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3 years ago

Hands-On Project 15-4 Make Network Cables

Computers and Technology

1 answer:

Mars2501 [29]3 years ago

8 0

Answer:

1. You can use either standard (T-568A or T-568B) - but must keep the standard throughout the installation (can't mix A with B).

The B standard is preferred in the US. A is mostly used overseas (Europe)

For A the proper wiring sequence is:

1 - Orange/White

2 - Orange

3 - Green / White

4 - Blue

5 - Blue / White

6 - Green

7 - Brown / White

8 - Brown

2. Cross over connections use a different pinout for one side of the cable. The second side is wired normally.

Crossover connections are only used for device to device connections. ( as in the case of connecting two NIC cards together directly). They will not work if plugged into a switch or other network device.

Normal side - use the pinout from question #1

Crossover side:

1 Green / White

2 Green

3 Orange / White

4 Brown / White

5 Brown

6 Orange

7 Blue

8 Blue / White

Explanation:

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A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st

Dima020 [189]

Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

$^nC_r = \frac{n!}{(n-r)!r!}$

Where $n = 15\ and\ r = 4$

$^{15}C_4 = \frac{15!}{(15-4)!4!}$

$^{15}C_4 = \frac{15!}{11!4!}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{32760}{24}$

$^{15}C_4 = 1365$

Hence, there are 1365 ways 

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

From the given parameters; 3 out of 15 is detective;

So;

$p = 3/15$

$p = 0.2$

$q = 1 - p$

$q = 1 - 0.2$

$q = 0.8$

Solving further using binomial;

$(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n$

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

$Probability = ^nC_3pq^3$

Substitute 4 for n, 0.2 for p and 0.8 for q

$Probability = ^4C_3 * 0.2 * 0.8^3$

$Probability = \frac{4!}{3!1!} * 0.2 * 0.8^3$

$Probability = 4 * 0.2 * 0.8^3$

$Probability = 0.4096$

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

Probability that at least one is defective + Probability that at none is defective = 1

Probability that none is defective is calculated as thus;

$Probability = q^n$

Substitute 4 for n and 0.8 for q

$Probability = 0.8^4$

$Probability = 0.4096$

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

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3 years ago

Energy requirements of analog and digital signals.

jek_recluse [69]

Analog signals require more energy.

Analog=Constant flowing modulated

Digital is bursts of on/off binary data.

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3 years ago

Read 2 more answers

A calculator is not a computer because

Lady_Fox [76]

$\huge\fbox\red{❥answer}$

By definition a computer is an electronic device used to store and compute data. So, by definition a calculator stores and interprets data/numbers, and that makes it a computer. However, a computer can also use programs, and manipulate stored data to complete those programs.

hope it helps!! :D

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3 years ago

Differences between placeholders and textbook in PowerPoint

Igoryamba

Answer:

fifthly, if it is an empty box. it disappear completely. sixth the size of a text box changes depending on the length of a text that we type . this is in contrast with a place holder box where changes are made of the font of the text

Explanation:

mark me a brainlist

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3 years ago

What are the importance of switches in our electron device

svp [43]

This way we can turn it on or off so we don't waste power.

Hope I helped! Just tell me if i'm wrong!

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3 years ago