In a plane, points p and q are 20 inches apart. if point r is randomly chosen from all the points in the plane that are 20 inches from p, what is the probability that r is closer to p than it is to q?

Step-by-step explanation:

All the points in the plane that are 20 inches from P constituing a circle with center in P and with radius 20 inch

We need find the angle in this circle by which the point R is closer to P than it is to Q. The limit situation occurs when the distances from P to R and from R to Q are equals and have the value 20 inches. In this situation the points P, R y Q form a equilater triangle with angles of value 60°.

Thus, the point R is closer to P than it is to Q in 240° of the circle, except 60° above of the line PQ and 60° below the line PQ.

Then the probability is

3201

step-by-step explanation:

y < -3

step-by-step explanation:

-3y - 2 > 7

-3y -2 + 2 > 7+2

-3y > 9

-3y/-3 < 9/-3   (change the inequality sign when multiplying or dividing by a negative number)

y < -3