Translate the following C program to Pep/9 assembly language.
#include
const int limit = 5;
int main() {
int number;
scanf("%d", &number);
while (number < limit) {
number++;
printf("%d ", number);
}
return 0;
}

Answer:

Storage space

Explanation:

Answer:

As a student I answered for students benefits only

Explanation:

I am explaining into list form so that everyone can understand it easily.

Benefits:

By sticking to this time plan, you have a better chance of tackling the task than getting to it with no predefined time allocated to it.

Proper time management enables you to prioritize tasks and tackle them first.

Instead of getting caught up in multitasking, you focus on one activity at a time for a specified duration.

Managing your time well allows you to get your work done on time. This elicits a sense of confidence and accomplishment in your capabilities. Getting through a long to-do list can also evoke these feelings.

Answer: Its B. 1,000,000 per second

Explanation: One hertz is one tick per second. A computer with a one million hertz processer would have 1,000,000 ticks per second.

Answer:

b(ii) prices[1]= 20.50;  // do not declare with double again

b (iii) scanf("%lf", &prices[4]);  //prices[4] is last element

d) See below

Explanation:

/* This program uses the string.h header file

The strncat() function in this file concatenates two strings. It takes three arguments, dest and source strings and number of characters to concatenate

*/

#include <stdio.h>
#include <string.h>

int main() {
char s1[10] = "Happy";
char s2[10] = " Birthday";
char s3[10]="";

strncat(s3, s1, 5);  //concatenate 5 characters of s1 to end of s3 and                            
                                    //store result in s3; s = "Happy"
strncat(s3, s2, 10); //concatenate s2 with current contents of s3
                                      //s3 becomes "Happy Birthday
printf("%s", s3);

return 0;

}

Option c is correct. Node.insertBefore(new, child) is the method that includes or embeds a new node directly ahead of a child node.

The Node interface's insertBefore() method places a node as a child of a given parent node ahead of a reference node.

InsertBefore() transfers the specified node to the new place if it already exists in the document. (That is, before adding it to the given new parent, it will automatically remove it from its current parent.)

As a result, a node cannot be present in two places in the document at once. The full contents of the DocumentFragment are moved into the child list of the chosen parent node if the given child is a DocumentFragment. the node that newNode is added before. In the event that this is null, newNode is appended to the end of the node's child nodes.

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Answer:

yo traciahenny how are you :)

The formula which is used to determine a company's customer retention rate is: 3. (The number of customers at the end of the period - the number of new customers acquired during the period)/ the number of customers at the start of the period x 100.

Customer retention rate can be defined as a measure of the number of customers that a business organization (company or firm) is able to successfully retain over a particular period of time and it is typically expressed as a percentage.

Mathematically, the customer retention rate of a business organization (company or firm) can be calculated by using this formula:

CRR = [(CE - CN)/CS] × 100

Where:

In conclusion, we can reasonably infer and logically deduce that the customer retention rate of a company simply refers to the percentage of existing customers that a business organization (company or firm) is able to retain after a given period of time.

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The time that is required for the packet to move over the the distance d at the speed s is  0.028 seconds.

The speed that It would take to move it in transmission is given as 0.008 seconds.

The details that we would have to solve the problem that we have here are given below

The distance = 5000 km, we convert to meters = 5000 * 1000

The speed of the propagation is given as (s) 2.5 x 108 m/s

The propagation delay is given as d / s

= 5000000 / 2.5 x 10^8

= 0.02 this is the value of the propagation

Next we have to find the length of the packet. L = 1000 bytes

this is equal to 8000 bits

The transmission rate is 1 Mbps = 1000000 bits

The transmission delay would be = L/R

= 8000 / 1000000

= 8 / 1000

= 0.008 seconds in transmission

The total required time that is to be gotten would be

Propagation + transmission

= 0.02 + 0.008

= 0.028

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Answer:

(99)₁₀  = (1100011)₂

Explanation:

I believe the question is asking you to convert 99 in base 10 to binary

If so, the above would be the answer easily computed using a programming calculator

If I have mis-interpreted your question, my apologies and you can report this answer

A program that, given a square matrix, identifies the largest number and what position it has, and indicates how many even and odd numbers it has, is given below:

// C++ implementation to arrange

// odd and even numbers

#include <bits/stdc++.h>

using namespace std;

// function to arrange odd and even numbers

void arrangeOddAndEven(int arr[], int n)

{

  int oddInd = 1;

   int evenInd = 0;

   while (true)

   {

       while (evenInd < n && arr[evenInd] % 2 == 0)

           evenInd += 2;

       while (oddInd < n && arr[oddInd] % 2 == 1)

           oddInd += 2;

      if (evenInd < n && oddInd < n)

           swap (arr[evenInd], arr[oddInd]);

       else

           break;

   }

}

// function to print the array

void printArray(int arr[], int n)

{

   for (int i = 0; i < n; i++)

       cout << arr[i] << " ";

}

// Driver program to test above

int main()

{

   int arr[] = { 3, 6, 12, 1, 5, 8 };

   int n = sizeof(arr) / sizeof(arr[0]);

   cout << "Original Array: ";

  printArray(arr, n);

   arrangeOddAndEven(arr, n);

   cout << "\nModified Array: ";

   printArray(arr, n);

   return 0;

}

Original Array: 3 6 12 1 5 8

Modified Array: 6 3 12 1 8 5

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Using the knowledge of computational language in JAVA it is possible to describe Generate a random integer from 1 to 100, Prompt the user to guess the number and  Display "Too high" or "Too low".

Writting the code:

import java.util.Random;

import java.util.Scanner;

public class GFG {

   public static void main(String[] args)

   {

       // stores actual and guess number

       int answer, guess;

         // maximum value is 100

       final int MAX = 100;

       // takes input using scanner

       Scanner in = new Scanner(System.in);

       // Random instance

       Random rand = new Random();

       boolean correct = false;

       // correct answer

       answer = rand.nextInt(MAX) + 1;

       // loop until the guess is correct

       while (!correct) {

           System.out.println(

               "Guess a number between 1 and 100: ");

           // guess value

           guess = in.nextInt();

           // if guess is greater than actual

           if (guess > answer) {

               System.out.println("Too high, try again");

           }

           // if guess is less than actual

           else if (guess < answer) {

               System.out.println("Too low, try again");

           }

           // guess is equal to actual value

           else {

               System.out.println(

                   "Yes, you guessed the number.");

               correct = true;

           }

       }

       System.exit(0);

   }

}

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Since Dana is using a cross-functional team to improve collaboration. She is using the Agile method of project development.

Agile is one that lets teams offer value to their clients more quickly and with fewer difficulties through an iterative approach to project management and software development.

Note that an agile team produces work in manageable, small-scale increments rather than staking all on a "big bang" launch.

Since the Agile technique divides a project into many parts and uses these phases to manage the project. Continuous improvement at every stage and ongoing collaboration with stakeholders are required. Teams cycle through a process of planning, carrying out, as well as assessing once the job starts.

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Answer:

c = 'z'

while c >= 'a':

 print(c)

 c = c h r(ord(c)-5)

Explanation:

Keep in mind, you could have also chosen to do:

print("z\nu\np\nk\nf\na\n")

but it seems the character arithmetic is what is asked for...

Answer: python

i just did a test on this

In this exercise we have to use the knowledge of C++ to write a program that is possible Define a function with no parameters that reads integers from input until integer 0 is read.

#include <iostream>

using namespace std;

bool

{

   int num;

   bool =true;

   do

   {

       cin>>num;

       if(num>=0 && num!=1)

           =false;

   }while(num!=1);

   return;

}

int main()

{

   bool allNegative;

   allNegative=

   if(allNegative==true)

       cout<<"All match";

   else

       cout<<"All not match";

   return 0;

}

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One disadvantage each for list, array, and queue data structures are:

It is difficult to quickly search the queue. You must begin at the very end and might need to keep another queue. Don't even consider utilizing a queue if you have data that you want to make searchable in the future. It is difficult to add or remove items from the queue's middle.

Note that An array doesn't verify its borders: In the C programming language, we are unable to determine whether the values placed in an array exceed its size or not.

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Answer:

create drones is an example of how agriculture uses computer programming

num = int(input("User Input: "))

mylist = []

while(num !=0):

   mylist.append(num)

   num = int(input("User Input: "))

mylist.reverse()

for i in mylist:

   print(i,end = ' ')

What is a program?

A program is a set of instructions that a computer follows to complete a specific task.

This question is answered in Python, and it employs 0 as a sentinel value, which means that the program stops prompting the user for input when the user enters 0.

This line requests input from the user

num = int(input("User Input: "))

This line declares a list that is empty

mylist = []

This loop is repeated as long as user enters input other than 0

while(num !=0):

This appends user input to the list

   mylist.append(num)

This prompts user for another input

   num = int(input("User Input: "))

This reverses the list

mylist.reverse()

The following iteration the list in reverse order

for i in mylist:

   print(i,end = ' ')

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Option c is correct. It tests whether the condition returned by the callback function holds for at least one item in array.

Every element in the array is put through the supplied test by the provided function using the every() method. It gives back a Boolean result. Until it locates the element for which callback function returns a false result, the every method calls the given callback function once for each element present in the array.

The every method immediately returns false if such an element is discovered. Every returns true if callback function does not return a truthy value for any of the items.

Only for array indexes with assigned values is callbackFn triggered. For empty slots in sparse arrays, it is not called.

The value of the element, the index of the element, and the Array object being traversed are passed as the three arguments to callbackFn.

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Bidirectional microphones have a very constricted pickup angle, making them useful for separating a single voice or mechanism in the presence of other sound sources, as lengthy as nothing is directly behind the mic. They are useful for picking up two sources that are next to each other.

The pressure gradient principle underpins this type of microphone. The sound can travel through the membrane on both sides.

When the sound on the front and backside of the diaphragm differs, the microphone sends a signal.

Bidirectional microphones have a particularly narrow pickup angle, which makes them advantageous for distinguishing a single voice or framework in the involvement of other sound sources, as long as nothing is directly behind the mic.

They are useful for detecting two sources that are close together.

Thus, in this setting a bidirectional microphone can be used.

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The correct matching of the tiles to their correct boxes is given below:

Ctrl + X

cuts selected text

runs the spelling and grammar check

F7

opens the Help menu

F1

Ctrl + V

pastes text saved in memory

This refers to the device or program that is used to input data or information, make edits, process and format them, and then provide output to the finished product.

Hence, we can see that from the given question, it is clearly stated that Mike is writing an essay using Microsoft Word on a Windows computer and he needs to use the keyboard shortcuts.

With this in mind, the word processor he is using is known as MS Word or Microsoft Word and it has some keyboard shortcuts that can help Mike write, edit and format his essay so it would be in the right and proper way for him to output.

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Answer:

stored-program computer is a computer that stores program instructions in electronically or optically accessible memory.

Microsoft's Windows is the most popular desktop operating system and as such is the true statement.

Check more about Microsoft  windows below.

Windows from Microsoft is seen as the Computer operating systems' global market share from 2012 through 2022, broken down by month.

Note that the desktop, tablet, as well as console OS markets are all dominated by Microsoft's Windows, which has a 70.68 percent market share worldwide.

Therefore, The most widely used desktop and laptop operating system is Windows 10. The most widely used PC operating system is Windows. An operating system is a piece of software that runs on a computer and controls how software programs and computer resources are used.

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INFOCON 1 is the prerequisite for cyber defence that creates a protection priority focus on just crucial and essential functions.

Information operations condition, or INFOCON 1, is a circumstance where an information system has been successfully attacked and the effects on Department of Defense  missions are clear, yet the computer network's defence system is on high alert.

INFOCON 1 basically assures that a computer network security system is as aware or ready as possible to deal with intrusion approaches that cannot be detected or countered at a lower readiness level.

In conclusion, INFOCON 1 is a need for cyber protection that places a priority on protecting only vital and necessary functions.

Thus, INFOCON 1 is the prerequisite for cyber defence.

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A tool which Kyla can use to store the document with a different name and makes changes in the new document is: B. Save As.

A document simply refers to a computer resource that is designed and developed to enable end users to easily store data as a single unit on a computer storage device.

Generally speaking, all computer documents can be identified by a title (name), date modified, size, and type such as the following;

A Save as command can be defined as a type of command associated with the file menu of a software application or document and it causes a copy of the current file to be created and stored to a different location, file name, and/or file type.

In this context, we can reasonably infer and logically deduce that Kyla  should save the document by using the "Save as" command.

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Answer:

Love

Comfortable

Trust

Explanation:

Three reasons why you should always be with your family is that most times they love you, and you love them. Being with them all of the time makes you more comfortable around them, being through their best and their worst. Lastly being with them so much, you get to know them, and with knowledge comes trust of knowledge. You already most times have trust with your family.

The program statements that complete the program are

for(x =0 ; x < NUM_CITIES; x++){

   if (citiesInMichigan[x] == inCity){

       foundIt = true;

   }

}

if(foundIt == false){

   cout<<"Not a city in Michigan";

}

The complete program in C++ language is as follows:

Note that the comments are used to explain each action

#include<iostream>

#include<string>

using namespace std;

int main(){

// Declare variables

string inCity; // name of city to look up in array

const int NUM_CITIES = 10;

// Initialized array of cities

string citiesInMichigan[] = {"Acme", "Albion", "Detroit", "Watervliet", "Coloma", "Saginaw", "Richland", "Glenn", "Midland", "Brooklyn"};

bool foundIt = false; // Flag variable

int x; // Loop control variable

// Get user input

cout << "Enter name of city: ";

cin >> inCity;

// Write your loop here

for(x =0 ; x < NUM_CITIES; x++){

   // Write your test statement here to see if there is

   // a match. Set the flag to true if city is found.

   if (citiesInMichigan[x] == inCity){

       foundIt = true;

   }

}

// Test to see if city was not found to determine if

// "Not a city in Michigan" message should be printed.

if(foundIt == false){

   cout<<"Not a city in Michigan";

}

return 0;

} // End of main()

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Answer:

I'm generally puzzled

Explanation:

A C program segment for prompting a user to enter an integer and save it in an option variable is given below:

#include <stdio.h>

#include <ctype.h>

int main(void)

{

   char ch;

   float num1,num2,answer;

   printf("Enter the operation of your choice:\n");

   printf("a. add     s. subtract\n");

   printf("m. multiply q. divide\n");

   printf("q. quit\n");

   while ((ch = getchar())!='q')

   {

       ch=tolower(ch);

       //Ignore whitespace

       if (ch=='\n')

           continue;

       else

       {

           switch(ch)

           {

               //Addition part

               case 'a':

                   //First number

                   printf("Enter first number: ");

                   //Check to see if input is a number

               while (scanf("%f",&num1)==0)

               {

                   printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");

                   scanf("%*s");

               }

               //Second number

              printf("Enter second number: ");

               while (scanf("%f",&num2)==0)

               {

                   printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");

                   scanf("%*s");

               }

               //Do math for respective operation

              answer = num1 + num2;

               //Print out result

               printf("%.3f + %.3f = %.3f\n", num1,num2,answer);

               break;

           //Subtraction part

           case 's':

              printf("Enter first number: ");

               while (scanf("%f",&num1)==0)

               {

                   printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");

                   scanf("%*s");

               }

               printf("Enter second number: ");

               while (scanf("%f",&num2)==0)

               {

                   printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");

                   scanf("%*s");

               }

               answer = num1 - num2;

               printf("%.3f - %.3f = %.3f\n", num1,num2,answer);

               break;

           //Multiplication part

           case 'm':

               printf("Enter first number: ");

               while (scanf("%f",&num1)==0)

               {

                   printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");

                   scanf("%*s");

               }

               printf("Enter second number: ");

               while (scanf("%f",&num2)==0)

               {

                   printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");

                   scanf("%*s");

               }

               answer = num1 * num2;

               printf("%.3f * %.3f = %.3f\n", num1,num2,answer);

               break;

           //Division part

           case 'd':

               printf("Enter first number: ");

               while (scanf("%f",&num1)==0)

               {

                  printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");

                   scanf("%*s");

               }

               printf("Enter second number: ");

               while (scanf("%f",&num2)==0)

               {

                   printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");

                   scanf("%*s");

               }

              //Check for if number is a zero

               while (num2==0)

               {

                   printf("Please enter a non-zero number, such as 2.5, -1.78E8, or 3: ");

                   while (scanf("%f",&num2)==0)

                   {

                       printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");

                       scanf("%*s");

                   }

               }

               answer = num1 / num2;

               printf("%.3f / %.3f = %.3f\n", num1,num2,answer);

               break;

           //For if a non-valid operation is entered

           default:

               printf("That is not a valid operation.\n");

               break;

       }

   }

  printf("Enter the operation of your choice:\n");

   printf("a. add      s. subtract\n");

   printf("m. multiply q. divide\n");

   printf("q. quit\n");

}

printf("Bye.\n");

return 0;

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