General admission to the State Fair of Texas is $16.00. For guests to ride the various rides, they must use coupons that cost $0
1 answer:
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Answer:
a) The Venn diagram is presented in the attached image to this answer.
b) 0.82
c) 0.16
Step-by-step explanation:
a) The Venn diagram is presented in the attached image to this answer.
n(U) = 100%
n(S) = 48%
n(B) = 66%
n(H) = 38%
n(S n B) = 30%
n(B n H) = 22%
n(S n H) = 28%
n(S n B n H) = 12%
The specific breakdowns for each subgroup is calculated on the Venn diagram attached.
b) The probability that a randomly selected student likes basketball or hockey.
P(B U H)
From the Venn diagram,
n(B U H) = n(S' n B n H') + n(S' n B n H) + n(S n B n H') + n(S n B n H) + n(S n B' n H) + n(S' n B' n H) = 26 + 10 + 18 + 12 + 16 + 0 = 82%
P(B U H) = 82/100 = 0.82
c) The probability that a randomly selected student does not like any of these sports.
P(S' n B' n H')
n(S' n B' n H') = n(U) - [n(S' n B n H') + n(S' n B n H) + n(S n B n H') + n(S n B n H) + n(S n B' n H) + n(S' n B' n H) + n(S n B' n H')]
n(S' n B' n H') = 100 - (26 + 10 + 18 + 12 + 16 + 0 + 2) = 100 - 84 = 16%
P(S' n B' n H') = 16/100 = 0.16
Answer:
There are 35990 ways in which 2 or 3 nodes fail
Step-by-step explanation:
Given : A computer network has 60 switching nodes.
To Find : In how many ways can 2 or 3 nodes fail?
Solution:
We are supposed to find no. of ways can 2 or 3 nodes fail.
So, we will use combination here .
No. of ways can 2 or 3 nodes fail=
Formula :
No. of ways can 2 or 3 nodes fail=
=
Hence there are 35990 ways in which 2 or 3 nodes fail
Answer:
side_1 = 10
side_2 = 3.16
side_3 = 9.49
Step-by-step explanation:
pitagora's theorem
