The given point (-3/5 , y) lies in the third quadrant.
It is also given that the point lies on a unit circle.
For a point (x,y) lying on a unit circle a and y are defined as:
x = cos θy = sin θSo, we can say for the point (-3/5 , y) the value -3/5 is equal to cos θ
sec θ is the reciprocal of cos θ.
So, sec θ = -5/3
Using Pythagorean identity we can first find sin θ.

Since the point lies in 3rd quadrant, both sin and cos will be negative.
So, now we can write:
Answers:sec θ = -5/3cot θ = 3/4
Answer:
1) Is more representative
Step-by-step explanation:
The problem with his selection is that maybe there are few students participating in certain sport and those students maybe do quite more excercise than the rest (or quite less). This will modify the results because the sample he selected is biased. This problem wont be solved by method 3 or 4, because he is still selecting students that may modify heavily the results with a high probability
This problem will also appear if he choose a sample by class. Maybe, in a class there are quite few students, and selecting from class will make those students appear quite more often than, lets say, a 7th grade student selected at random, therefore the selection is biased in this case as well.
If he has a list with all seventh grade students, each student is equallly likely to be selected and as a consequence, the the results wont be biased. Approach 1 is the best one.
B. Slope is rise over run