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Vitek1552 [10]
3 years ago
8

Does Someone own a fennec on Rocket League and can trade it and is on the Switch?

Computers and Technology
2 answers:
wlad13 [49]3 years ago
8 0

Answer:

YES

Explanation:

Elanso [62]3 years ago
4 0

Answer:

no

Explanation:

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A teacher wants to create a list of students in her class. Using the existing Student class in this exercise. Create a static Ar
aleksandr82 [10.1K]

Answer:

Check the explanation

Explanation:

CODE:-

import java.util.ArrayList;

class Student

{

  private String name;

  private int grade;

  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;

  }

 

  /*Don't change the code in this method!

  This method will print out all the Student names in the classList Array

  */

  public static String printClassList()

  {

      String names = "";

      for(Student name: classList)

      {

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

      }

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

  }

}

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);

      Student smith = new Student("Smith", 11);

      Student kane = new Student("Kane", 10);

      Student virat = new Student("Virat", 12);

      Student abd = new Student("ABD", 12);

      Student root = new Student("Root", 12);

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

  }

}

Kindly check the attached image below.

4 0
3 years ago
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
The Beaufort Wind Scale is used to characterize the strength of winds. The scale uses integer values and goes from a force of 0,
TEA [102]

Answer:

ranforce = randi([0, 12]);

if (ranforce == 0)

     disp('There is no wind')

else  if(ranforce>0 && ranforce <7)

     disp('There is a breeze')

else  if(ranforce>6 && ranforce <10)

     disp('This is a gale')

else  if(ranforce>9 && ranforce <12)

     disp('It is a storm')

else  if(ranforce==12)

     disp('Hello, Hurricane!')

end

Explanation:

<em>Replace all switch case statements with if and else if statements.</em>

<em>An instance is:</em>

<em>case {7,8,9}</em>

<em>is replaced with</em>

<em>else  if(ranforce>9 && ranforce <12)</em>

<em>All other disp statements remain unchanged</em>

5 0
3 years ago
HELLLP
mojhsa [17]

Explanation:

The output of this program is 5 7, because the first time bruce is printed, his value is 5, and the second time, his value is 7. The comma at the end of the first print statement suppresses the newline after the output, which is why both outputs appear on the same line.

Here is what multiple assignment looks like in a state diagram:



With multiple assignment it is especially important to distinguish between an assignment operation and a statement of equality. Because Python uses the equal sign (=) for assignment, it is tempting to interpret a statement like a = b as a statement of equality. It is not!

First, equality is symmetric and assignment is not. For example, in mathematics, if a = 7 then 7 = a. But in Python, the statement a = 7 is legal and 7 = a is not.

Furthermore, in mathematics, a statement of equality is always true. If a = b now, then a will always equal b. In Python, an assignment statement can make two variables equal, but they don’t have to stay that way:

a = 5

4 0
3 years ago
Read 2 more answers
Match the order in which you should develop a plan:
Advocard [28]

Develop

Present

Approve

Brainliiest please?

4 0
3 years ago
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