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masya89 [10]
3 years ago
6

Hel help help help help​

Mathematics
1 answer:
Nadya [2.5K]3 years ago
4 0

Answer:A

Step-by-step explanation:

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Weights were recorded for all nurses at a particular hospital, the mean weight for an individual nurse was 135 lbs. with a stand
Aloiza [94]
Given:
population mean, μ =135
population standard deviation, σ = 15
sample size, n = 19

Assume a large population, say > 100,
we can reasonably assume a normal distribution, and a relatively small sample.
The use of the generally simpler formula is justified.

Estimate of sample mean
\bar{x}=\mu=135

Estimate of sample standard deviation
\s=\sqrt{\frac{\sigma^2}{n}}
=\sqrt{\frac{15^2}{19}}=3.44124  to 5 decimal places.

Thus, using the normal probability table,
P(125
=P(\frac{125-135}{3.44124}
=P(-2.90593
=P(Z
=P(Z

Therefore 
The probability that the mean weight is between 125 and 130 lbs 
P(125<X<130)=0.0731166-0.0018308
=0.0712858



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In George’s neighborhood there are 33 cats and 33 dogs roaming free.What percent of the animals roaming free are dogs?
Alexxx [7]
First, we'll figure out how many animals there are in total by adding up the cats and the dogs. In this case, we have 33+33=66 total animals. Then we need to figure out what fraction of these 66 animals are dogs:
\frac{33}{60}=\frac{1}{2}

So half of the animals are dogs. To write that as a percent, we need the fraction to be over 100, so:
\frac{1}{2}=\frac{50}{100}

50% of the animals are dogs.
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Answer:

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Step-by-step explanation:

the answer is 2262.0646

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