To solve this problem, we need to find the union of two sets: \( (A \cap B) \) and \( (B \cap C) \).
Set \( A = \{2, 4, 6, 8, 10, 12\} \) and Set \( B = \{8, 10, 12, 14, 16, 18\} \).
The intersection \( A \cap B \) includes all elements common to both sets A and B.
\( A \cap B = \{8, 10, 12\} \).
Set \( B = \{8, 10, 12, 14, 16, 18\} \) and Set \( C = \{7, 8, 9, 10, 11, 12, 13\} \).
The intersection \( B \cap C \) includes all elements common to both sets B and C.
\( B \cap C = \{8, 10, 12\} \).
The union of \( A \cap B \) and \( B \cap C \) combines all elements from both intersections, without repeating any elements.
\( (A \cap B) \cup (B \cap C) = \{8, 10, 12\} \cup \{8, 10, 12\} = \{8, 10, 12\} \).
Since both intersections are identical, the union remains the same set of elements.
Conclusion: The result of the operation \( (A \cap B) \cup (B \cap C) \) is the set \(\{8, 10, 12\}\). Therefore, the correct option is
{8, 10, 12}
.
| $X_i$ | 5 | 6 | 8 | 10 |
| $F_i$ | 8 | 10 | 10 | 12 |
| X | 0 | 1 | 2 | 3 | 4 | 5 |
| P(X) | 0 | K | 2K | 3K | 4K | 5K |
