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PolarNik [594]
3 years ago
6

A study reported that finger rings increase the growth of bacteria on health-care workers' hands. Research suggests that 31 perc

ent of health-care workers who wear rings have bacteria on one or both hands, and 27 percent of health-care workers without rings have bacteria on one or both hands. Suppose that independent random samples of 100 health-care workers wearing rings and 100 health-care workers not wearing rings are selected. What is the standard deviation of the sampling distribution of the difference in the sample proportions (wear rings minus does not wear rings) of health-care workers having bacteria on one or both hands?
Mathematics
2 answers:
AysviL [449]3 years ago
7 0

Answer:

Help me!!!! Really !

Step-by-step explanation:

raketka [301]3 years ago
6 0

Answer:

Despite some evidence that suggest that finger ring use is associated with higher bacterial

colonization, healthcare providers continue to wear finger rings in a healthcare setting. The aim

of this systematic review was to synthesize the evidence to date regarding whether finger ring

use increases bacterial colonization of healthcare providers' hands. Articles that studied the

association finger ring use with hand hygiene and bacterial colonization were searched using

PubMed, Google Scholar, and Georgia State University's online library systems. The overall

results of this review suggest that finger ring use does increase the bacterial colonization on the

hands of HCPs but not with significant difference when compared to no finger ring use;

therefore, further research needs to be conducted to decide whether finger rings should be used in

a healthcare setting or not.

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Example 2: Combining Use of the Multiplication and Addition Rules
bulgar [2K]

Answer:

The probability of getting 6's on both cubes is \frac{1}{36}.

The probability that the total score is at least 11 is \frac{1}{12}.

Step-by-step explanation:

Consider the provided information.

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Part (A) Both cubes show 6’s.

Probability of getting 6 on red cube is \frac{1}{6}

Probability of getting 6 on blue cube is \frac{1}{6}

Thus, the probability of getting 6's on both cubes is:

P(\text{Both 6's})=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}

Hence, the probability of getting 6's on both cubes is \frac{1}{36}.

Part (B) The total score is at least 11.

The possible number of outcomes in which total score is at least 11 is:

Red shows 6 and Blue shows 5.

Blue shows 6 and Red shows 5.

Blue shows 6 and Red shows 6.

Thus, the probability of total score is at least 11.

P(\text{Total is at least 11})=\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}\\P(\text{Total is at least 11})=\frac{1}{12}

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3 years ago
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A = 12(8)
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3 years ago
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