Question

which regression model best fits the data set? (2, 13), (4, 8), (5, 7.5), (7, 7) (1 point) linear quadratic exponential growth exponential decay

Explanation:

Step1: Analyze the trend

As \(x\) (the first - value in the ordered pairs) increases (\(2\to4\to5\to7\)), \(y\) (the second - value in the ordered pairs) decreases (\(13\to8\to7.5\to7\)).
Exponential growth is when \(y\) increases as \(x\) increases, so we can rule out exponential growth.
For a quadratic function \(y = ax^{2}+bx + c\), the graph is a parabola. If \(a>0\), it opens upwards (first decreasing then increasing) and if \(a < 0\), it opens downwards (first increasing then decreasing). Here, \(y\) is only decreasing as \(x\) increases.
For an exponential decay function \(y=Ae^{-kx}+C\) (\(A,k>0\)), the rate of change is not constant. For a linear function \(y=mx + b\), the rate of change \(m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) is constant.
Let's calculate the slopes between consecutive points:
The slope between \((2,13)\) and \((4,8)\) is \(m_{1}=\frac{8 - 13}{4 - 2}=\frac{-5}{2}=-2.5\)
The slope between \((4,8)\) and \((5,7.5)\) is \(m_{2}=\frac{7.5 - 8}{5 - 4}=\frac{-0.5}{1}=-0.5\)
The slope between \((5,7.5)\) and \((7,7)\) is \(m_{3}=\frac{7 - 7.5}{7 - 5}=\frac{-0.5}{2}=-0.25\)

Although the slopes are not exactly the same (due to possible measurement or data - entry "noise" in a real - world or textbook - constructed data set), the general trend is that \(y\) changes in a non - parabolic (not first increasing then decreasing or vice - versa) and non - exponential - growth way. The most "straight - line - like" (in terms of the overall decreasing trend without the curvature of a parabola or the non - constant - ratio of an exponential) trend is linear.

Answer:

linear