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(C) Let $A$ be the set of Americans who like cheese and $B$ be the set of Americans who like apples.Let the total population of Americans be $100$.Giv

Vedclass · Accepted Answer

$39 \le x \le 63$

Vedclass · Answer

(C) Let $A$ be the set of Americans who like cheese and $B$ be the set of Americans who like apples.Let the total population of Americans be $100$.Given: $n(A) = 63$ and $n(B) = 76$.Using the formula for the union of two sets: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.$n(A \cup B) = 63 + 76 - n(A \cap B) = 139 - n(A \cap B)$.Since $n(A \cup B) \le 100$,we have $139 - n(A \cap B) \le 100$,which implies $n(A \cap B) \ge 39$.Also,the intersection of two sets is a subset of each set,so $n(A \cap B) \le n(A)$ and $n(A \cap B) \le n(B)$.Therefore,$n(A \cap B) \le 63$ and $n(A \cap B) \le 76$. The tighter bound is $n(A \cap B) \le 63$.Combining these,we get $39 \le n(A \cap B) \le 63$.Since $x = n(A \cap B)$,we have $39 \le x \le 63$.

$A$ survey shows that $63\%$ of the Americans like cheese whereas $76\%$ like apples. If $x\%$ of the Americans like both cheese and apples,then

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