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Alenkasestr [34]
2 years ago
7

What is a reason for zipping a file?A. to encrypt themB. to return them to their original sizeC. so they display properly on a W

eb pageD. to combine multiple files into one
Computers and Technology
1 answer:
Morgarella [4.7K]2 years ago
5 0

A reason for zipping a file is:

  • D. to combine multiple files into one.

<h3>What is Zipping a File for?</h3>

When a person converts any file to a Zip, it may have one of the following reasons:

  • Reduce the size of the files.
  • Merge files into one.
  • Facilitate the sending of files by mail or others.

Therefore, among the given options, it is possible that a person has as objective to combine multiple files into one, in this way you will be able to find in a single file the information concerning a specific topic and send or transport it more easily.

If you want to learn more about Files, you can visit the following link: brainly.com/question/4461652

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Give a big-O estimate for the number additions used in this segment of an algorithm.
rjkz [21]
Big-O notation is a way to describe a function that represents the n amount of times a program/function needs to be executed. 

(I'm assuming that := is a typo and you mean just =, by the way) 

In your case, you have two loops, nested within each other, and both loop to n (inclusive, meaning, that you loop for when i or j is equal to n), and both loops iterate by 1 each loop.

This means that both loops will therefore execute an n amount of times. Now, if the loops were NOT nested, our big-O would be O(2n), because 2 loops would run an n amount of times.

HOWEVER, since the j-loop is nested within i-loop, the j-loop executes every time the i-loop <span>ITERATES. 
</span>
As previously mentioned, for every i-loop, there would be an n amount of executions. So if the i-loop is called an n amount of times by the j loop (which executes n times), the big-O notation would be O(n*n), or O(n^2). 

(tl;dr) In basic, it is O(n^2) because the loops are nested, meaning that the i-loop would be called n times, and for each iteration, it would call the j-loop n times, resulting in n*n runs.

A way to verify this is to write and test program the above. I sometimes find it easier to wrap my head around concepts after testing them myself.
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