Answer :
Let's analyze the scatter plot data provided in the form of [tex]\( (x, y) \)[/tex] pairs:
[tex]\[ (1, 11), (4, 10), (5, 8), (8, 7), (10, 6) \][/tex]
To determine which equation best represents the trend line for the data, we need to derive the slope and intercept of the linear trend line that fits the data.
Here are the potential equations to be compared:
1. [tex]\( y = \frac{3}{5}x + 12 \)[/tex]
2. [tex]\( y = -\frac{3}{5}x - 12 \)[/tex]
3. [tex]\( y = -\frac{3}{5}x + 12 \)[/tex]
First, let's recall the formula for a linear equation:
[tex]\[ y = mx + b \][/tex]
where:
- [tex]\( m \)[/tex] is the slope of the trend line,
- [tex]\( b \)[/tex] is the y-intercept of the trend line.
From the data provided and the true solution, the derived slope ([tex]\( m \)[/tex]) and y-intercept ([tex]\( b \)[/tex]) for the best fit trend line are calculated as:
[tex]\[ \text{slope} = -0.573170731707317 \][/tex]
[tex]\[ \text{intercept} = 11.609756097560972 \][/tex]
Now, let's consider the potential equations one by one and compare their characteristics to the trend line parameters we have:
### Equation 1: [tex]\( y = \frac{3}{5}x + 12 \)[/tex]
- Slope ([tex]\( m \)[/tex]): [tex]\( \frac{3}{5} = 0.6 \)[/tex]
- Intercept ([tex]\( b \)[/tex]): 12
This slope value (0.6) does not match the calculated slope (-0.573), and the intercept (12) also differs from the calculated intercept (11.61).
### Equation 2: [tex]\( y = -\frac{3}{5}x - 12 \)[/tex]
- Slope ([tex]\( m \)[/tex]): [tex]\( -\frac{3}{5} = -0.6 \)[/tex]
- Intercept ([tex]\( b \)[/tex]): -12
Here, the slope (-0.6) is closer to the calculated slope (-0.573), but the intercept (-12) is very different from the calculated intercept (11.61).
### Equation 3: [tex]\( y = -\frac{3}{5}x + 12 \)[/tex]
- Slope ([tex]\( m \)[/tex]): [tex]\( -\frac{3}{5} = -0.6 \)[/tex]
- Intercept ([tex]\( b \)[/tex]): 12
The slope (-0.6) is close to the calculated slope (-0.573), and the intercept (12) is reasonably close to the calculated intercept (11.61).
Given that we need the trend line equation that best represents the data, the closest match to the true calculated slope and intercept is Equation 3. Therefore,
[tex]\[ y = -\frac{3}{5}x + 12 \][/tex]
is the equation that best represents the trend line for the given data.
[tex]\[ (1, 11), (4, 10), (5, 8), (8, 7), (10, 6) \][/tex]
To determine which equation best represents the trend line for the data, we need to derive the slope and intercept of the linear trend line that fits the data.
Here are the potential equations to be compared:
1. [tex]\( y = \frac{3}{5}x + 12 \)[/tex]
2. [tex]\( y = -\frac{3}{5}x - 12 \)[/tex]
3. [tex]\( y = -\frac{3}{5}x + 12 \)[/tex]
First, let's recall the formula for a linear equation:
[tex]\[ y = mx + b \][/tex]
where:
- [tex]\( m \)[/tex] is the slope of the trend line,
- [tex]\( b \)[/tex] is the y-intercept of the trend line.
From the data provided and the true solution, the derived slope ([tex]\( m \)[/tex]) and y-intercept ([tex]\( b \)[/tex]) for the best fit trend line are calculated as:
[tex]\[ \text{slope} = -0.573170731707317 \][/tex]
[tex]\[ \text{intercept} = 11.609756097560972 \][/tex]
Now, let's consider the potential equations one by one and compare their characteristics to the trend line parameters we have:
### Equation 1: [tex]\( y = \frac{3}{5}x + 12 \)[/tex]
- Slope ([tex]\( m \)[/tex]): [tex]\( \frac{3}{5} = 0.6 \)[/tex]
- Intercept ([tex]\( b \)[/tex]): 12
This slope value (0.6) does not match the calculated slope (-0.573), and the intercept (12) also differs from the calculated intercept (11.61).
### Equation 2: [tex]\( y = -\frac{3}{5}x - 12 \)[/tex]
- Slope ([tex]\( m \)[/tex]): [tex]\( -\frac{3}{5} = -0.6 \)[/tex]
- Intercept ([tex]\( b \)[/tex]): -12
Here, the slope (-0.6) is closer to the calculated slope (-0.573), but the intercept (-12) is very different from the calculated intercept (11.61).
### Equation 3: [tex]\( y = -\frac{3}{5}x + 12 \)[/tex]
- Slope ([tex]\( m \)[/tex]): [tex]\( -\frac{3}{5} = -0.6 \)[/tex]
- Intercept ([tex]\( b \)[/tex]): 12
The slope (-0.6) is close to the calculated slope (-0.573), and the intercept (12) is reasonably close to the calculated intercept (11.61).
Given that we need the trend line equation that best represents the data, the closest match to the true calculated slope and intercept is Equation 3. Therefore,
[tex]\[ y = -\frac{3}{5}x + 12 \][/tex]
is the equation that best represents the trend line for the given data.