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murzikaleks [220]
3 years ago
9

If a woman takes an early pregnancy test, she will either test positive, meaning that the test says she is pregnant, or test neg

ative, meaning that the test says she is not pregnant. Suppose that if a woman really is pregnant, there is a 98% chance that she will test positive. Also, suppose that if a woman really is not pregnant, there is a 99% chance that she will test negative. Assume that 1000 women are to be pregnancy tested and that exactly 100 of them are pregnant. If a randomly chosen woman from this group tests positive for pregnancy, then what is the probability that she really is pregnant?
Mathematics
1 answer:
k0ka [10]3 years ago
3 0

Answer:

0.0098

Step-by-step explanation:

Probability of being pregnant:100/1000

=1/10.

98% chance:=98/100 ×1/10

=98/1000

Therefore the probability that she really is pregnant is: 0.098

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Answer:

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Step-by-step explanation:

5/9 are girls. The 9 represents the total.

Set up a proportion and you have

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3708-2060= 1648

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Answer:

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Step-by-step explanation:

1) distribute so you get 6+0.75e=2-1.25e

2) move the variable to one side so you get 6+2e=2

3) subtract the 6: 2e=-4

4) divide by 2: e= -2

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Bumek [7]

Answer:

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3 years ago
An internet service provider changes an initial fee of $65 and $18 per hour of labor for office internet installation. Which equ
OlgaM077 [116]
<h2>Answer:</h2><h2>The total cost (C) charged by the internet service provider = $65 + $18(n)</h2>

Step-by-step explanation:

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The cost per hour be $18

Cost per two hours = 18 * 2 = $36

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