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sdas [7]
3 years ago
11

If 9 out of 10 doctors in a small city recommend afternoon naps, and there are 8 doctors in the city who don’t recommend naps, h

ow many doctors in the city do recommend afternoon naps?
Mathematics
1 answer:
Hitman42 [59]3 years ago
3 0

Answer:

  72

Step-by-step explanation:

1 out of 10 do not recommend naps, so 9 times as many do recommend naps.

  9 × (8 doctors) = 72 doctors . . . . . recommend naps

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Step-by-step explanation:

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after spending 7/9 of a three hours examination, a student left the exams hall. How many minutes has he to complete the duration
Montano1993 [528]
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According to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation, there were over 71 million wildif
suter [353]

Answer:

The probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is P=0.412.

Step-by-step explanation:

We know the population proportion π=0.8, according to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation.

If we take a sample from this population, and assuming the proportion is correct, it is expected that the sample's proportion to be equal to the population's proportion.

The standard deviation of the sample is equal to:

\sigma_s=\sqrt{\frac{\pi(1-\pi)}{N}}=\sqrt{\frac{0.8*0.2}{500}}=0.018

With the mean and the standard deviaion of the sample, we can calculate the z-value for 0.79 and 0.81:

z=\frac{p-\pi}{\sigma}=\frac{0.79-0.80}{0.018} =\frac{-0.01}{0.018} = -0.55\\\\z=\frac{p-\pi}{\sigma}=\frac{0.81-0.80}{0.018} =\frac{0.01}{0.018} = 0.55

Then, the probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is:

P(0.79

3 0
3 years ago
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