Answer: Third Option

Step-by-step explanation:
The equation of a line in the pending intersection form has the following form:

Where m is the slope of the line and b is the intersection with the y axis.
Observe in the graph that the data form a decreasing line. Then the adjustment line must have a negative slope
.
The first and the second option have positive slopes, therefore we discard them.
Notice in the scatter diagram that the intersection of the line with the y-axis (x = 0) is above 90.
The line of the fourth option has a value of
.
Therefore the line that best fits the data is the third option

Note that the line has a slope
and a value of 