Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
Answer:
25
Step-by-step explanation:
We'll compare it with the study of sets, the most noticeable operation of which is:
- n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
where,
n(A ∪ B) is the Union Set, i. e, the set that contains all the elements
n(A) is a subset of the Union Set
n(B) is another subset of the Union Set, and
n(A ∩ B) is the Intersection Set ,i.e, the set contains common elements from both A and B sets.
In the question:
- n(A ∪ B) = ? (the total number of students in the class who are into the above mentioned sports)
Let set A contains the students who play basketball and set B, the students who play Volleyball.
- n(A) = 25
- n(B) = 20
10 students play both of them, i. e.,
- n(A ∩ B) = 10 (as 10 students have common sports - Volleyball and Basketball)
Using the above operation:
=> n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
=> n(A ∪ B) = 25 + 20 - 10
=> n(A ∪ B) = 35
Final Step to the Answer:
The total number of students in the class
= students into the given sports + students who don't play any of them
- Total number of students in the class is 60
- Total number fo students playing basketball and volleyball is 35
The students who play neither = 60 - 35
= 25
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.