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3 years ago

A program in which students learn a specific skill for a particular job is called a

Computers and Technology

2 answers:

Ne4ueva [31]3 years ago

7 0

A certification program.

Zigmanuir [339]3 years ago

7 0

Answer:

<h2>Certification Program.</h2>

Explanation:

A certification program is defined as a set of indicators to give a training program. This is often offered by companies to their members to prove that they have all skills needed, also, it's aimed to give member new skills that the company needs due to rapid changes in the market.

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C++ code only

nordsb [41]

Using the computational language C++ it is possible to write a code is possible to create a function that stores items like this:

<h3>Writting the code in C++: </h3>

//Header files

#include <iostream>

#include <string>

#include <fstream>

using namespace std;

//Class inventoryItem

class inventoryItem

{

public:

//Variable declaration

string itemName;

string itemNumber;

double itemPrice;

int itemQuantity;

};

//implementation of menuItems() function.

void menuItems()

{

// Display the Menu

cout << "=============================\n";

cout << " 1) Add an item to memory\n"

<< " 2) Search memory for an item \n"

<< " 3) List what's in memory\n"

<< " 4) Total value on hand\n"

<< " 5) Save to a file\n"

<< " 6) Read from a file\n"

<< " 0) Exit Program" << endl;

}

//main method.

int main()

{

//variable declaration.

int mnuChoice, cn = 0, d;

string n;

double sum = 0, c;

int temp;

string a, b;

string inputBuffer;

inventoryItem items[100];

ifstream inventoryFile("inventory.txt");

if (inventoryFile)

{

//store the data from the file to memory.

while (!(inventoryFile.eof()))

{

inventoryFile >> items[cn].itemName;

inventoryFile >> items[cn].itemNumber;

inventoryFile >> inputBuffer;

items[cn].itemPrice = atof(inputBuffer.c_str());

inventoryFile >> inputBuffer;

items[cn].itemQuantity = atoi(inputBuffer.c_str());

cn++;

}

inventoryFile.close();

ifstream inventoryFile("inventory.txt");

ofstream fout;

fout.open("inventory.txt");

do

{

menuItems();

cout << cn << " items stored." << endl;

// Prompt and read choice form the user

cout << "Enter your choice: ";

cin >> mnuChoice;

switch (mnuChoice)

{

//Add a new inventory item to the data in memory.

case 1:

// Prompt and read name of the item from the user

cout << "Enter name of the item: ";

cin >> inputBuffer;

items[cn].itemName = inputBuffer;

// Prompt and read itemNumber from the user

cout << "Enter the itemNumber: ";

cin >> inputBuffer;

items[cn].itemNumber = inputBuffer;

// Prompt and read itemPrice from the user

cout << "Enter the itemPrice: ";

cin >> inputBuffer;

items[cn].itemPrice = atof(inputBuffer.c_str());

// Prompt and read itemQuantity from the user

cout << "Enter the itemQuantity: ";

cin >> inputBuffer;

items[cn].itemQuantity = atoi(inputBuffer.c_str());

cn++;

break;

//Search the inventory for an item by inventory number.

case 2:

temp = 0;

cout << "Enter inventory itemNumber: ";

cin >> n;

for (int i = 0; i < cn&&temp != 1; i++)

{

// If found, print the stored info

if (items[i].itemNumber.compare(n) == 0)

{

// Display the stored information

cout << "Item Name" << "\t" << "Item Number" << "\t" << " Item Price" << "\t" << " Item Quantity" << endl;

cout << items[i].itemName << "\t " << items[i].itemNumber << " \t" << items[i].itemPrice << "\t " << items[i].itemQuantity << endl;

temp = 1;

}

// otherwise print 'not found'.

if (temp == 0)

{

cout << "Not found!!!" << endl;

}

break;

//List the inventory on the screen

case 3:

cout << "The data in the memory is:" << endl;

cout << "Item Name" << "\t" << "Item Number" << "\t" << " Item Price" << "\t" << " Item Quantity" << endl;

for (int i = 0; i < cn; i++)

{

cout << items[i].itemName << "\t " << items[i].itemNumber << "\t " << items[i].itemPrice << " \t" << items[i].itemQuantity << endl;

}

break;

// Calculate the total value on hand.

case 4:

for (int i = 0; i < cn; i++)

{

// The sum is the accumulation of all the (quantities * prices).

sum += items[i].itemPrice*items[i].itemQuantity;

}

cout << "sum value on hand: " << sum << endl;

break;

// Save the inventory data in the given file

case 5:

for (int i = 0; i < cn; i++)

{

if (i == cn - 1)

{

fout << items[i].itemName << " \t" << items[i].itemNumber << " \t" << items[i].itemPrice << " \t" << items[i].itemQuantity;

}

else

{

fout << items[i].itemName << " \t" << items[i].itemNumber << " \t" << items[i].itemPrice << " \t" << items[i].itemQuantity << endl;

}

break;

// Read the data from the file

case 6:

while (!(inventoryFile.eof()))

{

inventoryFile >> a;

inventoryFile >> b;

inventoryFile >> inputBuffer;

c = atof(inputBuffer.c_str());

inventoryFile >> inputBuffer;

d = atoi(inputBuffer.c_str());

// Dsiplays the data in the file

cout << a << " " <

}

break;

}

// Check the menu choice is not equal to 0,

//if it is equal exit from the loop

} while (mnuChoice != 0);

// File closed

fout.close();

inventoryFile.close();

}

// If the inventory file does not exist

// and then display the error message.

else

{

cout << "File does not exist!" << endl;

}

return 0;

}

See more about C++ code at brainly.com/question/17544466

#SPJ1

5 0

2 years ago

Return the "centered" average of an array of ints, which we'll say is the mean average of the values, except ignoring the larges

denpristay [2]

Answer:

The code to this question can be defined as follows:

public double centeredAverage(ArrayList<Integer> nums) //defining a method centeredAverage that accepts an array nums

{

if ((nums == null) || (nums.size() < 3))//define if block for check simple case value

{

return 0.0;//return float value

}

double sum = 0; //defining a double variable sum

int smallest = nums.get(0);//defining integer variable and assign value

int largest = nums.get(0);//defining integer variable and assign value

for (int i = 0; i < nums.size(); i++) //defining for loop for stor ith value

{

int value = nums.get(i);//defining integer value variable to hold nums value

sum = sum + value;//use sum variable for add value

if (value < smallest)//defining if block to check value less then smallest

smallest = value;//hold value in smallest variable

else if (value > largest)//defining else if block that checks value greater them largest value

largest = value; //hold value in largest variable

}

sum = sum - largest - smallest; // use sum to decrease the sum largest & smallest value

return (sum / (nums.size() - 2)); //use return keyword for return average value

}

Explanation:

In the question, data is missing, that's why the full question is defined in the attached file please find it.

In the given code a method "centeredAverage" is used that accepts an array in its parameter and inside the method, if block is used that checks the cash value if it is true, it will return a float value.

In the next step, two integers and one double variable is defined, that use the conditional statement for check hold value in its variable and at the last use return keyword for return average value.

3 0

4 years ago

Select the correct answer.

Mekhanik [1.2K]

The best answer is A

5 0

3 years ago

Read 2 more answers

What does it mean for a heap to be complete?

Dmitry [639]

Answer:

A complete heap is a heap in which the all the nodes are completely filled and the level of the each node are equal to each other. So, if the heap satisfied this condition then, the heap are complete.

In the computer science, heap is the special type of tree based on the data structure. Binary heap is one of the type of heap and it become complete binary tree when, it satisfied the ordering of heap property.

There are basically two types of heap ordering that are:

1) MAX heap property: In the max heap ordering property each node are less than or equal to its parent node value.

2) MIN heap property: In the min heap property each node is greater and equal to its particular parent value.

3 0

4 years ago

If you're driving a car and know there's a motorcyclist around you, you should______. A. Not change your driving B. Check your b

Sedaia [141]

b check your blind spots frequently i guess not sure

5 0

3 years ago