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Stells [14]
3 years ago
6

Write steps to Delete data from ‘Datagridview’

Computers and Technology
1 answer:
julsineya [31]3 years ago
8 0

Explanation:

private void btnDelete_Click(object sender, EventArgs e)

{

if (this.dataGridView1.SelectedRows.Count > 0)

{

dataGridView1.Rows.RemoveAt(this.dataGridView1.SelectedRows[0].Index);

}

}

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Most people entering the construction industry learn a specific craft through: An apprenticeship program
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"how do we store information in long term memory?"
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By saving it in the hard drive.

Hope I helped!

8 0
3 years ago
Coupon collector is a classic statistic problem with many practical applications. The problem is to pick objects from a set of o
ahrayia [7]

Answer:

Here is the JAVA program:

public class Main {  //class name

public static void main(String[] args) {   //start of main method

//sets all boolean type variables spades, hearts diamonds and clubs to false initially

   boolean spades = false;  

   boolean hearts = false;

   boolean diamonds = false;

   boolean clubs = false;  

   String[] deck = new String[4];  //to store card sequence

   int index = 0;  //to store index position

   int NoOfPicks = 0;  //to store number of picks (picks count)

   while (!spades || !hearts || !diamonds || !clubs) {   //loop starts

       String card = printCard(getRandomCard());  //calls printCard method by passing getRandomCard method as argument to it to get the card

       NoOfPicks++;   //adds 1 to pick count

       if (card.contains("Spades") && !spades) {  //if that random card is a card of Spades and spades is not false

           deck[index++] = card;  //add that card to the index position of deck

           spades = true;  //sets spades to true

       } else if (card.contains("Hearts") && !hearts) {  //if that random card is a card of Hearts and hearts is not false

           deck[index++] = card;  

           hearts = true;   //sets hearts to true

       } else if (card.contains("Diamond") && !diamonds) {  //if that random card is a card of Diamond and diamonds is not false

           deck[index++] = card;

           diamonds = true;  //sets diamonds to true

       } else if (card.contains("Clubs") && !clubs) {  if that random card is a card of Clubs and clubs is not false

           deck[index++] = card;

           clubs = true;         }     }   //sets clubs to true

   for (int i = 0; i < deck.length; i++) {  //iterates through the deck i.e. card sequence array

       System.out.println(deck[i]);     }  //prints the card number in deck

   System.out.println("Number of picks: " + NoOfPicks);  }   //prints number of picks

public static int getRandomCard() {  //gets random card

   return (int) (Math.random() * 52); }   //generates random numbers of 52 range

public static String printCard(int cardNo) {   //displays rank number and suit

   String[] suits = { "Spades", "Hearts", "Diamonds", "Clubs", };  //array of suits

   String[] rankCards = { "Ace", "2", "3", "4", "5", "6", "7", "8", "9", "10",

           "Jack", "Queen", "King" };   //array of rank

  int suitNo = cardNo / 13;  //divides card number by 13 and stores to suitNo

 int rankNo = cardNo % 13;   //takes modulo of card number and 13 and store it to rankNo

   return rankCards[rankNo] + " of " + suits[suitNo];  }}  //returns rankCard at rankNo index and suits at suitNo index

Explanation:

The program is explained in the comments attached with each line of code. The screenshot of the program along with its output is attached.

8 0
3 years ago
Even closed systems are never perfectly closed. <br>true false​
Oxana [17]

Answer:

True

Explanation:

You can learn through many different textbooks that a closed system is always closed.

3 0
3 years ago
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter
trapecia [35]

Answer:

5 Letters

Explanation:

So we need 12 distinct codes made of a single letter or a pair of letters.

What would be the least number of letters?

Lets try with 3 letters A, B and C

The possible combinations are: A, B, C, AB, AC, BC

These are 6 codes and we need 12 so lets try more A, B, C and D

The possible combinations are: A, B, C, D, AB, AC, AD, BC, BD, CD

These are 10 codes and we need 12 so lets try more A, B, C, D and E

The possible combinations are:

A, B, C, D, E, AB, AC, AD, AE, BC, BD, BE, CD, CE, DE

Finally we got 15 distinct codes which are more than 12 so the least number of letters needed are 5.

Using formula:

Four letters = 4C1 + 4C2 = 4 + 6 = 10

Five letters = 5C1 + 5C2 = 5 + 10 = 15

5 0
3 years ago
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