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Colt1911 [192]
3 years ago
10

A network of Bennetton retail sales agents select the Bennetton designs that they feel will appeal to Bennetton retail customers

. The responsibility for picking the next "hot style" is in the hands of this group. This is an example of a _____ structure.
Computers and Technology
2 answers:
denpristay [2]3 years ago
7 0

Answer:

It is an example of a matrix structure

Explanation:

The matrix structure is a structure in which jurisdiction, controls and duties are carried and uphold by a group of employees instead of just the manager.

In other words, employees have dual reporting relationships; they can give reports to the functional manager and they can also give reports to the product manager.

The Bennetton design selected by the network of Bennetton retail sales agents made activities to be easily coordinated because, they have collectively performed the duty as a unit and this is one of the advantages of this form of the matrix organizational structure.

Semenov [28]3 years ago
3 0

Answer:

The correct answer is letter "B": committee.

Explanation:

Committee structures are implemented in organizations in an attempt to have different components of an organization providing an opinion on certain matters of the operations of the business. In most cases, the committee is composed of members of the same department of the company so it will be easier for them to gather and make decisions. In such organizations, top managers are not the only ones choosing the set of actions the entity will take.

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4.2 lesson practice help plzs
sertanlavr [38]

The output is 21 because c is incremented by 3 until it surpasses 10 and the value of c is added to sum each time the loop runs.

I hope this helps!

8 0
2 years ago
​Case-Based Critical Thinking Questions​Case 9-1Melissa, a computer science engineering student, is learning the basics of progr
Sunny_sXe [5.5K]

Answer:

Option D:  monthName.splice(2, 3, "Mar", "Apr", "May");

Explanation:

splice is a JavaScript array method which can be used to insert or remove item(s).

It has general syntax:  splice( i, n , item1, item2, item3, .... , itemX)

  • i : index position of the array where the item should be inserted or removed
  • n: number of item(s) should be removed starting from index-i
  • item1,item2,....itemX :  item(s) to be inserted

If we have an original monthName array with elements:

["January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"]

monthName.splice(2, 3, "Mar", "Apr", "May")  will

  • remove 3 items from index position-2 ("March", "April" and "May")
  • and insert three new items, "Mar", "Apr" and "May".

New array:

["January", "February", "Mar", "Apr", "May", "June", "July", "August", "September", "October", "November", "December"]

4 0
2 years ago
What variable type is the output of a logical operator in labview?
Stella [2.4K]

Answer:

Boolean data types

Explanation:

consist of only two values i.e. true and false. It is a logical data type providing the output in the form of 0 or 1 specifying false and true respectively. The Boolean data type is indicated by green data wires. LabView stores the Boolean data as 8-bit values.

5 0
2 years ago
Develop a plan to design and finally implement a set of functions using C++ that would implement the IEEE standard. Phase 1 will
AnnZ [28]

Answer:

See explaination code

Explanation:

#include<iostream>

#include<math.h>

using namespace std;

typedef union {

float number;

struct

{

// Order is important.

// Here the members of the union data structure

// use the same memory (32 bits).

// The ordering is taken

// from the LSB to the MSB.

unsigned int mantissa : 23;

unsigned int exponent : 8;

unsigned int sign : 1;

} Raw;

} MyFloat;

void printBinary(int n, int i)

{

// Prints the binary representation

// of a number n up to i-bits.

int k;

for (k = i - 1; k >= 0; k--) {

if ((n >> k) & 1)

cout << "1";

else

cout << "0";

}

}

void decToHex(int n){

// char array to store hexadecimal number

char hexaDeciNum[100];

// counter for hexadecimal number array

int i = 0;

while(n!=0)

{

// temporary variable to store remainder

int temp = 0;

// storing remainder in temp variable.

temp = n % 16;

// check if temp < 10

if(temp < 10)

{

hexaDeciNum[i] = temp + 48;

i++;

}

else

{

hexaDeciNum[i] = temp + 55;

i++;

}

n = n/16;

}

// printing hexadecimal number array in reverse order

for(int j=i-1; j>=0; j--)

cout << hexaDeciNum[j];

}

void floatBinary(float f){

long double binaryTotal, binaryFrac = 0.0, frac, fracFractor = 0.1;

long int integer, binaryInt = 0;

long int p = 0, rem, temp;

//separate the integer part from the input floating number

integer = (int)f;

//separate the fractional part from the input floating number

frac = f - integer;

//loop to convert integer part to binary

while (integer != 0) {

rem = integer % 2;

binaryInt = binaryInt + rem *pow(10, p);

integer = integer / 2;

p++;

}

//loop to convert fractional part to binary

while (frac != 0) {

frac = frac * 2;

temp = frac;

binaryFrac = binaryFrac + fracFractor * temp;

if (temp == 1)

frac = frac - temp;

fracFractor = fracFractor / 10;

}

cout << binaryInt + binaryFrac;

}

int findDecimal(float number){

int nfloor = number;

float nfloat = number - nfloor;

int nfloatfloor;

do {

nfloat *= 10;

nfloatfloor = nfloat;

} while (nfloat > nfloatfloor);

return nfloatfloor;

}

void first(float number){

if(number < 0)

cout << "SIGN BIT IS (1) SINCE NUMBER IS NEGATIVE" << endl;

else

cout << "SIGN BIT IS (0) SINCE NUMBER IS POSITIVE" << endl;

}

void second(float number){

cout << "INTEGER PART IN BASE-10:" << int(number) << " AND IN BINARY:";

printBinary(int(number),16);

cout << endl;

}

void third(float number){

cout << "DECIMAL PART IN BASE-10:" << findDecimal(number) << " AND IN BINARY:";

printBinary(findDecimal(number),16);

cout << endl;

}

void fourth(float number){

cout << "ENTERED NUMBER IN BASE-10:" << number << " AND IN BINARY:";

floatBinary(number);

cout << endl;

}

void fifth(MyFloat myfloat){

cout << "MANTISA IN BINARY:";

printBinary(myfloat.Raw.mantissa,32);

}

void sixth(MyFloat myfloat){

cout << "EXPONENT IN BASE-10:" << myfloat.Raw.exponent << " AND IN BINARY:";

printBinary(myfloat.Raw.exponent,8);

cout << endl;

}

void seventh(MyFloat myfloat){

cout << myfloat.Raw.sign << " | ";

printBinary(myfloat.Raw.exponent,8);

cout << " | ";

printBinary(myfloat.Raw.mantissa,32);

cout << endl;

}

void eigth(MyFloat myfloat){

cout << myfloat.Raw.sign << " | ";

decToHex(myfloat.Raw.exponent);

cout << " | ";

decToHex(myfloat.Raw.mantissa);

cout << endl;

}

int main(){

float number;

cout << "PLEASE ENTER A NUMBER TO DISPLAY THE IEEE 754 FLOATING POINT OPTIONS" << endl;

cin >> number;

MyFloat myfloat;

myfloat.number = number;

cout << "PLEASE CHOOSE ONE OF THE FOLLOWING OPERATIONS" << endl;

cout << " 1. DISPLAY THE SIGN BIT VALUE" << endl;

cout << " 2. DISPLAY THE INTEER PART IN BOTH BASE-10 AND CINARY FORMATS" << endl;

cout << " 3. DISPLAY THE DECIMAL PART IN BOTH BASE-10 AND BINARY FORMATS" << endl;

cout << " 4. DISPLAY THE NUMBER ENTERED IN BOTH BASE-10 AND BINARY FORMATS" << endl;

cout << " 5. DISPLAY THE MANTISA IN BINARY FORMATS" << endl;

cout << " 6. DISPLAY THE EXPONENT IN BORH BASEE-10 AND BINARY FORMATS" << endl;

cout << " 7. DISPLAY THE IEEE 754 SINGLE PRECISION BINARY LAYOUT" << endl;

cout << " 8. DISPLAY THE IEEE 754 SINGLE PRECISION BINARY LAYOUT" << endl;

int choice;

cin >> choice;

switch(choice){

case 1:first(number);

break;

case 2:second(number);

break;

case 3:third(number);

break;

case 4:fourth(number);

break;

case 5:fifth(myfloat);

break;

case 6:sixth(myfloat);

break;

case 7:seventh(myfloat);

break;

case 8:eigth(myfloat);

break;

default:cout << "ENTER VALID CHOICE" << endl;

}

}

Refer to attachment please for onscreen look.

6 0
2 years ago
Write pseudocode to solve the following problem: You are given an array A[1 . . . n] whose each element is a point of the plane
bixtya [17]

Answer:

Answer explained below

Explanation:

void bubbleSort(int X[], int Y[], int n)

{

   int i, j;

   for (i = 0; i < n-1; i++)      

     

   // Last i elements are already in place

   for (j = 0; j < n-i-1; j++)

       if (X[j] > X[j+1])

{

swap(X[j],X[j+1])

swap(Y[j],Y[j+1]);

}

       if (X[j] == X[j+1]&&Y[j]<Y[j+1])

{

swap(X[j],X[j+1])

swap(Y[j],Y[j+1]);

}

}

Since the above algorithm contains 2 nested loops over n.

So, it is O(n^2)

5 0
3 years ago
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