10-5y=-30 solve for y
2 answers:
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The probability according to the given scenario is:
(a) <em>0.0395</em>
(b) <em>0.95</em>
(c) <em>0.076</em>
The given values are:
(a)
The probability that the b/w has tuberculosis as well as having (+) test will be:
→
By substituting the values, we get
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→
(b)
The probability that a person does not have tuberculosis will be:
→
→
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(c)
The person does not have tuberculosis as well as have a (+) test, the probability will be:
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44-2*5=34
44+3*5=59
See the photo
44+3*5=59
See the photo