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motikmotik
4 years ago
11

The test to detect the presence of a liver disorder is 98% accurate for a person who has the disease and 97% accurate for a pers

on who does not have the disease. If 3.5% of the people in a given population actually have the disorder, what is the probability that a randomly chosen person tests positive?
0.0343



0.035



0.06325



0.02895
Mathematics
2 answers:
Daniel [21]4 years ago
7 0
<span>The answer is 0.06325. The probability of detecting the true presence is 98% = 0.98. The probability of not detecting the true presence is: 100%-98% = 2% = 0.02. The probability of detecting the true absence is 97% = 0.97. The probability of not detecting the true absence is: 100%-97% = 3% = 0.03. The probability of having the disorder is: 3.5% = 0.035. The probability of not having the disorder is 100% - 3.5% = 96.5% = 0.965. The probabilty of having a disorder and detecting the true presence is: 0.035 * 0.98 = 0.0343. The probability of not having the disorder and detecting the true absence is: 0.965 * 0.03 = 0.02895. The probability that a randomly chosen person tests positive is: the probabilty of having a disorder and detecting the true presence + the probability of not having the disorder and detecting the true absence = 0.0343 + 0.02895 = 0.06325.</span>
evablogger [386]4 years ago
6 0

Answer:0.6

Step-by-step explanation:

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Please find the perimeter and area of these shapes
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Answer:

ABC shaded area = 36\pi - 72   cm²

ABC shaded area perimeter = 6\pi +12\sqrt{2}    cm

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Step-by-step explanation:

<u>Shape ABC</u>

Assuming you want the area and perimeter of the shaded part of the shape only...

<u>Area</u>

Area of a sector = \dfrac12r^2\theta (where r is the radius and \theta<em> </em>

⇒ area of a sector = \dfrac12 \times 12^2\times \dfrac{\pi}{2} =36\pi  \ \textsf{cm}^2

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Therefore, area of shaded area = area of sector - area of triangle

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<u>Perimeter</u>

Arc length = r\theta (where r is the radius and \theta<em> </em>

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<u>Shape ABCD</u>

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⇒ area of large semicircle ABC = \dfrac12 \times \pi \times 2^2=2\pi  \ \textsf{cm}^2

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<u>Perimeter</u>

1/2 circumference = \pi r

⇒ perimeter = 2\pi +2+\pi=3 \pi+2 \ \textsf{cm}

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