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aivan3 [116]
3 years ago
7

If a thread is not finished running, perhaps because it had to wait or it was preempted, it is typically restarted on the same p

rocessor that previously ran it. what is this known as? cooperative multitasking multitasking processor affinity preemptive multitasking
Computers and Technology
1 answer:
andrew11 [14]3 years ago
4 0
<span>If a thread is not finished running, perhaps because it had to wait or it was preempted, it is typically restarted on the same processor that previously ran it.  This is this known as </span>processor affinity. 
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Now that the classList Array has been implemented, we need to create methods to access the list items.
natima [27]

The following code will be applied to Create the following static methods for the Student class

<u>Explanation:</u>

/* Note: Array index starts from 0 in java so ,, user ask for 1 index then the position will be i-1 , below program is based on this concept if you dont want this way just remove -1 from classList.get(i-1).getName(); ,and classList.add(i-1,student); */

/* ClassListTester.java */

public class ClassListTester

{

public static void main(String[] args)

{

//You don't need to change anything here, but feel free to add more Students!

Student alan = new Student("Alan", 11);

Student kevin = new Student("Kevin", 10);

Student annie = new Student("Annie", 12);

System.out.println(Student.printClassList());

System.out.println(Student.getLastStudent());

System.out.println(Student.getStudent(1));

Student.addStudent(2, new Student("Trevor", 12));

System.out.println(Student.printClassList());

System.out.println(Student.getClassSize());

}

}

/* Student.java */

import java.util.ArrayList;

public class Student

{

private String name;

private int grade;

//Implement classList here:

private static ArrayList<Student> classList = new ArrayList<Student>();

public Student(String name, int grade)

{

this.name = name;

this.grade = grade;

classList.add(this);

}

public String getName()

{

return this.name;

}

//Add the static methods here:

public static String printClassList()

{

String names = "";

for(Student name: classList)

{

names+= name.getName() + "\n";

}

return "Student Class List:\n" + names;

}  

public static String getLastStudent() {

// index run from 0 so last student will be size -1

return classList.get(classList.size()-1).getName();

}

public static String getStudent(int i) {

// array starts from 0 so i-1 will be the student

return classList.get(i-1).getName();

}

public static void addStudent(int i, Student student) {

// array starts from 0 so, we add at i-1 position

classList.add(i-1,student);

// remove extra student

classList.remove(classList.size()-1);

}  

public static int getClassSize() {

return classList.size();

}

}

7 0
4 years ago
Assume the method doSomething has been defined as follows: public static void doSomething (int[] values, int p1, int p2) { int t
Dmitriy789 [7]

Answer:

The answer to this question is option "b".

Explanation:

In the method definition, we perform swapping. To perform swapping we define a variable "temp" that swap values of variable p1 and p2 and store in array that is "values". and other options are not correct that can be defined as:

  • In option a, It does not insert any new value in array because we do not pass any value in function.
  • In option c, The function does not copy and assign a value in a new array because in this function we not assign any new array.  
  • In option d, It is incorrect because it can not move an element into high index position.
8 0
3 years ago
Instructions The population of town A is less than the population of town B. However, the population of town A is growing faster
weeeeeb [17]

Answer: The c++ program is given below.

#include <iostream>

using namespace std;

int main() {

float townA, townB, growthA, growthB, populationA, populationB;

    int year=0;    

cout<<"Enter present population of town A : ";

cin >> townA;  

cout<<endl<<"Enter present growth rate of town A : ";

cin >> growthA;

growthA = growthA/100;  

cout<<endl<<"Enter present population of town B : ";

cin >> townB;  

cout<<endl<<"Enter present growth rate of town B : ";

cin >> growthB;

growthB = growthB/100;  

do

{

    populationA = townA + (townA * growthA);

    populationB = townB + (townB * growthB);      

    townA = populationA;

    townB = populationB;      

    year++;      

}while(populationA < populationB);  

cout<<endl<<"After " <<year<< " years, population of town A is "<<populationA << " and population of town B is "<< population<<endl;

return 0;

}

Explanation:

All the variables for population and growth rate are declared with float datatype.

The user inputs the present population of both the towns.

The growth rate entered by the user is the percentage growth.

For example, town A has 10% growth rate as shown in the output image.

The program converts 10% into float as 10/100.

growthA = growthA/100;

growthB = growthB/100;

The above conversion is done to ease the calculations.

The year variable is declared as integer and initialized to 0.

The growth in population is computed in a do-while loop. After each growth is calculated, the year variable is incremented by 1.

The loop continues until population of town A becomes greater than or equal to population of town B as mentioned in the question.

Once the loop discontinues, the final populations of town A and town B and the years needed for this growth is displayed.

The new line is introduced using endl keyword.

The function main has return type int hence, 0 is returned at the end of the program.

6 0
4 years ago
At which layer of the osi model is the entire message referred to as the payload
erik [133]

I guess the correct answer is application

An applicatiοn layеr is an abstractiοn layеr that spеcifiеs thе sharеd cοmmunicatiοns prοtοcοls and intеrfacе mеthοds usеd by hοsts in a cοmmunicatiοns nеtwοrk. Thе applicatiοn layеr abstractiοn is usеd in bοth οf thе standard mοdеls οf cοmputеr nеtwοrking: thе Intеrnеt Prοtοcοl Suitе (TCP/IP) and thе ΟSI mοdеl. Althοugh bοth mοdеls usе thе samе tеrm fοr thеir rеspеctivе highеst lеvеl layеr, thе dеtailеd dеfinitiοns and purpοsеs arе diffеrеnt.

6 0
3 years ago
The list below represents the contents of a computer's main memory. I've left every other byte blank. Assume that the letters at
inysia [295]

Answer:

3) A Single linked list is a sequence of elements in which every element has link to its next element in the sequence.

DATA LINK

DATA stores actual value , LINK stores address of next node

As per information given in question, letters at the even addresses are items in linked list and the odd addresses will be used as links.

Even Address Odd Address

12 (Stores 't') 13 (Used as link)

14 (Stores 'm') 15 (Used as link)

16 (Stores 'a') 17 (Used as link)

18 (Stores 'r') 19 (Used as link)

20 (Stores 's') 21 (Used as link)

Numbers represented by circle are addresses of respective nodes. Here Front or Head has address 16. Which represents the Head Node.

Following image represents the word "smart" with respective nodes and their addressing.

​

Numbers represented by circle are addresses of respective nodes.

The head pointer is: 20

7 0
3 years ago
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