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:
x=-5
Step-by-step explanation:
opening the bracket on the LHS, we have
-8x-40= -3x+x-7-3
-8x-40= -2x-10
collecting like terms
-8x+2x= -10+40
-6x= 30
divide both sides by -6
x= -5
(368 points) / (32 games) = 11.5 points/game
(32 games) / (368 points) = 0.08696 game/point
ANSWER
My answer is in the photo above
sin(x) = 2cos(x)
tan(x) = 2
tan⁻¹[tan(x)] = tan⁻¹(2)
x ≈ 63.4
sin(2x) = sin[2(63.4)]
sin(2x) = sin(126.8)
sin(2x) ≈ 0.801