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ratelena [41]
3 years ago
5

10. In step 4 of the CSMA/CA protocol, a station that successfully transmits a frame begins the CSMA/CA protocol for a second fr

ame at step 2, rather than at step 1. What rationale might the designers of CSMA/CA have had in mind by having such a station not transmit the second frame immediately (if the channel is sensed idle)?

Computers and Technology
1 answer:
Neporo4naja [7]3 years ago
7 0

Answer:

It is done to avoid collision among the channel.

Explanation:

CSMA/CD stands for carrier sense multiple access/collision detection.

It is a media access protocol that tells the devices how to respond when a channel is used simultaneously by two devices. The main purpose is to manage the transmission when a collision is encountered.

In the given scenario when the channel is detected idle, the station does not transmits second frame in order to avoid collision. CSMA/CD protocol is designed so that after transmission of first frame, second frame transmission is started at step 2 instead of step 1.

Following is given the flow chart that depicts the working of the protocol CSMA/CD:

<h2> I hope it will help you!</h2><h2 /><h2 />

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Variables set equal to patterns are said to be:_______.
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Answer:

b. instantiated.

Explanation:

In computer programming, to be instatiated or when variables are instatiated; this simply refers to the acts of making example out of a group variables or a form of template. For example in a situation of class of objects.

Hence, in this case, the correct answer to the question above is the option B, in which Variables set equal to patterns are said to be INSTATIATED

5 0
3 years ago
Write a q basic program to find the sum of all the even numbers from 1 to 50​
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Answer:

CLS

M=2

SUM=0

DO WHILE M<=50

SUM=SUM+M

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LOOP

PRINT " The sum of all even numbers between 1 and 50 is"; sum

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3 years ago
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

Sample size(n) = 250

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

= P(x < 120)

= P(  p^​  < 0.48 )

= P(z < \frac{0.48-0.53}{0.0316}​)      (z=\frac{p^-p}{SE}​)  

= P(z < -1.58)

= 0.0571      ( From normal table )

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Masteriza [31]
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Answer:

c

Explanation:

6 0
4 years ago
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