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Dominik [7]
3 years ago
8

Which criterion can be used to select the appropriate type of network media for a network?

Computers and Technology
1 answer:
OverLord2011 [107]3 years ago
3 0

There are various kinds of network media and each one of them has different features and benefits. Some have different characteristics and not all of them are meant to perform the same purpose. While choosing network media, a number of criteria must be met. These criteria include:

The distance the selected medium can successfully carry a signal.

The environment where the selected medium is to be installed

Explanation:

The distance the selected medium can successfully carry a signal – The type of network media you choose depends on the distance it can cover before you install signal boosters or repeaters. Repeaters are used to regenerate signals that are required to cover greater distances. A UTP cable as compared to fiber is more prone to interference like noise and the distance it covers before a signal repeater is installed is shorter. Fiber cable is considered the best for distances as it can cover longer distances.

Environment should also be considered when choosing network media - If your network is setup in a place where there is too much electromagnetic field being generated, then you may be required to purchase fiber optic cables. Fiber optics are immune to such interference.

Learn about network media

brainly.com/question/9279069

#LearnWithBrainly

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Fifty-three percent of U.S households have a personal computer. In a random sample of 250 households, what is the probability th
aleksley [76]

Answer:

The correct Answer is 0.0571

Explanation:

53% of U.S. households have a PCs.

So, P(Having personal computer) = p = 0.53

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np(1-p) = 250 * 0.53 * (1 - 0.53) = 62.275 > 10

So, we can just estimate binomial distribution to normal distribution

Mean of proportion(p) = 0.53

Standard error of proportion(SE) =  \sqrt{\frac{p(1-p)}{n} } = \sqrt{\frac{0.53(1-0.53)}{250} } = 0.0316

For x = 120, sample proportion(p) = \frac{x}{n} = \frac{120}{250} = 0.48

So, likelihood that fewer than 120 have a PC

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= P(  p^​  < 0.48 )

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= 0.0571      ( From normal table )

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3 years ago
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