In a class of 50 students,27 are athletes 9 are musicians and 6 of the musicians are athletes . If one student is randomly selected from the class , find the probability of choosing an athlete or a musician

0.6

Step-by-step explanation:

Total population of class = 50

There are 27 athletes, and 9 musicians However, 6 of the musicians are also athletes.

Therefore, totals for athletes and musicians are:

27 + 9 -6 = 30

Thus, the probability of choosing an athlete or a musician is 30/50= 0.6

We could also use the formula:

P(A or M) = P(A) + P(M) - P(A and M)

P(A or M)=27/50+9/50-6/50

P(A or M)=30/50=0.6

I hope this helps!

here you hope this .

step-by-step explanation:

hypotenuse = √(15² + 8²)

=  √(225 +64)

=  √(289) = 17

sin(x) = cos(y) = 8/17

they are identical so long as you are doing the sine of one angle and the cosine of the only other angle that isn't the right angle.

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]