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stich3 [128]
3 years ago
15

A study concluded that among people with a certain​ virus, 99.2​% of tests conducted were​ (correctly) positive, while for peopl

e without the​ virus, 97.9​% of the tests were​ (correctly) negative. If 32​% of patients actually carry the​ virus, what's the probability that a patient testing negative is truly free of the​ virus?
Mathematics
1 answer:
Dima020 [189]3 years ago
3 0

Answer:

Following are the solution to the question:

Step-by-step explanation:

The probability of virus  P(V) =0.32  

No virus  P(NV) =0.66

Patient testing of negative P(TN) = virus, and negative testing+no virus and negative testing

=0.32(1-0.992)+0.66 \times 0.979 \\\\=0.00256+0.64614\\\\=0.6487

This is why the likelihood of a patient being free of the virus is negative.

= \frac{P(NV) \times\  tested \  negative }{P(TN)}  

= \frac{0.66 \times 0.979}{0.6487}\\\\= \frac{0.64614}{0.6487} \\\\=0.996053646

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On the other hand,

F (x) +G (x) = 3x ^ 2 +1 +(2x - 3) = 3x ^ 2+ 1+ 2x-3 = 3x ^ 2 +2x-2

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We also have:

F (-2) = 3 (-2) ^ 2+ 1 = 3 (4)+ 1 = 12+ 1 = 13

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ANswer:

Option A, A, B, C

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