Asked by BlUbArRy!2#
13.
Is the relationship shown by the data linear? If so, model the data with an equation.
x y
1 –4
7 –7
13 –10
19 –13
The relationship is linear; y – 1 = –2(x + 4).
The relationship is not linear.
The relationship is linear; y + 4 = -1/2 (x – 1).
The relationship is linear; y + 4 = 2(x – 1).
Is the relationship shown by the data linear? If so, model the data with an equation.
x y
1 –4
7 –7
13 –10
19 –13
The relationship is linear; y – 1 = –2(x + 4).
The relationship is not linear.
The relationship is linear; y + 4 = -1/2 (x – 1).
The relationship is linear; y + 4 = 2(x – 1).
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To determine if the relationship between the data points is linear and to model it with an equation, we can analyze the changes in the values of \(x\) and \(y\).
Here is the given data:
\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
1 & -4 \\
7 & -7 \\
13 & -10 \\
19 & -13 \\
\hline
\end{array}
\]
Now, let's compute the differences in the \(x\) and \(y\) values to see if the relationship is linear.
### Calculate the differences
- From \(x = 1\) to \(x = 7\): \(\Delta x = 7 - 1 = 6\) and \(\Delta y = -7 - (-4) = -3\)
- From \(x = 7\) to \(x = 13\): \(\Delta x = 13 - 7 = 6\) and \(\Delta y = -10 - (-7) = -3\)
- From \(x = 13\) to \(x = 19\): \(\Delta x = 19 - 13 = 6\) and \(\Delta y = -13 - (-10) = -3\)
### Calculate the slope
The change in \(y\) is consistently \(-3\) for every increase of \(6\) in \(x\). Therefore, the slope \(m\) can be calculated as follows:
\[
m = \frac{\Delta y}{\Delta x} = \frac{-3}{6} = -\frac{1}{2}
\]
### Use point-slope form to derive the equation
Using the point \( (1, -4) \) with the slope \( m = -\frac{1}{2} \):
Using the point-slope formula:
\[
y - y_1 = m(x - x_1)
\]
Substituting into the formula:
\[
y + 4 = -\frac{1}{2}(x - 1)
\]
This equation matches one of the options presented. Thus, the correct modeling of the data reflects:
**The relationship is linear; \( y + 4 = -\frac{1}{2}(x - 1) \).**
Here is the given data:
\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
1 & -4 \\
7 & -7 \\
13 & -10 \\
19 & -13 \\
\hline
\end{array}
\]
Now, let's compute the differences in the \(x\) and \(y\) values to see if the relationship is linear.
### Calculate the differences
- From \(x = 1\) to \(x = 7\): \(\Delta x = 7 - 1 = 6\) and \(\Delta y = -7 - (-4) = -3\)
- From \(x = 7\) to \(x = 13\): \(\Delta x = 13 - 7 = 6\) and \(\Delta y = -10 - (-7) = -3\)
- From \(x = 13\) to \(x = 19\): \(\Delta x = 19 - 13 = 6\) and \(\Delta y = -13 - (-10) = -3\)
### Calculate the slope
The change in \(y\) is consistently \(-3\) for every increase of \(6\) in \(x\). Therefore, the slope \(m\) can be calculated as follows:
\[
m = \frac{\Delta y}{\Delta x} = \frac{-3}{6} = -\frac{1}{2}
\]
### Use point-slope form to derive the equation
Using the point \( (1, -4) \) with the slope \( m = -\frac{1}{2} \):
Using the point-slope formula:
\[
y - y_1 = m(x - x_1)
\]
Substituting into the formula:
\[
y + 4 = -\frac{1}{2}(x - 1)
\]
This equation matches one of the options presented. Thus, the correct modeling of the data reflects:
**The relationship is linear; \( y + 4 = -\frac{1}{2}(x - 1) \).**
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