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defon
2 years ago
7

what is the answer here 90 student went to the zoo, 3 had hamburger milk and cake; had 5 milk and hamburger; 10 had cake and mil

k; 8 had cake and hamburger; 14 had hamburger; 37 had cake; 20 had milk nothing
Mathematics
1 answer:
dangina [55]2 years ago
4 0

Answer:

a) 37

b) 2

c) 17

d) 8

Step-by-step explanation:

90 Students went to the zoo. 3 had hamburger, milk and cake; 5 had milk and hamburger, 10 had cake and milk; 8 had cake and hamburger; 24 had hamburger; 38 had cake; 20 had milk. How many had a. nothing b. cake only c. milk only d. hamburger only

Solution:

Let h represent students that ate hamburger, m represent students that had milk and c represent students that had cake.

Given that:

n(h ∩ m ∩ c) = 3, n(m ∩ h) = 5, n(c ∩ m) = 10, n(c ∩ h) = 8, n(h) = 24, n(c) = 38, n(m) = 20

The number of students that had nothing = n(h  ∪ m ∪ C)'

The number of students that had only milk = n(m ∩ h' ∩ C')

The number of students that had only cake = n(m' ∩ h' ∩ C)

The number of students that had only hamburger = n(m' ∩ h ∩ C')

a) n(m ∩ h' ∩ C') = n(m) - n(m ∩ h) - n(c ∩ m) - n(h ∩ m ∩ c) = 20 - 5 - 10 - 3 = 2

n(m' ∩ h ∩ C') = n(h) - n(m ∩ h) - n(c ∩ h) - n(h ∩ m ∩ c) = 24 - 5 - 8 - 3 = 8

n(m' ∩ h' ∩ C) = n(m) - n(m ∩ c) - n(c ∩ h) - n(h ∩ m ∩ c) = 38 - 10 - 8 - 3 = 17

n(m ∩ h' ∩ C') + n(m' ∩ h ∩ C') + n(m' ∩ h' ∩ C) + n(h  ∪ m ∪ C)' + n(h ∩ m ∩ c) + n(m ∩ h) + n(c ∩ m) + n(c ∩ h) = 90

2 + 8 + 17 + 5 + 10 + 8 + 3 + n(h  ∪ m ∪ C)' = 90

53 + n(h  ∪ m ∪ C)' = 90

n(h  ∪ m ∪ C)' = 37

b) n(m' ∩ h' ∩ C) = 17

c) n(m ∩ h' ∩ C') = 2

d) n(m' ∩ h ∩ C') = 8

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We are asked what happens if the number of trees per acre was doubled, but we don't know what the original number was. Let's make it x=60

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