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Nookie1986 [14]
3 years ago
15

PLEASE HELP!! WILL MARK BRAINLIEST!!

Computers and Technology
1 answer:
Kobotan [32]3 years ago
7 0

Answer:

Wow, that’s pretty good question.

I would probably get a vpn because it gives free secure WiFi. And it’s most reliable.

Hope I helped!

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Write code which takes a user input of a String and an integer. The code should print each letter of the String the number of ti
Alexus [3.1K]

import java.util.Scanner;

public class JavaApplication70 {

   public static void main(String[] args) {

       Scanner scan = new Scanner(System.in);

       System.out.println("Input a String:");

       String txt = scan.nextLine();

       System.out.println("Input an integer:");

       int num = scan.nextInt();

       String newTxt = "";

       int w = 0;

       for (int i = txt.length()-1; i >= 0; i--){

           char c = txt.charAt(i);

           while (w < num){

               newTxt += c;

               w++;

           }

           w = 0;

       }

       System.out.println(newTxt);

   }

   

}

I hope this helps!

7 0
3 years ago
Read 2 more answers
Why was Windows 1.0 considered an operating environment rather than an operating system?
wolverine [178]

Windows 1.0 was only a shell program

The operating system for Windows 1.0 was MS-DOS

4 0
3 years ago
Read 2 more answers
Assume that the classes listed in the Java Quick Reference have been imported where appropriate. Unless otherwise noted in the q
mash [69]

Answer:

See explaination

Explanation:

import java.util.*;

class UserName{

ArrayList<String> possibleNames;

UserName(String firstName, String lastName){

if(this.isValidName(firstName) && this.isValidName(lastName)){

possibleNames = new ArrayList<String>();

for(int i=1;i<firstName.length()+1;i++){

possibleNames.add(lastName+firstName.substring(0,i));

}

}else{

System.out.println("firstName and lastName must contain letters only.");

}

}

public boolean isUsed(String name, String[] arr){

for(int i=0;i<arr.length;i++){

if(name.equals(arr[i]))

return true;

}

return false;

}

public void setAvailableUserNames(String[] usedNames){

String[] names = new String[this.possibleNames.size()];

names = this.possibleNames.toArray(names);

for(int i=0;i<usedNames.length;i++){

if(isUsed(usedNames[i],names)){

int index = this.possibleNames.indexOf(usedNames[i]);

this.possibleNames.remove(index);

names = new String[this.possibleNames.size()];

names = this.possibleNames.toArray(names);

}

}

}

public boolean isValidName(String str){

if(str.length()==0) return false;

for(int i=0;i<str.length();i++){

if(str.charAt(i)<'a'||str.charAt(i)>'z' && (str.charAt(i)<'A' || str.charAt(i)>'Z'))

return false;

}

return true;

}

public static void main(String[] args) {

UserName person1 = new UserName("john","smith");

System.out.println(person1.possibleNames);

String[] used = {"harta","hartm","harty"};

UserName person2 = new UserName("mary","hart");

System.out.println("possibleNames before removing: "+person2.possibleNames);

person2.setAvailableUserNames(used);

System.out.println("possibleNames after removing: "+person2.possibleNames);

}

}

4 0
3 years ago
What is Sleep mode? Check all of the boxes that apply.
9966 [12]

It is a power-saving mode for a computer.

It is a power-saving mode designed for laptops.

6 0
3 years ago
Considering the following algorithm, which of the following requirements are satisfied?
Alisiya [41]

Answer:

b) Bounded Waiting

Explanation:

int currentThread = 1;

bool thread1Access = true;

bool thread2Access = true;

thread1 { thread2 {

While (true) {

                   While (true)

                                   {

                     while(thread2Access == true)

                                       {

                                      while(thread1Access == true)

                                       {

                                            If (currentThread == 2) {

                                              If (currentThread == 1)

                                                {        

                                                  thread1Access = false; thread2Access = false;

                                                  While (currentThread == 2);

                                                 While (currentThread == 1);

                                                  thread1Access = true; thread2Access = true;

} }

/* start of critical section */ /* start of critical section */

currentThread = 2 currentThread = 1

… ...

/* end of critical section */ /* end of critical section */

thread1Access = false; thread2Access = false;

… ...

} }

} }

} }

It can be seen that in all the instances, both threads are programmed to share same resource at the same time, and hence this is the bounded waiting. For Mutual exclusion, two threads cannot share one resource at one time. They must share simultaneously. Also there should be no deadlock. For Progress each thread should have exclusive access to all the resources. Thus its definitely the not the Progress. And hence its Bounded waiting.

4 0
3 years ago
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