Since the 90% is 63. divide it by 9. then you would get value of 7. then multiple it by 10.
which is equal to 70. so 100% is 70.
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%*
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Divide 45 by 150 then multiply by 100:
45 / 150 = 0.3
0.3 x 100 = 30%
The answer is 30%
Answer: The person next to you ate more popcorn.
Step-by-step explanation:
3 divided by 4 is 0.75, which is equivalent to 75%. 4 divided by 5 is 0.8, which is equivalent to 80%. Now that you've calculated the percentages of how much they ate, you can assume that 80 > 75 (80 is greater than 75).
Answer:
Following are the solution to the given choices:
Step-by-step explanation:
Using chebyshev's theorem
:
In point a)
In point b)
In point c)
In point d)
using standard normal variate