A survey shows that 63% of the Americans like | vedclass

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

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$39 \le x \le 63$

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(C) Let $A$ denote the set of Americans who like cheese and $B$ denote the set of Americans who like apples.Let the total population be $100$.Then $n(A) = 63$ and $n(B) = 76$.We know that $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.Since $n(A \cup B) \le 100$,we have $n(A) + n(B) - n(A \cap B) \le 100$.$63 + 76 - n(A \cap B) \le 100 \implies 139 - n(A \cap B) \le 100 \implies n(A \cap B) \ge 39$.Also,$A \cap B \subseteq A$ and $A \cap B \subseteq B$,so $n(A \cap B) \le n(A)$ and $n(A \cap B) \le n(B)$.Thus,$n(A \cap B) \le 63$ and $n(A \cap B) \le 76$,which implies $n(A \cap B) \le 63$.Combining these,we get $39 \le n(A \cap B) \le 63$.Therefore,$39 \le x \le 63$.

$A$ 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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