If X and Y are two sets such that X cup Y has | vedclass

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(A) We are given that $n(X \cup Y) = 18$,$n(X) = 8$,and $n(Y) = 15$.Using the set theory formula for the union of two sets:$n(X \cup Y) = n(X) + n(Y)

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$5$

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(A) We are given that $n(X \cup Y) = 18$,$n(X) = 8$,and $n(Y) = 15$.Using the set theory formula for the union of two sets:$n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$Rearranging the formula to solve for $n(X \cap Y)$:$n(X \cap Y) = n(X) + n(Y) - n(X \cup Y)$Substituting the given values:$n(X \cap Y) = 8 + 15 - 18$$n(X \cap Y) = 23 - 18$$n(X \cap Y) = 5$Therefore,the number of elements in $X \cap Y$ is $5$.

If $X$ and $Y$ are two sets such that $X \cup Y$ has $18$ elements,$X$ has $8$ elements,and $Y$ has $15$ elements,how many elements does $X \cap Y$ have?

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