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riadik2000 [5.3K]
2 years ago
15

A technician is troubleshooting a small network with cable Internet service in which a user is receiving a message in her web br

owser stating a site cannot be found. While interviewing users, it is determined that every host on the network is experiencing the same issue with all websites. Which of the following is the first step in troubleshooting this issue?
a. Use a cable tester to test the patch cable for the first user who reported the issue.
b. Reboot the soho router connecting the lan to the cable modem.
c. Call the internet service provider and report the problem.
d. Replace the patch cable from the soho router to the cable modem.
e. Connect a computer directly to the cable modem and attempt to access the internet.
Computers and Technology
1 answer:
expeople1 [14]2 years ago
6 0

Answer:

Option E is correct.

Explanation:

A professional is fixing a cable internet access network by which a person in their internet browser receives a text saying that a link could not be reached. It would be known that each and every server along the network is having the individual problem across whole websites when interviewing a person.

So, connect the system to the wired connection directly, & try to access the internet for troubleshooting the following issue.

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Use the loop invariant (I) to show that the code below correctly computes the product of all elements in an array A of n integer
NeTakaya

Answer:

Given Loop Variant P = a[0], a[1] ... a[i]

It is product of n terms in array

Explanation:

The Basic Step: i = 0, loop invariant p=a[0], it is true because of 'p' initialized as a[0].

Induction Step: Assume that for i = n - 3, loop invariant p is product of a[0], a[1], a[2] .... a[n - 3].

So, after that multiply a[n - 2] with p, i.e P = a[0], a[1], a[2] .... a[n - 3].a[n - 2].

After execution of while loop, loop variant p becomes: P = a[0], a[1], a[2] .... a[n -3].a[n -2].

for the case i = n-2, invariant p is product of a[0], a[1], a[2] .... a[n-3].a[n-2]. It is the product of (n-1) terms. In while loop, incrementing the value of i, so i=n-1

And multiply a[n-1] with p, i.e P = a[0].a[1].a[2].... a[n-2].

a[n-1]. i.e. P=P.a[n-1]

By the assumption for i=n-3 loop invariant is true, therefore for i=n-2 also it is true.

By induction method proved that for all n > = 1 Code will return product of n array elements.

While loop check the condition i < n - 1. therefore the conditional statement is n - i > 1

If i = n , n - i = 0 , it will violate condition of while loop, so, the while loop will terminate at i = n at this time loop invariant P = a[0].a[1].a[2]....a[n-2].a[n-1]

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What Information Technology is Walt Thomas responsible for?
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Assume the variable date has been set to a string value of the form mm/dd/yyyy, for example 09/08/2010. (Actual numbers would ap
dmitriy555 [2]

Answer:

The code is attached

Explanation:

  1. a date in the string format 'mm/dd/yyyy' is assigned to variable dateStr
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Many people describe computers as “complex” machines. What can this mean
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Consider an array inarr containing atleast two non-zero unique positive integers. Identify and print, outnum, the number of uniq
Delvig [45]

Answer:

Program.java  

import java.util.Scanner;  

public class Program {  

   public static boolean isPalindrome(String str){

       int start = 0, end = str.length()-1;

       while (start < end){

           if(str.charAt(start) != str.charAt(end)){

               return false;

           }

           start++;

           end--;

       }

       return true;

   }  

   public static int countPalindromePairs(String[] inarr){

       int count = 0;

       for(int i=0; i<inarr.length; i++){

           for(int j=i+1; j<inarr.length; j++){

               StringBuilder sb = new StringBuilder();

               sb.append(inarr[i]).append(inarr[j]);

               if(isPalindrome(sb.toString())){

                   count++;

               }

           }

       }

       return count == 0 ? -1 : count;

   }

   public static void main(String[] args) {

       Scanner sc = new Scanner(System.in);

       String line = sc.next();

       String[] inarr = line.split(",");

       int count = countPalindromePairs(inarr);

       System.out.println("RESULT: "+count);

   }

}

Explanation:

OUTPUT:

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