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
Costs<span> $.59c(p) = .59p. C(p) has unit $/lbs, </span>p=<span>lbs</span>
Answer:
y = 2x^2 - 12x + 27
Step-by-step explanation:
<u>Step 1: Distribute the power</u>
y = 2(x - 3)² + 9
y = 2(x^2 - 6x + 9) + 9
y = 2x^2 - 12x + 18 + 9
y = 2x^2 - 12x + 27
Answer: y = 2x^2 - 12x + 27
Answer:
0.42* 1265=x
Step-by-step explanation:
There are 1265 students in Cypress Middle School. So we already know one part of the equation. 42% of these students are in PE. So we know that 42% of these 1265 students are in PE. Of these means you probably have to multiply 42% and 1265. 42% = 0.42. So to write an equation we take the 0.42 and 1265 and put it together. Therefore, 0.42 times 1265=x.
Answer:
0, for q ≠ 0 and q ≠ 1
Step-by-step explanation:
Assuming q ≠ 0, you want to find the value of x such that ...
q^x = 1
This is solved using logarithms.
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x·log(q) = log(1) = 0
The zero product rule tells us this will have two solutions:
x = 0
log(q) = 0 ⇒ q = 1
If q is not 0 or 1, then its value is 1 when raised to the 0 power. If q is 1, then its value will be 1 when raised to <em>any</em> power.
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<em>Additional comment</em>
The applicable rule of logarithms is ...
log(a^b) = b·log(a)
