1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Naddika [18.5K]
3 years ago
12

As a highly flexible structure, __________ can be quickly configured to adapt as required due to changes.

Computers and Technology
1 answer:
tresset_1 [31]3 years ago
7 0

As a highly flexible structure, a <u>matrix organization</u> can be quickly configured to adapt as required due to changes.

<h3>What is a matrix organizational structure?</h3>

A matrix organizational structure can be defined as a type of work structure where reporting relationships between employees and the top executive (employers) are set up as a matrix rather than the conventional hierarchy approach, which makes it highly flexible and adaptable to subsequent changes.

<h3>The types of matrix organizational structure.</h3>

In business management, there are three types of matrix organizational structure and these include:

  • Weak matrix structure.
  • Balanced matrix structure.
  • Strong matrix structure.

Read more on matrix organization here: brainly.com/question/7437866

You might be interested in
Write a program that create Employee class with fields id,name and sal and create Employee object and store data and display tha
sashaice [31]

Answer:

Here is the C++ program for Employee class with fields id,name and sal.

#include <iostream>  // to use input output functions

#include <string>  //to manipulate and use strings

using namespace std;   // to access objects like cin cout

class Employee {  //class Employee

private:  

/* the following data members are declared as private which means they can only be accessed by the functions within Employee class */

  string name;  //name field

  int id; //id field

  double sal;   //salary field

public:    

  Employee();  // constructor that initializes an object when it is created

/* setName, setID and setSalary are the mutators which are the methods used to change data members. This means they set the values of a private fields i.e. name, id and sal */

  void setName(string n)  //mutator for name field

     { name = n; }        

  void setId(int i)  //mutator for id field

     { id = i; }        

  void setSalary(double d)  //mutator for sal field

     { sal = d; }  

/* getName, getID and getSalary are the accessors which are the methods used to read data members. This means they get or access the values of a private fields i.e. name, id and sal */

  string getName()  //accessor for name field

     { return name; }        

  int getId()  //accessor for id field

     { return id; }        

  double getSalary()  //accessor for sal field

     { return sal; }  };  

Employee::Employee() {  //default constructor where the fields are initialized

  name = "";  // name field initialized

  id = 0;  // id field initialized to 0

  sal = 0;   }   // sal field initialized to 0

void display(Employee);  

// prototype of the method display() to display the data of Employee

int main() {  //start of the main() function body

  Employee emp;  //creates an object emp of Employee class

/*set the name field to Abc Xyz which means set the value of Employee class name field to Abc Xyz  through setName() method and object emp */

  emp.setName("Abc Xyz");  

/*set the id field to 1234 which means set the value of Employee class id field to 1234  through setId() method and object emp */

  emp.setId(1234);

/*set the sal field to 1000 which means set the value of Employee class sal field to 1000  through setSalary() method and object emp */

  emp.setSalary(1000);    

  display(emp);  }   //calls display() method to display the Employee data

void display(Employee e) {  // this method displays the data in the Employee //class object passed as a parameter.

/*displays the name of the Employee . This name is read or accessed through accessor method getName() and object e of Employee class */

  cout << "Name: " << e.getName() << endl;  

/*displays the id of the Employee . This id is read or accessed by accessor method getId() and object e */

  cout << "ID: " << e.getId() << endl;

/*displays the salary of the Employee . This sal field is read or accessed by accessor method getSalary() and object e */

  cout << "Salary: " << e.getSalary() << endl;  }

Explanation:

The program is well explained in the comments mentioned with each statement of the program.

The program has a class Employee which has private data members id, name and sal, a simple default constructor Employee(), mutatator methods setName, setId and setSalary to set the fields, acccessor method getName, getId and getSalary to get the fields values.

A function display( ) is used to display the Employee data i.e. name id and salary of Employee.

main() has an object emp of Employee class in order to use data fields and access functions defined in Employee class.

The output of the program is:

Name: Abc Xyz                                                                                                      

ID: 1234                                                                                                                  

Salary: 1000

The program and its output are attached.

5 0
3 years ago
If the transmitting device would like for the receiving device to push currently buffered data to the application on the receivi
Andreyy89

Answer:

The right approach will be "PSH".

Explanation:

  • The PSH flag seems to be an acceptable solution through TCP which always encourages the submit implementation to immediately transfer data but only whenever the buffer isn't complete.
  • The implementation helps to formulate this same flag towards true for something like the connection, as well as TCP will instantly begin transmitting the relevant information.
3 0
3 years ago
Suppose a computer has 16-bit instructions. The instruction set consists of 32 different operations. All instructions have an op
Kazeer [188]

Answer:

2^7= 128

Explanation:

An instruction format characterizes the diverse part of a guidance. The fundamental segments of an instruction are opcode and operands. Here are the various terms identified with guidance design:  Instruction set size tells the absolute number of guidelines characterized in the processor.  Opcode size is the quantity of bits involved by the opcode which is determined by taking log of guidance set size.  Operand size is the quantity of bits involved by the operand.  Guidance size is determined as total of bits involved by opcode and operands.

6 0
4 years ago
What is the difference in the way computers store numeric data and non-numeric data?
Elan Coil [88]
Both numeric and non-numeric data have different ASCII codes, that will be the first difference. Second, numeric data are stored in float, double, and int variables while non-numeric data are stored in char. However, both can be stored in a String variable if and only if the desired data to be stored will not be used for computuation using mathematical operations such as addition.

To store a non-numeric data in an array, the array must first be declared as a String variable array while a numeric data can be stored in an array once the array which will store the numeric data is declared as an integer, double, or float variable.
3 0
3 years ago
Data Structure in C++
agasfer [191]

The code .cpp is available bellow

#include<iostream>

using namespace std;

//declaring variables

void merge(int* ip, int sz, int* opt, bool opt_asc); //merging

int* mergesort(int* ip, int sz);

void mergesort(int *ip, int sz, int* opt, bool opt_asc);

void merge(int* ip, int sz, int* opt, bool opt_asc)

{

  int s1 = 0;

  int mid_sz = sz / 2;

  int s2 = mid_sz;

  int e2 = sz;

  int s3 = 0;

  int end3 = sz;

  int i, j;

   

  if (opt_asc==true)

  {

      i = s1;

      j = e2 - 1;

      while (i < mid_sz && j >= s2)

      {

          if (*(ip + i) > *(ip + j))

          {

              *(opt + s3) = *(ip + j);

              s3++;

              j--;

          }

          else if (*(ip + i) <= *(ip + j))

          {

              *(opt + s3) = *(ip + i);

              s3++;

              i++;

          }

      }

      if (i != mid_sz)

      {

          while (i < mid_sz)

          {

              *(opt + s3) = *(ip + i);

              s3++;

              i++;

          }

      }

      if (j >= s2)

      {

          while (j >= s2)

          {

              *(opt + s3) = *(ip + j);

              s3++;

              j--;

          }

      }

  }

  else

  {

      i = mid_sz - 1;

      j = s2;

      while (i >= s1 && j <e2)

      {

          if (*(ip + i) > *(ip + j))

          {

              *(opt + s3) = *(ip + i);

              s3++;

              i--;

          }

          else if (*(ip + i) <= *(ip + j))

          {

              *(opt + s3) = *(ip + j);

              s3++;

              j++;

          }

      }

      if (i >= s1)

      {

          while (i >= s1)

          {

              *(opt + s3) = *(ip + i);

              s3++;

              i--;

          }

      }

      if (j != e2)

      {

          while (j < e2)

          {

              *(opt + s3) = *(ip + j);

              s3++;

              j++;

          }

      }

  }

   

  for (i = 0; i < sz; i++)

      *(ip + i) = *(opt + i);

}

int* mergesort(int* ip, int sz)

{

  int* opt = new int[sz];

   

  mergesort(ip, sz, opt, true);

  return opt;

}

void mergesort(int *ip, int sz, int* opt, bool opt_asc)

{

  if (sz > 1)

  {

      int q = sz / 2;

      mergesort(ip, sz / 2, opt, true);

      mergesort(ip + sz / 2, sz - sz / 2, opt + sz / 2, false);

      merge(ip, sz, opt, opt_asc);

  }

}

int main()

{

  int arr1[12] = { 5, 6, 9, 8,25,36, 3, 2, 5, 16, 87, 12 };

  int arr2[14] = { 2, 3, 4, 5, 1, 20,15,30, 2, 3, 4, 6, 9,12 };

  int arr3[10] = { 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 };

  int *opt;

  cout << "Arays after sorting:\n";

  cout << "Array 1 : ";

  opt = mergesort(arr1, 12);

  for (int i = 0; i < 12; i++)

      cout << opt[i] << " ";

  cout << endl;

  cout << "Array 2 : ";

  opt = mergesort(arr2, 14);

  for (int i = 0; i < 14; i++)

      cout << opt[i] << " ";

  cout << endl;

  cout << "Array 3 : ";

  opt = mergesort(arr3, 10);

  for (int i = 0; i < 10; i++)

      cout << opt[i] << " ";

  cout << endl;

  return 0;

}

4 0
4 years ago
Other questions:
  • Write a program that calls fork(). Before calling fork(), have the main process access a variable (e.g., x) and set its value to
    5·1 answer
  • Memory chips are used in ____ storage devices to read and write data
    9·1 answer
  • What is a possible style of formatting your company could prefer?
    14·2 answers
  • How should you decide what to wear to an interview? What kind of things should be considered?
    5·1 answer
  • False when you tap or click the ‘decrease font size’ button, excel assigns the next lowest font size in the font size gallery to
    13·1 answer
  • After compiling source code, which command still needs to be run in order to copy the newly compiled binaries into a directory l
    5·1 answer
  • When designing a database, what should your first step be?
    5·1 answer
  • What are the similarities between data mining and data analytics
    11·1 answer
  • Which of the following statements is FALSE?
    5·1 answer
  • During which phase of system development would you acquire any necessary hardware and software?
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!