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Montano1993 [528]
3 years ago
11

Cathy works for a large organization. The organization has separate servers for managing different types of resources, such as d

ata, files, and printers. What are servers that manage only one type of resource called?
Servers that manage only one type of resource are called ________ servers.
Computers and Technology
1 answer:
Zolol [24]3 years ago
5 0

Answer:

It's called application server.

Explanation:

Application means it can be file server, data server, network server, mail servers, and many more.

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Please help with my Python code - Functions
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Explanation:

see attached for help

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Consider two vectors that are NOT sorted, each containing n comparable items. How long would it take to display all items (in an
Darina [25.2K]

Answer:

The correct answer to the following question will be "Θ(​n​2​)

". The further explanation is given below.

Explanation:

If we're to show all the objects that exist from either the first as well as the second vector, though not all of them, so we'll have to cycle around the first vector, so we'll have to match all the objects with the second one.

So,

This one takes:

= O(n^2)

And then the same manner compared again first with the second one, this takes.

= O(n^2)

Therefore the total complexity,

= Θ(​n​2​)

7 0
3 years ago
Write a program that generates 1,000 random integers between 0 and 9 and displays the count for each number. (Hint: Use a list o
masya89 [10]

Answer:

import random

count0, count1, count2, count3,

count4, count5, count6, count7,

count8, count9, i = [0 for _ in range(11)]

while i < 1000:

   number = random.randint(0,9)

   if number == 0:

       count0 = count0 + 1

   if number == 1:

       count1 = count1 + 1

   if number == 2:

       count2 = count2 + 1

   if number == 3:

       count3 = count3 + 1

   if number == 4:

       count4 = count4 + 1

   if number == 0:

       count5 = count5 + 1

   if number == 6:

       count6 = count6 + 1

   if number == 7:

       count7 = count7 + 1

   if number == 0:

       count8 = count8 + 1

   if number == 9:

       count9 = count9 + 1

   

   i = i+1

print("0's: "+ str(count0) + "\n"+ "1's: "+ str(count1) + "\n"+

"2's: "+ str(count2) + "\n"+ "3's: "+ str(count3) + "\n"+

"4's: "+ str(count4) + "\n"+ "5's: "+ str(count5) + "\n"+

"6's: "+ str(count6) + "\n"+ "7's: "+ str(count7) + "\n"+

"8's: "+ str(count8) + "\n"+ "9's: "+ str(count9) + "\n")

Explanation:

- Initialize variables to hold the count for each number

- Initialize <em>i</em> to control the while loop

- Inside the while loop, check for the numbers and increment the count values when matched.

- Print the result

5 0
3 years ago
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