There are N bulbs numbered from 1 to N, arranged in a row. The first bulb is plugged into the power socket and each successive b

Answer:

Below is the JAVA program:

import java.util.Scanner;   //to take input from user

public class Main {  //name of the class

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

      Scanner input = new Scanner(System.in);  //creates Scanner instance

      int integer = input.nextInt();  //defines and collects an integer for the number of words

      String stringArray[] = new String[integer]; //initializes an array to hold strings

      for (int i = 0; i < integer; i++) {  //iterates through the array

          stringArray[i] = input.next();         }  //collects strings

      int frequency[] = new int[integer];  //creates an array for holding frequencies

      int count;         //initializes variable to calculate frequency of each word

      for (int i = 0; i < frequency.length; i++) {  //iterates through frequency array

          count = 0;  //sets count to zero

          for (int j = 0; j < frequency.length; j++) {  //iterates through array

              if (stringArray[i].equals(stringArray[j])) {  //checks if element at ith index matches jth index

                  count++;                 }            }  //increments count

          frequency[i] = count;      }  //stores count in the frequency array

      for (int i = 0; i < stringArray.length; i++) {  //iterates through the string array

          System.out.println(stringArray[i] + " " + frequency[i]);         }    }    } //displays each word and its frequency

Explanation:

To illustrate the program, consider:

let integer = 3

for (int i = 0; i < integer; i++) is used to input strings into stringArray.

On the first pass:

i = 0

0<3

stringArray[i] = input.next(); takes a word and saves it to the ith index of stringArray. For example, if the user types "hey", it will be in stringArray[0].

Then, i increments to i = 1

On the second pass:

i = 1

1<3

stringArray[i] = input.next(); takes another word and assigns it to stringArray[1]. If the user enters "hi", then hi will be in stringArray[1]

Next, i becomes i = 2

On the third pass:

i = 2

2<3

stringArray[i] = input.next(); captures a word and places it in stringArray[2]. If the user types "hi", it goes to stringArray[2]

Then, i increments to i = 3

The loop terminates since i<integers is false.

The next outer loop for (int i = 0; i < frequency.length; i++) and inner loop for (int j = 0; j < frequency.length; will check each word and if (stringArray[i].equals(stringArray[j])) identifies any duplicates. Words that repeat will have their count incremented and the frequency array will record these values. For example, hey appears once while hi shows up twice, thus resulting in the final outputs:

hey 1

hi 2

hi 2

The visual representation of the program and its outcome based on the example provided is attached.

Response:

d. RAID 6

Clarification:

RAID is a technological method for data storage that integrates several physical hard drive components into a unified logical structure. Its primary purpose is to ensure both performance and data redundancy.

RAID 0 is focused on data striping, but it lacks redundancy.

RAID 1 enhances performance to nearly double but restricts disk space usage to around 50%.

RAID 5 offers both redundancy and improved performance, though it is constrained by smaller drive sizes.

RAID 6 provides redundancy as well but with a decrease in performance.

RAID 10 boosts both performance and data security.

Hence, RAID 6 is the optimal choice that emphasizes redundancy at the cost of speed.

The correct answer to this query is "Hard drive." Explanation: Given the context of the inquiry, it's suggested that the most likely cause of this issue is the improper connection of the video cable. An incorrectly connected video cable can lead to display disruptions, such as the screen flashing black. While overheating of the GPU could also be a potential cause, it typically results in visual artifacts rather than shifting displays. Therefore, "Hard drive" is indeed the right answer.

The primary issue was declaring int prod within the while loop, which caused prod to reset to 1 each time the loop executed.

Answer:

First I/O instruction format -> 256 ports

Second I/O instruction format -> 65536 ports

Explanation:

• The initial instruction format can address 256 ports, computed from eight bits for port addressing as 2^8 = 256 ports.

• The second instruction format allows for 65536 ports, which can be calculated using sixteen bits for addressing as 2^16 = 65536 ports.

Changing the opcode facilitates the selection between the first and second instruction format, allowing one input or output operation at a given moment.