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Alecsey [184]
3 years ago
10

An IT technician has manually configured an IP address on a laptop for a new employee. Each time the employee tries to connect t

o the LAN, the employee receives a duplicate IP address error message. What can the company do to prevent this situation from arising again? Group of answer choices?
Computers and Technology
1 answer:
ycow [4]3 years ago
3 0

Answer:

Use DHCP (which stands for Dynamic Host Configuration Protocol)

Explanation:

This is a common problem, using DHCP is debated amongst the IT world that it could serve as a petri dish for hackers to invade but realistically, the pros definitely outweigh the cons. But if you want to booster shot your security, you can always assign IP addresses to MAC addresses at your DHCP server but manually. Don't forget to backup the DHCP server multiple times are day.

You could also adopt the usage of using shorter DHCP default lease times. As well try to enable the IP address conflict detection in the DHCP server.

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Design a program that will receive a valid time in the 24-hour format (e.g. 2305) and convert to its equivalent 12-hour format (
coldgirl [10]

Answer:

import java.io.*;

public class Main

{

  public static void main(String[] args) throws IOException {

      BufferedReader bufferObject=new BufferedReader(new InputStreamReader(System.in));

      String stringObject=bufferObject.readLine();

      while(!stringObject.equals("99:99AM")){

      System.out.println(convertedTime(stringObject));

      stringObject=bufferObject.readLine();

      }

  }

  public static String convertedTime(String stringObject){

  String s=stringObject.substring(stringObject.length()-2);

  String[] timeObject=stringObject.substring(0,5).split(":");

  if(s.equals("AM")){

  if(timeObject[0].equals("12")) return "00"+timeObject[1];

  else return timeObject[0]+timeObject[1];

  }

  else{

  if(timeObject[0].equals("12")) return "12"+timeObject[1];

  else{

  int hours=Integer.valueOf(timeObject[0]);

  timeObject[0]=String.valueOf(12+hours);

  return timeObject[0]+timeObject[1];

  }

  }

 

  }

}

Explanation:

  • Inside the main method run a while loop until stringObject is not equal to the string "99:99AM".
  • Call the convertedTime method and display the results.
  • Use the same hours and minutes except for 12th hour If the time is in AM.
  • Use "00" instead of 12, if it is 12th hour.
  • Add hours to 12, if the time is in PM and don't change anything in case of 12.
8 0
3 years ago
Windows server 2012 r2 supports two types of folder shares. what are those two types?
larisa [96]
<span>Windows server 2012 r2 supports the following two types of folder shares: </span>Network File System and Server Message Blocks. Correct answer: D
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The Server Message Block<span> (SMB) Protocol is a </span>network file sharing protocol.
7 0
3 years ago
/* ELEN 1301 Programming Assignment #5. Name : Your name. Student ID : Your student ID #. Due date : Due date Purpose of the pro
cestrela7 [59]

Answer:

Here is the C++ program:

#include<iostream> //to use input output functions

#include<iomanip> //to use setprecision

using namespace std; //to identify objects cin cout

int main() { //start of main function

     int n1, n2, n3, n4, n5, n6, n7, min, max;  // declare variables for 7 numbers, minimum value and maximum value

     double sum= 0; //declare variable to hold sum of 7 numbers

     double average;  //declare variable to hold average of 7 numbers

     cout<<"Enter first number: "; //prompts user to enter 1st number

     cin>>n1; //reads first number from user

     max = n1; //sets the first number to maximum

     min=n1; //sets the first number to minimum

     cout<<"Enter second number: "; //prompts user to enter 2nd number

     cin>>n2; //reads second number from user

     if(n2<min){ //if second number is less than min

          min=n2;      } //sets min to n2

     if(n2>max){ //if n2 is greater than max

          max = n2;      } //sets max to n2

     cout<<"Enter third number: ";  //prompts user to enter 3rd number

     cin>>n3; //reads third number from user

     if(n3<min){ //checks if n3 is greater than min

          min=n3;      } //sets n3 to min

     if(n3>max){ //checks if n3 is greater than max

          max = n3;      }      //sets max to n3

    cout<<"Enter fourth number: ";//prompts user to enter 4th number

     cin>>n4; //reads fourth number from user

     if(n4<min){  //if n4 is less than min

          min=n4;      }  //sets min to n4

     if(n4>max){  //if n4 is greater than max

          max = n4;      }  //sets max to n4

     cout<<"Enter fifth number: "; //prompts user to enter 5th number

     cin>>n5; //reads fifth number from user

     if(n5<min){  //if n5 is less than min

          min=n5;     }  //sets min to n5

     if(n5>max){  //if n5 is greater than max

          max = n5;      }  //sets max to n5

     cout<<"Enter sixth number: "; //prompts user to enter 6th number

     cin>>n6; //reads sixth number from user

     if(n6<min){ // if n6 is less than min

          min=n6;      }  //sets min to n6

     if(n6>max){  //if n6 is greater than max

          max = n6;      }  //sets max to n6

     cout<<"Enter seventh number: ";//prompts user to enter 7th number

     cin>>n7; //reads seventh number from user

     if(n7<min){  //if n7 is less than minimum number

          min=n7;      }  //assigns n7 to min

     if(n7>max){  //if n7 is greater than the maximum number

          max = n7;      }  //assigns n7 to max

     sum = n1+n2+n3+n4+n5+n6+n7;  //adds 7 numbers

     average = sum/7;  //computes average of 7 numbers

     cout<<"The average is: "<<fixed<<setprecision(3)<<average<<endl; //displays average value up to show 3 digits below decimal point using setprecision method of iomanip library

     cout<<"The maximum number is: "<<max<<endl; //displays maximum number of 7 numbers

     cout<<"The minimum number is: "<<min<<endl;  //displays miimum number of 7 numbers    

     return 0; }

Explanation:

The program is well explained in the comments attached to each statement of program. For example if

n1 = 3

n2 = 9

n3 = 7

n4 = 6

n5 = 2

n6 = 5

n7 = 4

When n1 is read using cin then this number is set to max and min as:

min = 9

max = 3

Now when n2 is read, the first if condition checks if n2 is less than min and second if condition checks if n2 is greater than max. As n2 = 9 so it is not less than min so this if condition is false and n2 is greater than max i.e. 3 so this condition is true. So 9 is assigned to max.

min = 9

max = 9

Now when n3 is read the values of min and max become:

min = 7

max = 9

Now when n4 is read the values of min and max become:

min = 6

max = 9

Now when n5 is read the values of min and max become:

min = 2

max = 9

Now when n6 is read the values of min and max become:

min = 2

max = 9

Now when n7 is read the values of min and max become:

min = 2

max = 9

Now the statement       sum = n1+n2+n3+n4+n5+n6+n7;

executes which becomes:

sum = 3 + 9 + 7 + 6 + 2 + 5 + 4

sum = 36.0

Next program control moves to statement:

average = sum/7;

this becomes

average = 36/7

5.142857

Since this is to be displayed up to 3 decimal places so average = 5.143

the complete output of the program is attached.

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3 years ago
What are extremely fast and small amounts of memory used to quicken the implementation of computer programs by providing access
Marysya12 [62]

Answer:

Registers

Explanation:

5 0
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If you are working in a word-processing program and need to learn more about its features, the best place to get assistance is f
jek_recluse [69]

Answer:

Either the help section in the program or Google

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