In a survey of a town,it was found that 65 % | vedclass

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(C) Let $A$ be the set of people who watch news on $T.V.$ and $B$ be the set of people who read a newspaper.Given: $n(A) = 65 \%$,$n(B) = 40 \%$,and $

Vedclass · Accepted Answer

$20$

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(C) Let $A$ be the set of people who watch news on $T.V.$ and $B$ be the set of people who read a newspaper.Given: $n(A) = 65 \%$,$n(B) = 40 \%$,and $n(A \cap B) = 25 \%$.The percentage of people who watch news on $T.V.$ or read a newspaper is given by the union formula:$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$n(A \cup B) = 65 \% + 40 \% - 25 \% = 80 \%$.The percentage of people who neither watch news on $T.V.$ nor read a newspaper is:$100 \% - n(A \cup B) = 100 \% - 80 \% = 20 \%$.

In a survey of a town,it was found that $65 \%$ of the people surveyed watch the news on $T.V.$,$40 \%$ read a newspaper,and $25 \%$ read a newspaper and watch the news on $T.V.$ What percent of the people surveyed neither watch the news on $T.V.$ nor read a newspaper?

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