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FromTheMoon [43]
3 years ago
7

Im timed!!!!!!!!!!!!!!!!!! I NEED HELP ASAP THANK YOU SO MUCH

Computers and Technology
2 answers:
Troyanec [42]3 years ago
6 0

Answer:

card sorting

Explanation:

tell me if im wrong...

Svet_ta [14]3 years ago
3 0

Answer:

C.

Explanation:

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Consider the following method: static void nPrint(String a, int n) { System.out.println(a.charAt(n)); } What potential error cou
Juliette [100K]
<span>Briefly explain why it is necessary to critique a scientific argument before it is accepted.
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4 0
3 years ago
Due to the difficult economic times, increased global competition, demand for customization, and increased consumer sophisticati
Lunna [17]

Answer:

b. False

Explanation:

Difficult economic times, increased global competition, demand for customization, and increased consumer sophistication does not direct companies to hire more employees and outsource middle managers.

Rather, companies tend to invest new technologies, information systems to satisfy the needs in global competition, consumer sophistication and customization.

Middle managers outsourced cannot deal these issues effectively, therefore they need to be company employees.  

In difficult economic times, companies does not want to hire extra employees because of increased cost.

8 0
3 years ago
A customer reports that recently several files cannot be accessed. The service technician decides to check the hard disk status
Lyrx [107]

Answer:

Back up the user data to removable disk

Explanation:

Before you work on a computer, especially anything that has to do with files not accessible, this might need to format the system because it might either be a virus or other forms of malware. Since backup was done to a different logical partition on the disk, the first thing to do before performing any diagnostic procedures on the disk is to back up the user data to a removable disk in order not to lose the information in the system.

8 0
3 years ago
Create a program which will input data into a pipe one character at a time. Count the number of characters as they are written i
LenKa [72]

Answer:

a)

#include<stdio.h>

#include<stdlib.h>

#include<fcntl.h>

#include<unistd.h>

 

#define BUFSIZE 16

int main(){

   //pipe descriptors

   char msg[BUFSIZE];

   char buf[BUFSIZE];

   int pipefd[2];

   if(pipe(pipefd) == -1){

       //pipe creation error

       perror("pipe");

       exit(EXIT_FAILURE);

   }

   //pipe creation successfull

   //write four messages

   sprintf(msg , "apple");

   write(pipefd[1], msg, BUFSIZE);

   printf("Written %s\n", msg);

   sprintf(msg , "boy");

   write(pipefd[1], msg, BUFSIZE);

   printf("Written %s\n", msg);

   sprintf(msg , "cat");

   write(pipefd[1], msg, BUFSIZE);

   printf("Written %s\n", msg);

   sprintf(msg , "dog");

   write(pipefd[1], msg, BUFSIZE);

   printf("Written %s\n", msg);

   //read

   read(pipefd[0], buf, BUFSIZE);

   printf("Read: %s\n", buf);

   read(pipefd[0], buf, BUFSIZE);

   printf("Read: %s\n", buf);

   read(pipefd[0], buf, BUFSIZE);

   printf("Read: %s\n", buf);

   read(pipefd[0], buf, BUFSIZE);

   printf("Read: %s\n", buf);

   close(pipefd[0]);

   close(pipefd[1]);

   return 0;

}

(b)

#include<stdio.h>

#include<stdlib.h>

#include<fcntl.h>

#include<unistd.h>

#include<wait.h>

#define BUFSIZE 16

int main(){

   //pipe descriptors

   char msg[BUFSIZE];

   char buf[BUFSIZE];

   int pipefd[2];

   if(pipe(pipefd) == -1){

       //pipe creation error

       perror("pipe");

       exit(EXIT_FAILURE);

   }

   //pipe creation successfull

   //write four messages

   if(fork() == 0){

       //this is child

       //close unused write end

       close(pipefd[1]);

       read(pipefd[0], buf, BUFSIZE);

       printf("This is child process reading first message. Content is %s\n", buf);

       read(pipefd[0], buf, BUFSIZE);

       printf("This is child process reading second message. Content is %s\n", buf);

       read(pipefd[0], buf, BUFSIZE);

       printf("This is child process reading third message. Content is %s\n", buf);

       read(pipefd[0], buf, BUFSIZE);

       printf("This is child process reading fourth message. Content is %s\n", buf);

       close(pipefd[0]);

       //exit from child

       exit(EXIT_SUCCESS);

   }

   //this is parent process

   //close unused read end

   close(pipefd[0]);

   sprintf(msg , "apple");

   printf("This is parent process. Writing first message into pipe\n");

   write(pipefd[1], msg, BUFSIZE);

   sprintf(msg , "boy");

   printf("This is parent process. Writing second message into pipe\n");

   write(pipefd[1], msg, BUFSIZE);

   sprintf(msg , "cat");

   printf("This is parent process. Writing third message into pipe\n");

   write(pipefd[1], msg, BUFSIZE);

   sprintf(msg , "dog");

   printf("This is parent process. Writing fourth message into pipe\n");

   write(pipefd[1], msg, BUFSIZE);

   close(pipefd[1]);

   //wait for child

   wait(NULL);

   return 0;

}

(c)

#include<stdio.h>

#include<stdlib.h>

#include<fcntl.h>

#include<unistd.h>

#include<sys/wait.h>

#include<sys/types.h>

#include<signal.h>

typedef struct sigaction Sigaction;

unsigned long long size = 0;

void alarmhandler(int sig){

   //alarm fired writing blocked

   printf("Write blocked after %llu characters\n", size);

   exit(EXIT_SUCCESS);

}

int main(){

   //pipe descriptors

   int pipefd[2];

   if(pipe(pipefd) == -1){

       //pipe creation error

       perror("pipe");

       exit(EXIT_FAILURE);

   }

   //install handler

   sigset_t mask , prev;

   sigemptyset(&mask);

   sigaddset(&mask , SIGALRM);

   sigprocmask(SIG_BLOCK , &mask , &prev);

   Sigaction new_action;

   sigemptyset(&new_action.sa_mask);

   new_action.sa_flags = SA_RESTART;

   new_action.sa_handler = alarmhandler;

   sigaction(SIGALRM , &new_action , NULL);

   sigprocmask(SIG_SETMASK , &prev, NULL);

   while(1){

     

       //print size on multiple of 4

       if(size != 0 && size % 1024 == 0){

           printf("%llu characters in pipe\n", size);

       }

       //reset previous alarm

       alarm(0);

       //set new alarm for 5 seconds

       alarm(5);

       //write to pipe one character

       write(pipefd[1], "A", sizeof(char));

       size++;

   }

   return 0;

}

Explanation:

Output for a, b, c are pasted accordingly

4 0
3 years ago
Select the standardized features of Windows that make it easier to manage your tools:
anygoal [31]

Answer: The Standardized feature of Windows that make it easier to manage tools are:

1. Toolbar

2. Menu

3. Taskbar

Explanation: 1. Toolbar helps to easily show the various tools that can be used on Windows.

2. Menu helps to separate the different parts of the Windows for easy access.

3. Taskbar helps to show the various tools for doing a task.

5 0
3 years ago
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