Answer:
B and D
Step-by-step explanation:
they both make srnse to me
1
P(V|A) is not 0.95. It is opposite:
P(A|V)=0.95
From the text we can also conclude, that
P(A|∼V)=0.1
P(B|V)=0.9
P(B|∼V)=0.05
P(V)=0.01
P(∼V)=0.99
What you need to calculate and compare is P(V|A) and P(V|B)
P(V∩A)=P(A)⋅P(V|A)⇒P(V|A)=P(V∩A)P(A)
P(V∩A) means, that Joe has a virus and it is detected, so
P(V∩A)=P(V)⋅P(A|V)=0.01⋅0.95=0.0095
P(A) is sum of two options: "Joe has virus and it is detected" and "Joe has no virus, but it was mistakenly detected", therefore:
P(A)=P(V)⋅P(A|V)+P(∼V)⋅P(A|∼V)=0.01⋅0.95+0.99⋅0.1=0.1085
Try A I believe that’s right
The total weight of Mason is 177
<em><u>Solution:</u></em>
Given that, Mr.Mason takes two-thirds of his body weight and drinks that amount of gatorade in ounces per day
Mr.Mason drinks 118 ounces each day
Let "x" be the total weight of Mr.Mason
Therefore, we can say,
Two third of body weight = Amount of gatorade drank in ounces per day
Two third of x = 118

Thus total weight of Mason is 177
Answer:
a) The expected number of smokers in a random sample of 140 students from this university is 16.8 smokers.
b) No, it is unlikely that smoking habits and waking up early to go to the gym on Saturday are independent.
Step-by-step explanation:
To calculate the expected numbers of smokers in a sample with size n=140 and proportion p=12%, we use the expected value of the binomial distribution:

The expected number of smokers in a random sample of 140 students from this university is 16.8 smokers.
If we take a sample at the opening of the gym, the sample is not expected to be representative of the population of the students. Most of the students that go to the gym usually have healthy habits, so the proportions of smokers is expected to be lower than the average of the university population.