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Yuri [45]
3 years ago
11

The endocrine system is composed of many parts of the body. a)True b)False

Computers and Technology
2 answers:
Finger [1]3 years ago
4 0
It is a) true because its compouse by organs and body tissues
Elan Coil [88]3 years ago
4 0
<span>The endocrine system is composed of many parts of the body.
This is a true statement.
</span>
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He could use bold or bigger text just a suggestion.
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How does naming affect a variable’s creation and use?
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Each variable is named so it is clear which variable is being used at any time. It is important to use meaningful names for variables: ... The name given to each variable is up to the programmer, but ideally a variable name should have meaning, ie it should reflect the value that it is holding.

Variables make code more than a static set of instructions. They allow logic to occur, enabling developers to measure time, analyze data, and customize the program to the user. Variables are so important to the code that they deserve a good name that accurately describes their purpose

Explanation:

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You use_____ to view an XPS file
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3 years ago
What command utilizes the limpel-ziv (lz77) compression algorithm and achieves a compression ratio of 60-70 percent?
AURORKA [14]

The limpel-ziv (lz77) compression technique is used by the GNU zip (gzip) tool to produce a compression ratio of 60–70%.

<h3>How do methods for data compression function? Describe the LZW algorithm.</h3>

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4 0
1 year ago
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
3 years ago
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