In a town of 10,000 families,it was found tha | vedclass

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(B) Given total families $N = 10,000$.
$n(A) = 40\% 	ext{ of } 10,000 = 4,000$
$n(B) = 20\% 	ext{ of } 10,000 = 2,000$
$n(C) = 10\% 	ext{ of } 1

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(B) Given total families $N = 10,000$.
$n(A) = 40\% 	ext{ of } 10,000 = 4,000$
$n(B) = 20\% 	ext{ of } 10,000 = 2,000$
$n(C) = 10\% 	ext{ of } 10,000 = 1,000$
$n(A \cap B) = 5\% 	ext{ of } 10,000 = 500$
$n(B \cap C) = 3\% 	ext{ of } 10,000 = 300$
$n(A \cap C) = 4\% 	ext{ of } 10,000 = 400$
$n(A \cap B \cap C) = 2\% 	ext{ of } 10,000 = 200$
To find the number of families that buy newspaper $A$ only,we use the formula:
$n(A 	ext{ only}) = n(A) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C)$
$n(A 	ext{ only}) = 4,000 - 500 - 400 + 200$
$n(A 	ext{ only}) = 4,000 - 700 = 3,300$

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