Answer:
![y=53.75x+70](https://img.qammunity.org/2023/formulas/mathematics/college/r4i6s01p4sm71tjgmhge6dumkabqe97mt3.png)
Explanation:
From inspection of the given graph with added trendline:
- y-intercept ≈ (0, 70)
- another point on the trendline ≈ (8, 500)
Find the slope of the trendline by substituting the identified points into the slope formula:
![\implies \textsf{slope}\:(m)=(y_2-y_1)/(x_2-x_1)=(500-70)/(8-0)=53.75](https://img.qammunity.org/2023/formulas/mathematics/college/zu56u7ckthpeztxr0arcp896mcfgnmy963.png)
![\boxed{\begin{minipage}{6.3 cm}\underline{Slope-intercept form of a linear equation}\\\\$y=mx+b$\\\\where:\\ \phantom{ww}$\bullet$ $m$ is the slope. \\ \phantom{ww}$\bullet$ $b$ is the $y$-intercept.\\\end{minipage}}](https://img.qammunity.org/2023/formulas/mathematics/college/bpvlu5vjaqnseqj3h5nxcjen8omzu35wqi.png)
Substitute the found slope and y-intercept into the slope-intercept formula to create an equation for the trendline:
![\implies y=53.75x+70](https://img.qammunity.org/2023/formulas/mathematics/college/ni6ta51o2dhyohmnryq2239gdrj5gn1ju1.png)