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Elena L [17]
3 years ago
8

MEDICAL TESTS. Medical tests are used to indicate whether a person has a particular disease. The sensitivity of a test is the pr

obability that a person having the disease will test positive (indicating the person has the disease, i.e., it’s the probability the test will pick up the fact that a person has the disease). The specificity of a test is the probability that a person not having the disease will test negative. A test for a certain disease has been in use for many years. Based on the "track record" of this test, its sensitivity is 0.934 and its specificity is 0.968. Finally, in the population, 1 in 500 people has the disease. What is the probability that a person who tests positive actually has the disease?
Mathematics
1 answer:
OverLord2011 [107]3 years ago
4 0

Answer:

The probability that a person who tests positive actually has the disease is 5.52%.

Step-by-step explanation:

One out of 500 people have the disease. That means 0.2% of the population (0.002).

Of this proportion, according to the sensitivity of the test, 93,4% (0.934) of them will test positive (true positives).

Also, the rest of the population, that represents 99.8% (0.998), don't have the disease, but, according to the specificity of the test, (1-0.968)=0.032 will test positive (false positives).

We have:

True positives tests = 0.002 * 0.934 = 0.001868, and

False positives tests = 0.998 * 0.032 = 0.031936

We can calculate the probability that a person who tests positive actually have the the disease as the division between the true positives and the total amount of positives:

P = True positives / (True positives + False positives)

P = 0.001868 / (0.001868+0.031936) = 0.055259733  = 5.52%

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EX: 75/100
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Your answer should be 3/4! Make sure to show your work! good luck!
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8/x-5 - 9/x-4 = 5/x^2-9x+20
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If you're using the app, try seeing this answer through your browser:  brainly.com/question/2752942

_______________


Solve the equation:

\mathsf{\dfrac{8}{x-5}-\dfrac{9}{x-4}=\dfrac{5}{x^2-9x+20}\qquad\qquad(x\ne 5~and~x\ne 4)}


Reduce the fractions at the left side so that they have the same denominator:

\mathsf{\dfrac{8(x-4)}{(x-5)(x-4)}-\dfrac{9(x-5)}{(x-4)(x-5)}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32}{x^2-4x-5x+20}-\dfrac{9x-45}{x^2-4x-5x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32}{x^2-9x+20}-\dfrac{9x-45}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}\\\\\\
\mathsf{\dfrac{8x-32-(9x-45)}{x^2-9x+20}=\dfrac{5}{x^2-9x+20}}


Numerators must be equal:

\mathsf{8x-32-(9x-45)=5}\\\\
\mathsf{8x-32-9x+45=5}\\\\
\mathsf{8x-9x=5+32-45}\\\\
\mathsf{-x=-8}\\\\
\mathsf{x=8}\quad\longleftarrow\quad\textsf{this is the solution.}


I hope this helps. =)


Tags:  <em>rational equation fraction solution algebra</em>

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an airplane flies from city a to city b, which is 100 kilometers north and 150 kilometers west of city a. how far does the plane
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Step-by-step explanation:

so it travels 100 kilometers north and another 150 kilometers north so 100+150 is 250 kilometers

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