Let n U -700- n A -200- n B -300- n A - B -100- then n A - B -A- 400B- 600C- 500D- 300

The correct option is D 300By De Morgan's Second Law, we have n(A'∩B')=n(A∪B)'=n(U) n(A∪B)=n(U) n(A)+n(B) n(A∩B)=700 200+300 100⇒n(A'∩B')=300 nbsp; ...

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Standard X

Mathematics

Demorgan's law

Let n U =700,...

Question

$Let n (U) = 700, n (A) = 200, n (B) = 300, n (A \cap B) = 100, then n (A' \cap B') =$

400

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600

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500

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300

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Solution

The correct option is D 300
By De Morgan's Second Law, we have
$n (A' \cap B') = n (A \cup B)' = n (U) - n (A \cup B) = n (U) - (n (A) + n (B) - n (A \cap B)) = 700 - (200 + 300 - 100) \Rightarrow n (A' \cap B') = 300$

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Let n(U) = 700, n(A) = 200, n(B) = 300, n(A∩B)=100, then n(A'∩B')=

The correct option is D 300By De Morgan's Second Law, we have n(A'∩B')=n(A∪B)'=n(U)−n(A∪B)=n(U)−(n(A)+n(B)−n(A∩B))=700−(200+300−100)⇒n(A'∩B')=300

$Let n (U) = 700, n (A) = 200, n (B) = 300, n (A \cap B) = 100, then n (A' \cap B') =$

The correct option is D 300
By De Morgan's Second Law, we have
$n (A' \cap B') = n (A \cup B)' = n (U) - n (A \cup B) = n (U) - (n (A) + n (B) - n (A \cap B)) = 700 - (200 + 300 - 100) \Rightarrow n (A' \cap B') = 300$