Mathematics: Let the universal set, s, have 203 elements. a and b are subsets of s. set a contains 98 elements and set b contains 81 elements. if the total number of elements in either a or b is 173, how many elements are in a but not in b?

Let the universal set, s, have 203 elements. a

cici86

16.03.2022

There are 92 elements in A but not in B.

Step-by-step explanation:

Since we have given that

n(S) = 203

n(A) = 98

n(B) = 81

n(A∪ B) = 173

We first find the value of n(A∩B).

$n(A\cup B)=n(A)+n(B)-n(A\cap B)\\\\173=98+81-n(A\cap B)\\\\173=179-n(A\cap B)\\\\173-179=-n(A\cap B)\\\\6=n(A\cap B)$

We need to find the number of elements that are in A but not in B.

$n(A-B)=n(A)-n(A\cap B)\\\\n(A-B)=98-6\\\\n(A-B)=92$

Hence, there are 92 elements in A but not in B.

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20jmurphy82

16.03.2022

Step-by-step explanation:

it is given that total number of elements =203

set A contains 98 that is A=98

set B contain 81 elements B=81

It is given that either A or B is 173

we know the formula

$n\left ( A\cup B \right )=n(A)+n(B)-n(A\cap B)$

173=98+81 - $n\left ( A\cap B \right )$

$n\left ( A\cap B \right )$ = 6

so the elements in A but not B =98-6=92

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christylivingsowzxa2

16.11.2020

I think its B

Step-by-step explanation:

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