The probability that a high school student in in the marching band is 0.47
so, 
The probability that a student plays a varsity sport is 0.32
so, 
the probability that a student is in the marching band and plays a varsity sport is 0.24
so, we get
P(M∩V)=0.24
a student plays a varsity sport if we know she is in the marching band
so,
P(V|M)=(P(V∩M))/(P(M))
now, we can plug values
and we get
So,
the probability that a student plays a varsity sport if we know she is in the marching band is 0.51063......Answer
Answer:
1/3
Step-by-step explanation:
Probability of solving ten problems by Carl
Given
Total number of problems = 20
Number of problems studied by Carl = 15
Number of problems out of the 20 set of problem coming in test is 10
Case I - All ten problems come from the set of known 15 question
10/15 * 10/20 = 1/3
I have no clue what to answer 9
Answer:
66 ≤ f ≤100
Explanation
Mean= ( Σ x ) / n
Mean= sum of scores/ number of subject she took
Now, she already too 3 subject which sum is 85+83+86=254
Now we need to know range of score for her to have (grade) a mark between 80 and 89
Now let take the lower limit mean=80
The lowest score she can get is
Mean = ( Σx) / n
80=(85+83+86+f)/4
80×4= 254+f
Therefore, f= 320-254=66
Therefore the minimum score she can have to have a B is 66.
Then, let take the upper limit mean 89. i.e the maximum she can have so that she don't have an A grade.
Mean = ( Σx) / n
89=( 83+85+86+f)/4
89×4= 254+f
f= 356-254
f=102.
Therefore this shows that she cannot have an A grade in the exam. The maximum score for the exam is 100.
There the range of score is 66 ≤ f ≤100 to have a B grade
66 ≤ f ≤100 answer
Since she cannot score 102 in the examination.
