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# Please find the answer above question. ..................................................................................,........ ## 2 Answers

one year ago

n(F)=38

n(C)=20

n(B)=15

N(FᴒCᴒB)=3

N(FuCuB)=58

n(FuCuB)=n(F)+n(C)+n(B)- n(FᴒC) - n(BᴒC) - n(FᴒB) +n(FᴒBᴒC)

58=38+20+15 - N(FᴒC) - N(BᴒC)-N(FᴒB)+3

58=76 -N(FᴒC) - N(BᴒC)-N(FᴒB)+3

N(FᴒC) - N(BᴒC)-N(FᴒB)=76-58=18

a+3+c+3+b+3=18 (we ascertain this frm venn diagram)

a+b+c+9=18

a+b+c=9

therefore, 9 received medal inexactly 2 of the sports

one year ago
Dear student
Let x get exactly one mrdal and y get 2 medal and as stated 3 men get 3 medals.
therefore
x+2y+3*3=38+15+20=73
x+2y=64
also
x+y+3=58= total no. of participants
x+y=55
solving we get
y=9

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