It's 1 stops per person the expected stop is 10
I hope this helps
Answer:
65.22% probability that it is in compliance of both safety and sanitary standards
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
Number of restaurant in compliance of both safety and sanitary standards.
First I will find those not in compliance, using a Venn's set.
Set A: Violation of sanitary standards.
Set B: Violation of safety standards.
<em>5 were in violation of sanitary standards, a total of 6 were in violation of safety standards, and 3 were in violation of both.</em>
This means that ![A = 5, B = 6, (A \cap B) = 3](https://tex.z-dn.net/?f=A%20%3D%205%2C%20B%20%3D%206%2C%20%28A%20%5Ccap%20B%29%20%3D%203)
<em>At least one:</em>
![A \cup B = A + B - (A \cap B)](https://tex.z-dn.net/?f=A%20%5Ccup%20B%20%3D%20A%20%2B%20B%20-%20%28A%20%5Ccap%20B%29)
So
![A \cup B = 5 + 6 - 3 = 8](https://tex.z-dn.net/?f=A%20%5Ccup%20B%20%3D%205%20%2B%206%20-%203%20%3D%208)
8 are in violation of at least one of these standards.
So 23-8 = 15 are in compliance of both safety and sanitary standards.
What is the probability that it is in compliance of both safety and sanitary standards?
15 out of 23
15/23 = 0.6522
65.22% probability that it is in compliance of both safety and sanitary standards
What lesson is it? tell e the lesson and I can help ypu
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Answer:
each person has to pay 12.82