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defon
3 years ago
5

Which term describes a set of concepts and practices that provide detailed descriptions and comprehensive checklists, tasks, and

procedures for common IT practices?
Computers and Technology
1 answer:
devlian [24]3 years ago
3 0

Answer: Information Technology Infrastructure Library (ITIL)

Explanation:Information Technology Infrastructure Library (ITIL) is the system that contains the practice and information of the various IT services .It is created for the IT businesses that face the risk, cost-degradation and other hazards.

This library provides the services like risk management, improving IT environment and ethics, making stable customer relationship etc.These practices help the information technology organization to change and grow.

You might be interested in
Given six memory partitions of 100 MB, 170 MB, 40 MB, 205 MB, 300 MB, and 185 MB (in order), how would the first-fit, best-fit,
nlexa [21]

Answer:

We have six memory partitions, let label them:

100MB (F1), 170MB (F2), 40MB (F3), 205MB (F4), 300MB (F5) and 185MB (F6).

We also have six processes, let label them:

200MB (P1), 15MB (P2), 185MB (P3), 75MB (P4), 175MB (P5) and 80MB (P6).

Using First-fit

  1. P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
  2. P2 will be allocated to F1. Therefore, F1 will have a remaining space of 85MB from (100 - 15).
  3. P3 will be allocated F5. Therefore, F5 will have a remaining space of 115MB from (300 - 185).
  4. P4 will be allocated to the remaining space of F1. Since F1 has a remaining space of 85MB, if P4 is assigned there, the remaining space of F1 will be 10MB from (85 - 75).
  5. P5 will be allocated to F6. Therefore, F6 will have a remaining space of 10MB from (185 - 175).
  6. P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

The remaining free space while using First-fit include: F1 having 10MB, F2 having 90MB, F3 having 40MB as it was not use at all, F4 having 5MB, F5 having 115MB and F6 having 10MB.

Using Best-fit

  1. P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
  2. P2 will be allocated to F3. Therefore, F3 will have a remaining space of 25MB from (40 - 15).
  3. P3 will be allocated to F6. Therefore, F6 will have no remaining space as it is entirely occupied by P3.
  4. P4 will be allocated to F1. Therefore, F1 will have a remaining space of of 25MB from (100 - 75).
  5. P5 will be allocated to F5. Therefore, F5 will have a remaining space of 125MB from (300 - 175).
  6. P6 will be allocated to the part of the remaining space of F5. Therefore, F5 will have a remaining space of 45MB from (125 - 80).

The remaining free space while using Best-fit include: F1 having 25MB, F2 having 170MB as it was not use at all, F3 having 25MB, F4 having 5MB, F5 having 45MB and F6 having no space remaining.

Using Worst-fit

  1. P1 will be allocated to F5. Therefore, F5 will have a remaining space of 100MB from (300 - 200).
  2. P2 will be allocated to F4. Therefore, F4 will have a remaining space of 190MB from (205 - 15).
  3. P3 will be allocated to part of F4 remaining space. Therefore, F4 will have a remaining space of 5MB from (190 - 185).
  4. P4 will be allocated to F6. Therefore, the remaining space of F6 will be 110MB from (185 - 75).
  5. P5 will not be allocated to any of the available space because none can contain it.
  6. P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

The remaining free space while using Worst-fit include: F1 having 100MB, F2 having 90MB, F3 having 40MB, F4 having 5MB, F5 having 100MB and F6 having 110MB.

Explanation:

First-fit allocate process to the very first available memory that can contain the process.

Best-fit allocate process to the memory that exactly contain the process while trying to minimize creation of smaller partition that might lead to wastage.

Worst-fit allocate process to the largest available memory.

From the answer given; best-fit perform well as all process are allocated to memory and it reduces wastage in the form of smaller partition. Worst-fit is indeed the worst as some process could not be assigned to any memory partition.

8 0
3 years ago
What happens to formulas with relative cell references when they are copied using the fill handle?
Masteriza [31]
The correct answer is D
4 0
3 years ago
10 sentences about computer parts.
Ahat [919]
One computer part is the CPU, it’s a piece of hardware the last allows your computer to access and interact all the applications and programs. The first ever CPU chip was invented around 4 decades ago. The keyboard is another computer part and it allows the user to type letters and numbers. There are about 104 keys on a keyboard and there are different parts in it. Some of the parts include, control keys, function keys, navigation keys, numeric keypad, and so on. A mouse is another device used with the keyboard to position the cursor. It’s a hand held device that detects two-dimensional motion relative to a surface. This motion is typically translated Into the motion of a pointer on a display, which allows a smooth control of the graphical user. Memory is a device to store all of your information and saved data. The motherboard is the backbone that tied together the computers components at one spot.
6 0
2 years ago
This is your code. >>> A = ['dog''red'] >>> B = [' cat', 'blue']>>>C = ['fish', 'green'] You can impl
natima [27]

Answer:

The answer to this question is given below in the explanation section.

Explanation:

The given Python code is:

>>> A = ['dog''red']

>>> B = [' cat', 'blue']

>>>C = ['fish', 'green']

We can implement an array of this data by using the dictionary in Python.

Dictionary in python:

A dictionary in python stores (key-value) pairs. For example, we can implement the -given arrays A, B, and C in the dictionary as given below:

d={'dog : red', 'cat : blue', 'fish : green'}

print(d['dog']) # Get an entry from a dictionary; prints "red"

print('cat' in d)    # Check if a dictionary has a given key; prints "True"

d['fish'] = 'black'    # Set an entry in a dictionary

print(d['fish'])      # Prints "black"

# print(d['elephant'])  # KeyError: 'elephant' not a key of d

print(d.get('elephant', 'N/A'))  # Get an element with a default; prints "N/A"

print(d.get('fish', 'N/A'))   # Get an element with a default; prints "black"

del d['dog']        # Remove an element from a dictionary

print(d.get('dog', 'N/A')) # "dog" is no longer a key; prints "N/A"

3 0
2 years ago
Read 2 more answers
Write code for a function with the following prototype: /* Addition that saturates to TMin or TMax */ int saturating_add(int x,
Kazeer [188]

Answer:

See explaination

Explanation:

program code.

/* PRE PROCESSOR DIRECTIVES */

#include<stdio.h>

/* PRE-DEFINED VALUES FOR TMAX AND TMIN */

#define TMax 2147483647

#define TMin (-TMax -1)

/* saturating_add(int,int) METHOD IS CALLED HERE */

int saturating_add(int firstNumber, int secondNumber)

{

/*

FOR BETTER UNDERSTANDING, LETS TAKE TEST CASE,

WHERE firstNumber = 5 AND secondNumber = 10

*/

int w = sizeof(firstNumber) << 3;

/*

sizeof(firstNumber) VALUE IS 4, SO USING BINARY LEFT SHIFT OPERATOR TO THREE PLACES,

WE HAVE NOW VALUE 32, ASSIGNED TO w

*/

/* ADDITION IS CALCULATED => 15 */

int addition = firstNumber + secondNumber;

/*

MASK INTEGER VARIABLE IS TAKEN

mask BIT IS LEFT SHIFTED TO 31 PLACES => 2^31 IS THE NEW VALUE

*/

int mask = 1 << (w - 1);

/* FIRST NUMBER MOST SIGNIFICANT BIT IS CALCULATED BY USING AND OPERATOR */

int msbFirstNumber = firstNumber & mask;

/* SECOND NUMBER MOST SIGNIFICANT BIT IS CALCULATED BY USING AND OPERATOR */

int msbSecondNumber = secondNumber & mask;

/* MOST SIGNIFICANT BIT OF ADDITION IS CALCULATED BY USING AND OPERATOR */

int msbAddition = addition & mask;

/* POSITIVE OVERFLOW IS DETERMINED */

int positiveOverflow = ~msbFirstNumber & ~msbSecondNumber & msbAddition;

/* NEGATIVE OVERFLOW IS DETERMINED */

int negativeOverflow = msbFirstNumber & msbSecondNumber & !msbAddition;

/* THE CORRESPONDING VALUE IS RETURNED AS PER THE SATURATING ADDITION RULES */

(positiveOverflow) && (addition = TMax);

(negativeOverflow) && (addition = TMin);

return addition;

}

/* MAIN FUNCTION STARTS HERE */

int main(){

/* TEST CASE */

int sum = saturating_add(5, 10);

/* DISPLAY THE RESULT OF TEST CASE */

printf("The Sum Is : %d\n\n",sum);

}

7 0
3 years ago
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