A) Given the points (1,18.35) and (29,17.5), we can find the linear model with the following formulas:
[tex]\begin{gathered} \text{slope:} \\ m=\frac{y_2-y_1}{x_2-x_1}=\frac{71.5-18.35}{29-1}=\frac{-0.85}{28}=-0.03 \\ \text{equation of the line:} \\ y-y_1=m(x-x_1) \\ \Rightarrow y-18.35=-0.03(x-1)=-0.03x+0.03 \\ \Rightarrow y=-0.03x+0.03+18.35=-0.03x+18.38 \\ y=-0.03x+18.38 \end{gathered}[/tex]therefore, the linear model is y = -0.03x+18.38
B)We have the general cosine model:
[tex]y(t)=A+B\cos (\omega(t-\phi))[/tex]Where A is the vertical shift, B is the amplitude, w is the frequency and phi is the phase shift.
First, we can find the vertical shift with the following formula:
[tex]A=\frac{y_{\max }+y_{\min }}{2}[/tex]in this case, we have that the maximum value for y is 19.47 and the minimum value for y is16.18, then:
[tex]A=\frac{19.47+16.18}{2}=17.825[/tex]next, we can find the amplitud with the following formula:
[tex]B=y_{\max }-A[/tex]We have then:
[tex]B=19.47-17.825=1.645[/tex]Now, notice that the graph will repeat every 356 values for t, then, for the frequency we have the following expression:
[tex]\omega=\frac{2\pi}{356}=\frac{\pi}{178}[/tex]To find the phase shift, notice that for the point (172,19.47), we have the following:
[tex]\begin{gathered} y(172)=19.47 \\ \Rightarrow17.825+1.645\cos (\frac{\pi}{178}(172-\phi))=19.47 \\ \Rightarrow1.645\cos (\frac{\pi}{178}(172-\phi))=1.645 \\ \Rightarrow\cos (\frac{\pi}{178}(172-\phi))=1 \end{gathered}[/tex]notice that if the cosine equals 1, then its argument must equal to 0, then, we have:
[tex]\begin{gathered} \frac{\pi}{178}(172-\phi)=0 \\ \Rightarrow172-\phi=0 \\ \Rightarrow\phi=172 \end{gathered}[/tex]we have that the phase shift is phi = 172, then, the final cosine model is:
[tex]y(x)=17.825+1.465\cos (\frac{\pi}{178}(x-172))[/tex]