Question

In a survey of 1000 family showed that 750 families had T.V. and 500 family had ipad, and 80 family had none of T.V. and ipad. Find the number of family who had both T.V. and ipad.

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Answer 1

330 families had both a TV and an ipad.

Explanation:

In a survey of 1000 families, we are told that 80 families had neither a TV or an ipad. Therefore, the number of families that had a TV, an ipad, or both devices is:

`\sf 1000 - 80 = 920`

Let x be the number of families who had both a TV and an ipad:

Given 750 families had a TV, 500 families had an ipad, and 920 families had a TV, an ipad, or both, then:

`(750-x)+(500-x)+x=920`

Simplify and solve for x:

`750+500-x=920`

`1250-x=920`

`x=1250-920`

`x=330`

Therefore, 330 families had both a TV and an ipad.

Answer 2

330 families

Explanation:

Let A be the set of families that have a TV, and B be the set of families that have an iPad. Then we know that:

n(A)= 750 (the size of set A is 750)

n(B)= 500 (the size of set B is 500)

n(A ∪ B) = 1000 - 80 = 920 (the size of the union of sets A and B is 920)

We want to find n(A ∩ B) (the size of the intersection of sets A and B), which represents the number of families that have both a TV and an iPad.

We can use the following formula:

n(A ∪ B)=n(A)+ n(B)- n(A ∩ B)

n(A ∩ B)=n(A)+ n(B)- n(A ∪ B)

Plugging in the numbers we know:

n(A ∩ B)= 750 + 500 - 920 = 330

Therefore, there are 330 families that have both a TV and an iPad.