Answer: 0.31 or 31%
Let A be the event that the disease is present in a particular person
Let B be the event that a person tests positive for the disease
The problem asks to find P(A|B), where
P(A|B) = P(B|A)*P(A) / P(B) = (P(B|A)*P(A)) / (P(B|A)*P(A) + P(B|~A)*P(~A))
In other words, the problem asks for the probability that a positive test result will be a true positive.
P(B|A) = 1-0.02 = 0.98 (person tests positive given that they have the disease)
P(A) = 0.009 (probability the disease is present in any particular person)
P(B|~A) = 0.02 (probability a person tests positive given they do not have the disease)
P(~A) = 1-0.009 = 0.991 (probability a particular person does not have the disease)
P(A|B) = (0.98*0.009) / (0.98*0.009 + 0.02*0.991)
= 0.00882 / 0.02864 = 0.30796
*round however you need to but i am leaving it at 0.31 or 31%*
If you found this helpful please mark brainliest
The answer is B. It is shifted up by 6*1.5-2-2 with is 9-2-2=5.
184.2 is the answer to 4 significant figure
Answer:
They will Markup the price by $19.76.
Step-by-step explanation:
Markup percent = 38%
Wholesale Price = $52
Markup price can be calculated by simply multiplying the wholesale price by 38% or 0.38.
so
Markup price = 38% of 52
= 38/100 × 52
= 0.38 × 52
= $19.76
Therefore, they will Markup the price by $19.76.
