how many students both smoke and drink?
2006-12-17 10:42:22 · 17 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Drink and/or smoke = 100 -20
= 80
Drink or smoke = 60 +25
= 85
Drink and smoke = 85 - 80
= 5
2006-12-17 10:44:25 · answer #1 · answered by PIPI B 4 · 0⤊ 0⤋
5
2006-12-17 18:45:24 · answer #2 · answered by jay b 3 · 0⤊ 0⤋
5
2006-12-17 18:45:21 · answer #3 · answered by Anonymous · 0⤊ 0⤋
5
2006-12-17 18:44:40 · answer #4 · answered by teekshi33 4 · 0⤊ 0⤋
5
2006-12-17 18:44:33 · answer #5 · answered by teef_au 6 · 0⤊ 0⤋
20 neither=100-20=80, 60+25=85-80=5
2006-12-17 18:44:23 · answer #6 · answered by Zidane 3 · 0⤊ 0⤋
D = drink
S = smoke
/\ = inter
U = union
(D U S) - 20 = n(D) + n(S) - (D /\ S)
100 - 20 = 60 + 25 - (D /\ S)
80 - 85 = -(D/\S)
-(D/\S) = -5
(D/\S) = 5
Answer: Five students smoke and drink.
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2006-12-17 18:51:03 · answer #7 · answered by aeiou 7 · 0⤊ 0⤋
Call people who JUST smoke S
Call people who JUST drink D
Call people who drink AND smoke B
20 neither drink or smoke, so 80 drink and/or smoke
60 drink so D + B = 60
25 smoke so S + B = 25
80 drink and/or smoke so S + D + B = 80
D + 2B + S = 85 (1)
D + B + S = 80 (2)
Subtract (2) from (1)
B = 5
So 5 people both drink and smoke
2006-12-17 19:13:07 · answer #8 · answered by Tom :: Athier than Thou 6 · 0⤊ 0⤋
Probability of drinking = p(d) = .6
Probability of smoking = p(s) = .25
Probability of neither drinking or smoking = p(0) = .2
Probability of both drinking and smoking = p(dâ©s)
P(dUs) = p(d) + p(s) - p(dâ©s) = 1 - p(0) = 1 - .2 = .8
p(d) + p(s) - .8 = p(dâ©s)
p(dâ©s) = p(d) + p(s) - .8
p(dâ©s) = .6 + .25 - .8 = .05
Number of students who both smoke and drink:
100*(.05) = 5
2006-12-17 19:01:39 · answer #9 · answered by Northstar 7 · 0⤊ 0⤋
5 students both smoke and drink
2006-12-17 18:48:12 · answer #10 · answered by imamulleith 2 · 0⤊ 0⤋