Fill in the blanks to make each of the following a true statement :
$\varnothing^ {\prime}\cap A$
$A$
$\varnothing^{\prime} \cap A=U \cap A=A$
$\therefore \varnothing^{\prime} \cap A=A$
Taking the set of natural numbers as the universal set, write down the complements of the following sets:
$\{ x:x$ is a positive multiple of $3\} $
Given $n(U) = 20$, $n(A) = 12$, $n(B) = 9$, $n(A \cap B) = 4$, where $U$ is the universal set, $A$ and $B$ are subsets of $U$, then $n({(A \cup B)^C}) = $
If $n(U)$ = $600$ , $n(A)$ = $100$ , $n(B)$ = $200$ and $n(A \cap B )$ = $50$, then $n(\bar A \cap \bar B )$ is
($U$ is universal set and $A$ and $B$ are subsets of $U$)
$A \cap A^{\prime}=\ldots$
Let $A$ and $B$ be two sets then $(A \cup B)' \cup (A' \cap B)$ is equal to
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