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:
35.25
Step-by-step explanation:
Give the data set:
23 37 49 34 35 41 40 26 32 22 38 42
We are expected to calculate the midquartile of the given data set.
22 23 26 32 34 35 37 38 40 41 42 49
First step is to find the lower quartile which comprises of
22 23 26 32 34 35
Here the Q1 is (26+32)/2 = 58/2= 29
Second step to find the upper quartile which comprises of
37 38 40 41 42 49
Here the Q3 is (40+41) /2 = 81/2 = 41.5
Then to find the midquartile which is (Q1+Q3) /2 where Q1 is 29 and Q3 is 41.5
= (29+41.5)/2
= (70.5) /2 = 35.25
I think step five is correct but no sure
365*24*60*60=31536000s
So third option will be the answer