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Inessa05 [86]
3 years ago
6

The operations planning practice of inputting sales forecasts into computer software that accurately predicts the amount and tim

ing of materials needed to complete products and projects is ________________.
Computers and Technology
1 answer:
netineya [11]3 years ago
3 0

Answer:

"Materials requirement planning" is the correct answer for the above question.

Explanation:

  • The "Materials requirement planning" is a software system that is used to hold the record of the raw materials.
  • It is used to tell about the material which is present in the stocks. It also used to schedule the delivery.
  • This software is used to enhance the productivity of the company.
  • The above question asked about the software which is needed to keep the record of the raw material. This software is known as "Materials requirement planning".
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Which methods will remove filters? Check all that apply.
dezoksy [38]

Answer:

The last three options are the correct ones

Explanation:

6 0
3 years ago
Read 2 more answers
Which of these is a discipline?
ivolga24 [154]

Answer: Information technology

Nanotechnology

Explanation:

Database Managers are the individuals who responsible create databases for organizations and maintain existing ones.

Nanotechnology simply refers to a research field that has to do with the building of things and devices.

Information technology refers to the study of computers for the storage and sending of information.

A network technician is someone who maintains, and repairs computer systems and they also set up several Internet connections, and networks.

We should note that information technology and nanotechnology are both disciplines as they are studied in different colleges.

3 0
3 years ago
Create a program in c/c++ which accepts user input of a decimal number in the range of 1 -100. Each binary bit of this number wi
kirill115 [55]

Answer:

// here is code in C++.

#include <bits/stdc++.h>

using namespace std;

void switch_fun(int input)

{

//array to store binary of input number

   int bin[7];

   // convert the number into binary

   for(int i=0; input>0; i++)

       {

           bin[i]=input%2;

           input= input/2;

       }

   // print the switch number and their status

   cout<<"Switch number:\t 1\t 2\t 3\t 4\t 5\t 6\t 7 \n";

   cout<<"status:      ";

   for(int j=6;j>=0;j--)

   {

   // if bit is 1 print "ON" else print "OFF"

       if(bin[j]==1)

           cout<<"\t ON";

       else

           cout<<"\t OFF";

   }

}

int main()

{

   int n;

   cout<<"enter the number between 1-100 only:");

   // read a number in decimal

   cin>>n

   // validate the input

   while(n<1 ||n>100)

   {

     cout<<"wrong input!! Input must be in between 1-100:"<<endl;

     cout<<"enter the number again :";

     cin>>n;

   }

// call the function with input parameter

  switch_fun(n);

return 0;

}

Explanation:

Read a decimal number from user. If the number is not between 1 to 100 the it will again ask user to enter the number until the input is in between 1-100.Then it will call the fun() with parameter n. First it will convert the decimal to binary array of size 7.Then print the status of switch.If bit is 1 then print "ON" else it will print "OFF".

Output:

enter the number between 1-100 only:-5

wrong input!! Input must be in between 1-100:

enter the number again :125

wrong input!! Input must be in between 1-100:

enter the number again :45

Switch number:   1       2       3       4       5       6       7

Status:          OFF     ON      OFF     ON      ON      OFF     ON

8 0
4 years ago
Typically, what form do most database designers consider a database structure to be normalized?
klemol [59]
Hello,

Answer is Third 

Hope This Help!!!
6 0
3 years ago
I want an A! Write a pseudocode algorithm that asks the user for their grades in CSE 1321, along with their attendance, then cal
Vsevolod [243]

Answer:

The Java pseudocode and the Java code itself are given for better understanding

Explanation:

<u>Java pseudocode</u>

  1. preAvg(int test[], int qgrade):
  2. sum := 0
  3. for each testi in test:
  4. sum := sum + testi
  5. sum := sum + qgrade
  6. return sum / 5
  7. additional(int at):
  8. res = (at * 1.5) / 30
  9. return res
  10. total = preAvg(test, qgrade) + additional(at)

<u>Java code</u>

import java.util.Scanner;

public class CSE1322 {

  public static void main(String[] args) {

     

      Scanner scanner=new Scanner(System.in);

     

      int test[] = new int[4];

     

      for(int i=1;i<=4;i++){

          System.out.println("Enter your result for test"+i+": ");

          test[i-1] = scanner.nextInt();

      }

     

      System.out.println("Enter the average quiz grade: ");

      int qgrade = scanner.nextInt();

     

      System.out.println("Enter the classes you attended(out of 30): ");

      int at = scanner.nextInt();

     

      double avg = preAvg(test,qgrade);

      System.out.println("Your average before attendance is: "+avg);

     

      double add = additional(at);

      System.out.println("You receive an additional "+add+" points for attendance");

     

      double total = avg + add;

     

      System.out.println("Final grade is : "+total);

     

  }

  private static double additional(int at) {

      double res = ((double)at * 1.5)/30.0;

      return res;

  }

  private static double preAvg(int[] test, int qgrade) {

      double sum = 0;

     

      for(int i=0;i<4;i++) sum += test[i];

     

      sum += qgrade;

     

      return sum/5;

  }

}

4 0
4 years ago
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