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MissTica
2 years ago
12

Identify the names of the following:Please help me.​

Computers and Technology
1 answer:
patriot [66]2 years ago
5 0

Hi,

1st photo is <u>Printer</u>

2nd photo is <u>Hard </u><u>Disk</u>

3rd photo is <u>Printer's </u><u>internal </u><u>CPU</u>

<em>Hope</em><em> it</em><em> helps</em><em> you</em><em>.</em><em>.</em><em>.</em><em> </em><em>pls</em><em> mark</em><em> brainliest</em><em> if</em><em> it</em><em> </em><em>helps</em><em> you</em>

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Develop an sec (single error correction) code for a 16-bit data word. generate the code for the data word 0101000000111001. show
natita [175]

Answer:

010100000001101000101

Explanation:

When an error occurs in data bits, the SEC code is used to determine where the error took place. 5 check-bits are needed to generate SEC code for 16-bits data word. The check bits are:

C16=0, C8=0, C4=0,C2=0,C1=1

Therefore the SEC code is 010100000001101000101

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3 years ago
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How might your use of computers and knowledge of technology systems affect your personal and professional success?
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Amost all professions include theuse of computers. Knowing a lot about them and the way they work could help you use them when needed in a work environment. It would also make an employer more likely to hire you.
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3 years ago
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2. Consider a 2 GHz processor where two important programs, A and B, take one second each to execute. Each program has a CPI of
allsm [11]

Answer:

program A runs in 1 sec in the original processor and 0.88 sec in the new processor.

So, the new processor out-perform the original processor on program A.

Program B runs in 1 sec in the original processor and 1.12 sec in the new processor.

So, the original processor is better then the new processor for program B.

Explanation:

Finding number of instructions in A and B using time taken by the original processor :

The clock speed of the original processor is 2 GHz.

which means each clock takes, 1/clockspeed

= 1 / 2GH = 0.5ns

Now, the CPI for this processor is 1.5 for both programs A and B. therefore each instruction takes 1.5 clock cycles.

Let's say there are n instructions in each program.

therefore time taken to execute n instructions

= n * CPI * cycletime = n * 1.5 * 0.5ns

from the question, each program takes 1 sec to execute in the original processor.

therefore

n * 1.5 * 0.5ns = 1sec

n = 1.3333 * 10^9

So, number of instructions in each program is 1.3333 * 10^9

the new processor :

The cycle time for the new processor is 0.6 ns.

Time taken by program A = time taken to execute n instructions

=  n * CPI * cycletime

= 1.3333 * 10^9 * 1.1 * 0.6ns

= 0.88 sec

Time taken by program B = time taken to execute n instructions

= n * CPI * cycletime

= 1.3333 * 10^9 * 1.4 * 0.6

= 1.12 sec

Now, program A runs in 1 sec in the original processor and 0.88 sec in the new processor.

So, the new processor out-perform the original processor on program A.

Program B runs in 1 sec in the original processor and 1.12 sec in the new processor.

So, the original processor is better then the new processor for program B.

5 0
2 years ago
Which tool lists active tcp connections?
dimaraw [331]
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Type netstat -a to get a complete overview.
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3 years ago
You are a desktop administrator for Nutex Corporation. A user reports that he is unable to access network resources. You notice
Tju [1.3M]

Answer:

The right answer is wire crimper tool.

Explanation:

According to the scenario, the most appropriate answer is the wire crimping tool because the wire crimping tool is a tool that can strip the wire then cut it in proper shape and then crimp the wire in the RJ45 connector very effectively.

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A wire crimper effectively connects the end of the cable to an RJ45 connector.

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2 years ago
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