Why is a cable modem classified as a digital model?

Answer:

The principle behind how computer systems operate involves a primary machine-based function that remains invisible to us, serving as a control center that changes the input data into output. This central element known as the central processing unit (CPU) illustrates that the operation of computers is quite intricate.

Explanation:

Here's a C code snippet: #include <stdio.h> int main() { int year; printf("enter year:"); scanf("%d",&year); if(year>=2101) { printf("Distant future"); } else if(year>=2001&&year<=2100) { printf("21st century"); } else if(year>=1901&&year<=2000) { printf("20th century"); } else if(year<=1900) { printf("Long ago"); } return 0;} Output: enter year:2018 21st century. </stdio.h>

The phrases "or 1=1" and "or ''=''" are typical examples used to manipulate an SQL WHERE clause into evaluating as true.

Consequently, if you type in <span>' or ''=' as your password, you can authenticate if the query would be:

</span><span>select username,pass from users where username='you' and password='' or ''='<span>' limit 0,1;</span></span>

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.

IT'S CORRECT!! I LOOKED IT UP