9 < = 6 - y
9 - 6 < = -y
3 < = -y
-3 > = y or y < = -3
solutions include : -3,-4,-6
.375=3/8 (ignore this))))))))))))))))))))))))))))))))))))))))))))))))))))
Answer:
<em>Two possible answers below</em>
Step-by-step explanation:
<u>Probability and Sets</u>
We are given two sets: Students that play basketball and students that play baseball.
It's given there are 29 students in certain Algebra 2 class, 10 of which don't play any of the mentioned sports.
This leaves only 29-10=19 players of either baseball, basketball, or both sports. If one student is randomly selected, then the propability that they play basketball or baseball is:

P = 0.66
Note: if we are to calculate the probability to choose one student who plays only one of the sports, then we proceed as follows:
We also know 7 students play basketball and 14 play baseball. Since 14+7 =21, the difference of 21-19=2 students corresponds to those who play both sports.
Thus, there 19-2=17 students who play only one of the sports. The probability is:

P = 0.59
Answer:
A. { }
Step-by-step explanation:
Absolute value cannot be less than zero.
-4 < 0
There is no solution for x, ∈, R.
Hope this helps.
Answer:
4 because the edges are like vertices.