As an experienced tutor registered on UrbanPro, I can guide you through this problem. First, let's break down the sets:

• U = {x : x ∈ N, x ≤ 9} represents the set of natural numbers less than or equal to 9.
• A = {x : x is an even number, 0 < x < 10} represents the set of even numbers between 0 and 9.
• B = {2, 3, 5, 7} represents the set of prime numbers less than 10.

Now, we need to find the complement of the union of sets A and B, denoted as (A U B)'. The union of sets A and B (A U B) contains all elements that are in either set A or set B, or in both.

• A U B = {0, 2, 4, 6, 8, 3, 5, 7} (includes all even numbers from A and prime numbers from B)

Now, the complement of this union set (A U B)' contains all elements that are in the universal set U but not in the set (A U B).

• (A U B)' = U - (A U B)

Let's calculate:

U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

(A U B) = {0, 1, 2, 3, 4, 5, 6, 7, 8}

So, (A U B)' = {9}

Therefore, the set (A U B)' is {9}. This means it contains only the element 9, as it's the only element in the universal set U that is not in the union of sets A and B.

As an experienced tutor registered on UrbanPro, I'd approach this problem systematically. Firstly, let's denote:

• HH as the number of students who play hockey,
• BB as the number of students who play basketball,
• CC as the number of students who play cricket.

Given information:

• H=23H=23,
• B=15B=15,
• C=20C=20,
• H∩B=7H∩B=7,
• B∩C=5B∩C=5,
• H∩C=4H∩C=4,
• Total students=60Total students=60,
• Students not playing any game=15Students not playing any game=15.

Now, we can use the principle of inclusion-exclusion to find the answers.

(i) How many play hockey, basketball, and cricket? To find this, we'll use the formula: n(H∪B∪C)=n(H)+n(B)+n(C)−n(H∩B)−n(B∩C)−n(H∩C)+n(H∩B∩C)n(H∪B∪C)=n(H)+n(B)+n(C)−n(H∩B)−n(B∩C)−n(H∩C)+n(H∩B∩C)

n(H∪B∪C)=23+15+20−7−5−4+0=42n(H∪B∪C)=23+15+20−7−5−4+0=42

(ii) How many play hockey but not cricket? n(H∩C‾)=n(H)−n(H∩C)n(H∩C)=n(H)−n(H∩C) n(H∩C‾)=23−4=19n(H∩C)=23−4=19

(iii) How many play hockey and cricket but not basketball? n(H∩C∩B‾)=n(H∩C)−n(H∩B∩C)n(H∩C∩B)=n(H∩C)−n(H∩B∩C) n(H∩C∩B‾)=4−0=4n(H∩C∩B)=4−0=4

So, as an experienced tutor on UrbanPro, I'd explain these concepts to my students using clear and concise language, ensuring they understand the logic behind the calculations.

As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into the properties of sets to prove the given statement.

To prove: A−(A∩B)=A−BA−(A∩B)=A−B

Let's break it down step by step using set operations:

1. Left-hand side (LHS) of the equation: A−(A∩B)A−(A∩B)

   • This means we're taking all elements in set A that are not in the intersection of sets A and B.

2. Right-hand side (RHS) of the equation: A−BA−B

   • This means we're taking all elements in set A that are not in set B.

To prove the equality, we need to show that the LHS is equal to the RHS.

Proof:

1. Let's start with the LHS: A−(A∩B)A−(A∩B)

   • This means we're removing from set A all elements that are in both sets A and B.

2. Now, let's consider the RHS: A−BA−B

   • This means we're removing from set A all elements that are in set B.

3. If we compare the two operations:

   • On the LHS, we're removing elements from A that are in both A and B.
   • On the RHS, we're removing elements from A that are in set B.

4. It follows that whatever elements were in both A and B, they are removed in both operations.

   • So, A−(A∩B)A−(A∩B) and A−BA−B both remove the same elements from set A.

5. Therefore, A−(A∩B)=A−BA−(A∩B)=A−B, proving the equality.

This completes the proof, demonstrating that for all sets A and B, A−(A∩B)A−(A∩B) is indeed equal to A−BA−B. This principle is fundamental in set theory and can be useful in various mathematical and logical contexts.