If n(U) = 600,n(A) = 100,n(B) = 200,and n(A c | vedclass

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(B) According to De Morgan's Law,$n(\bar{A} \cap \bar{B}) = n(\overline{A \cup B})$.We know that $n(\overline{A \cup B}) = n(U) - n(A \cup B)$.First,c

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$350$

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(B) According to De Morgan's Law,$n(\bar{A} \cap \bar{B}) = n(\overline{A \cup B})$.We know that $n(\overline{A \cup B}) = n(U) - n(A \cup B)$.First,calculate $n(A \cup B)$ using the formula $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.Substituting the given values: $n(A \cup B) = 100 + 200 - 50 = 250$.Now,substitute this into the universal set equation: $n(\bar{A} \cap \bar{B}) = 600 - 250 = 350$.

If $n(U) = 600$,$n(A) = 100$,$n(B) = 200$,and $n(A \cap B) = 50$,then $n(\bar{A} \cap \bar{B})$ is (where $U$ is the universal set and $A$ and $B$ are subsets of $U$).

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