Sets- 9,10,11,12. I can't understand the question. Please help me with the answer.
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Answered by
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n(A∪B)=n(A)+n(B)-n(A∩B)
therefore, 30=15+n(B)-6
therefore, n(B)=21
therefore, 30=15+n(B)-6
therefore, n(B)=21
Madhushri:
Thanks!!
Answered by
1
n(P∪Q) = n(P) + n(Q) - n(P∩Q)
substituting the values
15 = 10 + n(Q) - 5
15 = 5 + n( Q)
n(Q) = 10
------------------------------------------------------------------------------------------------
n(T∪C) = n(T) + n(C) - n(T∩C)substituting the values
35 = 20 + 22 - n(T∩C)
35 = 44 - n(T∩C)
n(T∩C) = 9
------------------------------------------------------------------------------------------------------
n(P∪Q) = n(P) + n(Q) - n(P∩Q)
substituting the values
65 = 45 + 35 - n(P∩Q)
65 = 80 - n(P∩Q)
n(P∩Q) = 15
--------------------------------------------------------------------------------------------------
n(A∪B)=n(A)+n(B)-n(A∩B)
substituting the values
30 = 15 + n(B) - 6
30 = 9 + n(B)
n(B) = 21
---------------------------------------------------------------------------------------------------
substituting the values
15 = 10 + n(Q) - 5
15 = 5 + n( Q)
n(Q) = 10
------------------------------------------------------------------------------------------------
n(T∪C) = n(T) + n(C) - n(T∩C)substituting the values
35 = 20 + 22 - n(T∩C)
35 = 44 - n(T∩C)
n(T∩C) = 9
------------------------------------------------------------------------------------------------------
n(P∪Q) = n(P) + n(Q) - n(P∩Q)
substituting the values
65 = 45 + 35 - n(P∩Q)
65 = 80 - n(P∩Q)
n(P∩Q) = 15
--------------------------------------------------------------------------------------------------
n(A∪B)=n(A)+n(B)-n(A∩B)
substituting the values
30 = 15 + n(B) - 6
30 = 9 + n(B)
n(B) = 21
---------------------------------------------------------------------------------------------------
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