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Lubov Fominskaja [6]
4 years ago
13

IT professionals have a responsibility to educate employees about the risks of hot spots. Which of the following are risks assoc

iated with hot spots? Check all of the boxes that apply.
third-party viewing

unsecured public network

potential for computer hackers

unauthorized use
Computers and Technology
1 answer:
kodGreya [7K]4 years ago
7 0

Answer:

third partying and computer hackers.

Explanation:

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Programs for embedded devices are often written in assembly language. Some embedded processors have limited instructions, like M
oksano4ka [1.4K]

Answer:

#include <iostream>

using namespace std;

int main()

{

   char address[2];

   int tag, firstBit, secondBit, setNumber;

   int cache[4][2]={{1,5}, {2,4}, {3,2}, {6,0}};

   cout << "Enter the address as hex(in small letters: "<

   cin >> address;

   for (int i < 0; i < 8; i++){

       if (address[0] == '0'){

           tag = 0;

           firstBit = 0;

        } else if (address[0] == '1'){

           tag = 0;

           firstBit = 1;

        } else if (address[0] == '2'){

           tag =1;

           firstBit = 0;

        } else if (address[0] == '3'){

           tag = 1;

           firstBit = 1;

        } else if (address[0] == '4'){

           tag = 2;

           firstBit = 0;

        } else if (address[0] == '5'){

           tag = 2;

           firstBit = 1;

        }  else if (address[0] == '6'){

           tag = 3;

           firstBit = 0;

       } else if (address[0] == '7'){

           tag = 3;

           firstBit = 1;

       }  else if (address[0] == '8'){

           tag = 4;

           firstBit = 0;

       } else if (address[0] == '9'){

           tag = 4;

           firstBit = 1;

       }   else if (address[0] == 'A'){

           tag = 5;

           firstBit = 0;

       } else if (address[0] == 'B'){

           tag = 5;

           firstBit = 1;

       }  else if (address[0] == 'C'){

           tag = 6;

           firstBit = 0;

       } else if (address[0] == 'D'){

           tag = 6;

           firstBit = 1;

        }  else if (address[0] == 'E'){

           tag = 7;

           firstBit = 0;

       } else if (address[0] == 'F'){

           tag = 7;

           firstBit = 1;

       } else{

           cout<<"The Hex number is not valid"<< endl;

        }

   }

   if(address[1]>='0' && address[1]<'8'){

       secondBit = 0;

  }  else if(address[1]=='8'|| address[1]=='9'||(address[1]>='a' && address[1]<='f')){

       secondBit = 1;

   }  else{

       cout<<"The Hex number is not valid"<< endl;  

       return 0;

   }

   setNumber = firstBit * 2 + secondBit;

   if(cache[setNumber][0]==tag || cache[setNumber][1]==tag){

       cout<<"There is a hit";

   } else{

       cout<< "There is a miss";

   }

   return 0;

}

Explanation:

The C++ source code prompts the user for an input for the address variable, then the nested if statement is used to assign the value of the firstBit value given the value in the first index in the address character array. Another if statement is used to assign the value for the secondBit and then the setNumber is calculated.

If the setNumber is equal to the tag bit, Then the hit message is printed but a miss message is printed if not.

4 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.

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Elena L [17]
The answer is backlighting
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Which of the following is not one of the steps a company would take in an attempt to prevent a malfunction or failure of their p
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