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
Brilliant_brown [7]
2 years ago
12

Pls list up to five key lessons and knowledge areas that you have acquired in this course about operations management. How do yo

u believe they help you in your future professional career?
Try to describe your response in brief detail.
Engineering
1 answer:
Aneli [31]2 years ago
3 0

Explanation:

Production planning: Planning is ideal so that there are the right resources, at the right time and in the right quantity that can meet the production needs of a period.

Strategies: The strategic development of production is the area that will assist in organizational competitiveness and in meeting consumer demand and needs.

Product and service design: Development of new products and services and their improvement, innovations and greater benefits

Production systems: Study of physical arrangements so that production takes place effectively according to the ideal layout for each type of product or service.

Production capacity planning: Analysis of the short, medium and long term related to production, and identification if necessary to obtain more resources, increase in staff, machinery, etc., to meet present and future demands.

<em>Each area of ​​knowledge acquired will assist in the development of a professional career, as technical knowledge is essential in decision-making, provision, problem solving, the development of new ideas and innovation.</em>

You might be interested in
Create two arrays with 5 elements each: one will hold Strings and the second will hold integers. Write a program to ask the user
MrMuchimi

Answer:

#include <iostream>

#include <iomanip>

#include <string>

using namespace std;

int main() {

   string name[5];  

   int age[5];  

   int i,j;  

   for ( i = 0; i<=4; i++ ) {  

       cout << "Please enter student's name:";  

       cin >> name[i];  

       cout << "Please enter student's age:";  

       cin >> age[i];          

   }  

for (i=0;i<=4;i++){

   cout<<"Age of  "<< name[i]<<"  is  "<<age[i]<<endl;  

}

}

Output of above program is displayed in figure attached.

5 0
3 years ago
Air enters the combustor of a jet engine at p1=10 atm, T1=1000°R, and M1=0.2. Fuel is injected and burned, with a fuel/air mass
snow_lady [41]

Answer:

M2 = 0.06404

P2 = 2.273

T2 = 5806.45°R

Explanation:

Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.

Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,

To1 = (1.008)*(1000) = 1008 ºR

R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)

F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga

For the air q = cp(To2– To1)

(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2

Table A.3 of steam table gives P/P* = 2.273,

T/T* = 0.2066,

To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =

F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit

5 0
2 years ago
In part A you are asked to write the pseudocode for the program. In part B you are asked to write the syntax of the code for the
Naya [18.7K]

Answer:

C++.

Explanation:

#include <iostream>

#include <string>

using namespace std;

///////////////////////////////////////////////////////////////

int main() {

   string quote, book;

   int page;

   

   cout<<"What is your favorite quote from a book?"<<endl;

   getline(cin, quote);

   cout<<endl;

   /////////////////////////////////////////////

   cout<<"What book was that quote from?"<<endl;

   getline(cin, book);

   cout<<endl;

   /////////////////////////////////////////////

   cout<<"What page was that quote from?"<<endl;

   cin>>page;

   cout<<endl;

   /////////////////////////////////////////////

   int no_of_upper_characters = 0;

   for (int i=0; i<quote.length(); i++) {

       if (isupper(quote[i]))

          no_of_upper_characters++;

   }

   

   cout<<"No. of upper case characters: "<<no_of_upper_characters<<endl;

   /////////////////////////////////////////////

   int no_of_characters = quote.length();

   cout<<"No. of characters: "<<no_of_characters<<endl;

   /////////////////////////////////////////////

   bool isDog = false;

   for (int i=0; i<quote.length(); i++) {

       if (isDog == true)

           break;

       else if (quote[i] == 'd') {

           for (int j=i+1; j<quote.length(); j++) {

               if (isDog == true)

                   break;

               else if (quote[j] == 'o') {

                   for (int z=j+1; z<quote.length(); z++) {

                       if (quote[z] == 'g') {

                           isDog = true;

                           break;

                       }

                   }

               }

           }

       }

   }

   

   if (isDog == true)

       cout<<"This includes 'd' 'o' 'g' in the quote";

   //////////////////////////////////////////////

   return 0;

}

3 0
3 years ago
A direct contact heat exchanger (where the fluid mixes completely) has three inlets and one outlet. The mass flow rates of the i
lara31 [8.8K]

Answer:

Enthalpy at outlet=284.44 KJ

Explanation:

m_1=1 Kg/s,m_2=1.5 Kg/s,m_3=22 Kg/s

h_1=100 KJ/Kg,h_2=120 KJ/Kg,h_3=500 KJ/Kg

We need to Find enthalpy of outlet.

Lets take the outlet mass m and outlet enthalpy h.

So from mass conservation

m_1+m_2+m_3=m

   m=1+1.5+2 Kg/s

  m=4.5 Kg/s

Now from energy conservation

m_1h_1+m_2h_2+m_3h_3=mh

By putting the values

1\times 100+1.5\times 120+2\times 500=4.5\times h

So h=284.44 KJ

4 0
3 years ago
Complete function PrintPopcornTime(), with int parameter bagOunces, and void return type. If bagOunces is less than 3, print "To
weqwewe [10]

Answer:

#include <iostream>

using namespace std;

void PrintPopcornTime(int bagOunces) {

if(bagOunces < 3){

 cout << "Too small";

 cout << endl;

}

else if(bagOunces > 10){

 cout << "Too large";

 cout << endl;

}

else{

 cout << (6 * bagOunces) << " seconds" << endl;

}

}

int main() {

  PrintPopcornTime(7);

  return 0;

}

Explanation:

Using C++ to write the program. In line 1 we define the header "#include <iostream>"  that defines the standard input/output stream objects. In line 2 "using namespace std" gives me the ability to use classes or functions, From lines 5 to 17 we define the function "PrintPopcornTime(), with int parameter bagOunces" Line 19 we can then call the function using 7 as the argument "PrintPopcornTime(7);" to get the expected output.

8 0
3 years ago
Other questions:
  • What is EL Niño?
    9·1 answer
  • What is pixel's intensity ?​
    8·1 answer
  • Refers to the capability to keep moving forward on a specified grade.
    5·1 answer
  • A piston-cylinder device contains 0.8 kg of steam at 300°C and 1 MPa. Steam is cooled at constant pressure until one-half of the
    9·1 answer
  • Are you able to text without looking at your phone?
    10·1 answer
  • As the impurity concentration in solid solution of a metal is increased, the tensile strength:________.a) decreasesb) increasesc
    9·1 answer
  • What is the relationship between orifice diameter and pipe diameter​
    15·1 answer
  • What engine does the Mercedes 400e have?
    10·1 answer
  • Which of the following is true regarding screw gauges and shank?
    5·1 answer
  • There are some sections of the SDS that are not mandatory.
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!