c\). The data points provided are:
- (5, 13)
- (10, 10)
- (15, 7)
- (20, 5)
- (25, 3)
- (30, 1)
- (35, 0)
- (40, -1)
- (45, -2)
- (50, -3)
- (55, -3)
- (60, -4)
From these points, we will select three points to create the system of equations. Let's use:
1. \( (5, 13) \)
2. \( (20, 5) \)
3. \( (35, 0) \)
Using point (5, 13):
\[
13 = a(5^2) + b(5) + c
\]
\[
13 = 25a + 5b + c \quad \text{(1)}
\]
Using point (20, 5):
\[
5 = a(20^2) + b(20) + c
\]
\[
5 = 400a + 20b + c \quad \text{(2)}
\]
Using point (35, 0):
\[
0 = a(35^2) + b(35) + c
\]
\[
0 = 1225a + 35b + c \quad \text{(3)}
\]
Now, we have the following system of equations:
1. \( 25a + 5b + c = 13 \)
2. \( 400a + 20b + c = 5 \)
3. \( 1225a + 35b + c = 0 \)
To solve these equations, we can eliminate \(c\) by subtracting one equation from the other.
Subtracting equation (1) from equation (2):
\[
(400a + 20b + c) - (25a + 5b + c) = 5 - 13
\]
\[
375a + 15b = -8 \quad \text{(4)}
\]
Subtracting equation (2) from equation (3):
\[
(1225a + 35b + c) - (400a + 20b + c) = 0 - 5
\]
\[
825a + 15b = -5 \quad \text{(5)}
\]
Now we can solve equations (4) and (5).
First, let's isolate \(b\) in equation (4):
\[
15b = -8 - 375a
\]
\[
b = \frac{-8 - 375a}{15} \quad \text{(6)}
\]
Now substitute equation (6) into equation (5):
\[
825a + 15\left(\frac{-8 - 375a}{15}\right) = -5
\]
\[
825a - 8 - 375a = -5
\]
\[
450a - 8 = -5
\]
\[
450a = 3
\]
\[
a = \frac{1}{150} \approx 0.00667
\]
Now substitute \(a\) back into equation (6):
\[
b = \frac{-8 - 375\left(\frac{1}{150}\right)}{15}
\]
\[
b = \frac{-8 - 2.5}{15} = \frac{-10.5}{15} \approx -0.7
\]
Finally, substitute \(a\) and \(b\) back into one of the original equations to find \(c\), let's use equation (1):
\[
25\left(\frac{1}{150}\right) + 5(-0.7) + c = 13
\]
\[
\frac{25}{150} - 3.5 + c = 13
\]
\[
c = 13 + 3.5 - \frac{25}{150}
\]
\[
c \approx 16.6
\]
Now rounding the coefficients to three decimal places, our quadratic model can be summarized as:
\[
y \approx 0.007x^2 - 0.700x + 16.600
\]
Thus, the quadratic model is:
\[
y = 0.007x^2 - 0.700x + 16.600
\]
\[
y = ax^2 + bx + c
\]
where \(y\) is the wind chill temperature and \(x\) is the wind speed (in mph).
We can use the data points provided to create a system of equations that we will then solve for the coefficients \(a\), \(b\), and \(c\). The data points provided are:
- (5, 13)
- (10, 10)
- (15, 7)
- (20, 5)
- (25, 3)
- (30, 1)
- (35, 0)
- (40, -1)
- (45, -2)
- (50, -3)
- (55, -3)
- (60, -4)
From these points, we will select three points to create the system of equations. Let's use:
1. \( (5, 13) \)
2. \( (20, 5) \)
3. \( (35, 0) \)
Using point (5, 13):
\[
13 = a(5^2) + b(5) + c
\]
\[
13 = 25a + 5b + c \quad \text{(1)}
\]
Using point (20, 5):
\[
5 = a(20^2) + b(20) + c
\]
\[
5 = 400a + 20b + c \quad \text{(2)}
\]
Using point (35, 0):
\[
0 = a(35^2) + b(35) + c
\]
\[
0 = 1225a + 35b + c \quad \text{(3)}
\]
Now, we have the following system of equations:
1. \( 25a + 5b + c = 13 \)
2. \( 400a + 20b + c = 5 \)
3. \( 1225a + 35b + c = 0 \)
To solve these equations, we can eliminate \(c\) by subtracting one equation from the other.
Subtracting equation (1) from equation (2):
\[
(400a + 20b + c) - (25a + 5b + c) = 5 - 13
\]
\[
375a + 15b = -8 \quad \text{(4)}
\]
Subtracting equation (2) from equation (3):
\[
(1225a + 35b + c) - (400a + 20b + c) = 0 - 5
\]
\[
825a + 15b = -5 \quad \text{(5)}
\]
Now we can solve equations (4) and (5).
First, let's isolate \(b\) in equation (4):
\[
15b = -8 - 375a
\]
\[
b = \frac{-8 - 375a}{15} \quad \text{(6)}
\]
Now substitute equation (6) into equation (5):
\[
825a + 15\left(\frac{-8 - 375a}{15}\right) = -5
\]
\[
825a - 8 - 375a = -5
\]
\[
450a - 8 = -5
\]
\[
450a = 3
\]
\[
a = \frac{1}{150} \approx 0.00667
\]
Now substitute \(a\) back into equation (6):
\[
b = \frac{-8 - 375\left(\frac{1}{150}\right)}{15}
\]
\[
b = \frac{-8 - 2.5}{15} = \frac{-10.5}{15} \approx -0.7
\]
Finally, substitute \(a\) and \(b\) back into one of the original equations to find \(c\), let's use equation (1):
\[
25\left(\frac{1}{150}\right) + 5(-0.7) + c = 13
\]
\[
\frac{25}{150} - 3.5 + c = 13
\]
\[
c = 13 + 3.5 - \frac{25}{150}
\]
\[
c \approx 16.6
\]
Now rounding the coefficients to three decimal places, our quadratic model can be summarized as:
\[
y \approx 0.007x^2 - 0.700x + 16.600
\]
Thus, the quadratic model is:
\[
y = 0.007x^2 - 0.700x + 16.600
\]