Answer:
C. [tex]y=x+200[/tex]
Step-by-step explanation:
Line of best fit (trendline) : a line through a scatter plot of data points that best expresses the relationship between those points.
All the given options for the line of best fit are linear equations.
Therefore, we can add the line of best fit to the graph (see attached), remembering to have roughly the same number of points above as below the line.
Linear equation: [tex]y=mx+b[/tex]
(where [tex]m[/tex] is the slope and [tex]b[/tex] is the y-intercept)
From inspection of the line of best fit, we can see that the y-intercept (where x = 0) is approximately 200. So this suggests that either option C or option D is the solution.
We can also see that slope (gradient) is approximately 1, as for every +100 increase in x, y increases by +100.
[tex]\textsf{slope = rate of change}=\sf\dfrac{change \ in \ y}{change \ in \ x}=\dfrac{100}{100}=1[/tex]
Therefore, this suggests that C is the solution and that the closest approximation to the line of best fit is [tex]y=x+200[/tex]
