a)We can take the points (3,68) and (7,60) to find an equation of the line in slope-intercept form:
[tex]\begin{gathered} \text{slope:} \\ m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{60-68}{7-3}=\frac{-8}{4}=-2 \\ m=-2 \\ \text{Equation in slope intercept form:} \\ y-y_1=m(x-x_1) \\ \Rightarrow y-68=-2(x-3)=-2x+6 \\ \Rightarrow y=-2x+6+68=-2x+74 \\ y=-2x+74 \end{gathered}[/tex]
therefore, a possible model for the trend line is y=-2x+74
b)The model can have some points that represent the trend, but some others may be a little off, but when the model is represented using some of the points and trying to get the minimum distance from the line to the points that not necesarily are on the line, the model can represent with some significance level the trend line.
