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Vladimir [108]
3 years ago
8

Which of the following is not a search engine

Computers and Technology
1 answer:
Irina18 [472]3 years ago
7 0
It's Gopher and others areare search engines.
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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
Which of te following ranges of cells is correctly named
Natasha_Volkova [10]
D it is because its .a5.g1
3 0
3 years ago
Given a collection of n nuts and a collection of n bolts, arranged in an increasing order of size, give an O(n) time algorithm t
kari74 [83]

Answer:

See explaination

Explanation:

Keep two iterators, i (for nuts array) and j (for bolts array).

while(i < n and j < n) {

if nuts[i] == bolts[j] {

We have a case where sizes match, output/return

}

else if nuts[i] < bolts[j] {

this means that size of nut is smaller than that of bolt and we should go to the next bigger nut, i.e., i+=1

}

else {

this means that size of bolt is smaller than that of nut and we should go to the next bigger bolt, i.e., j+=1

}

}

Since we go to each index in both the array only once, the algorithm take O(n) time.

8 0
4 years ago
_____ is a web application server that provides the ability to connect web servers to multiple data sources.
o-na [289]
ColdFusion/JRun is a web application server that provides the ability to connect web servers to multiple data sources.ColdFusion/JRun  is an application server developed by Adobe, that is based on Sun Microsystems' Java 2 Platform, Enterprise Edition (J2EE). The other options are eliminated because: Ms Access is <span>Database Management System , FoxPro is database development system and dBase is also data management system.</span>
7 0
4 years ago
Who benefits the most from billing by the second for cloud resources, such as virtual machines?.
scoundrel [369]

Clients that run numerous virtual machines.

<h3>What is a virtual machine?</h3>
  • A virtual machine is the virtualization or emulation of a computer system in computing.
  • The functionality of a physical computer is provided by virtual machines, which are built on computer architectures.
  • Their use may necessitate specialized hardware, software, or a combination of both.
  • Virtual machines' primary function is the simultaneous operation of several operating systems on a single piece of hardware.
  • Without virtualization, running two different physical units would be necessary to run different operating systems, such as Windows and Linux.
  • Through the use of virtualization technologies, virtual machines are made possible.
  • Multiple virtual machines (VMs) can run on a single machine thanks to virtualization, which simulates virtual hardware using software.
  • The real machine is referred to as the host, and any virtual machines running on it as the guests.

To learn more about virtual machines, refer to:

brainly.com/question/27961159

#SPJ4

7 0
2 years ago
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