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ikadub [295]
4 years ago
14

You have received several trouble tickets from the employees in the warehouse for the stand-alone computers used to control vari

ous shipping machines because the computers are not booting when powered. Each time a technician resolves the booting issue the boot order is changed in the firmware. Each computer is required to have the USB ports disabled in the firmware to keep employees from connecting rogue devices.1. Which of the following steps should be taken to eliminate these trouble tickets?O Disconnect the USB ports from the motherboard.O Enable the supervisor password in the BIOS/UEFI setup.O Require all employees to use a unique Windows user account and password.O Install a lock on the computer case to prevent removal of the covers.
Computers and Technology
1 answer:
Arturiano [62]4 years ago
3 0

Answer:

Enable the supervisor password in the BIOS/UEFI setup.

Explanation:

Since each computer is required to have the USB ports disabled in the firmware, the most likely option is that some employees attempt to circumvent the restriction thereby affecting the Boot Order.

A supervisor password is used to restrict access to the BIOS. Users without a correct BIOS supervisor password cannot make changes to system settings.

However, if the Supervisor Password is enabled, they will not be able to make changes to the BIOS Setup.

All other options are not relevant to this particular ticket.

You might be interested in
Which line(s) immediately flush the output buffer?
Murljashka [212]

Answer:

lines 2,3 and 5 ( c )

Explanation:

The lines that immediately flush the output buffer are :  System.out.print("Two for the show\n"); .   System.out.println("Three to get ready"); and   System.out.flush();

making use of the new line character or printIn() statement the buffered output gets flushed therefore making autoflush true

but System.out.print() doesn't  use any flush technique so we need to flush the buffered output done by the print() statement hence we choose lines 2,3,5

3 0
4 years ago
Write a recursive method to return the number of uppercase letters in a String. You need to define the following two methods. Th
satela [25.4K]

Answer:

<u>Recursive function with two parameters that return the number of uppercase letters in a String</u>

public static int count(String str,int h)//Defining function

{

      if(str.charAt(0)>='A' && str.charAt(0)<='Z')//Checking the characters from A to Z

{

          h++; //incrementing the counter

          if(str.length()>=2){

              return count(str.substring(1),h);//recalling function

          }

      }

      if(str.length()>=2){

              return count(str.substring(1),h); //recalling function

      }

      return h;

  }

This is the recursive function with the name count of return type integer,having parameters str of string type and h of integer type.In this we are checking the characters at a particular position from A to Z.

<u>Recursive function with one parameter that return the number of uppercase letters in a String</u>

public static int count(String str)//Defining function

{

      if(str.charAt(0)>='A' && str.charAt(0)<='Z')//Checking the characters from A to Z

{

          count++; //incrementing the counter

          if(str.length()>=2){

              return count(str.substring(1));//recalling function

          }

      }

      if(str.length()>=2){

              return count(str.substring(1)); //recalling function

      }

      return count;

  }

This is the recursive function with the name count of return type integer,having parameters str of string type .In this we are checking the characters at a particular position from A to Z.

<u>Java program that return the number of uppercase letters in a String</u>

import java.util.*;

public class Myjava{

static int count =0;//Defining globally  

 

public static int count(String str,int h)//Defining function

{

      if(str.charAt(0)>='A' && str.charAt(0)<='Z')//Checking the characters from A to Z

{

          h++;

//incrementing the counter

          if(str.length()>=2){

              return count(str.substring(1),h);//recalling function

          }

      }

      if(str.length()>=2){

              return count(str.substring(1),h);

//recalling function

      }

      return h;

 

  }

  public static void main(String[] args)//driver function

  {

      System.out.println("Enter a string");//taking input

      Scanner scan = new Scanner(System.in);

      String s = scan.nextLine();

      int h =0;

      System.out.println("Counting the Uppercase letters: "+count(s,h));

  }

}

<u>Output</u>

Enter a string  WolFIE

Counting the Uppercase letters: 4

4 0
3 years ago
Given two complex numbers, find the sum of the complex numbers using operator overloading.Write an operator overloading function
Inessa05 [86]

Answer:

I am writing the program in C++ programming language.  

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

using namespace std;   // to identify objects like cin cout

class ProblemSolution {  //class name

private:  

// private data members that only be accessed within ProblemSolution class

int real, imag;

 // private variables for real and imaginary part of complex numbers

public:  

// constructor ProblemSolution() to initialize values of real and imaginary numbers to 0. r is for real part and i for imaginary part of complex number

ProblemSolution(int r = 0, int i =0) {  

real = r; imag = i; }  

/* print() method that displays real and imaginary part in output of the sum of complex numbers */

void print(){  

//prints real and imaginary part of complex number with a space between //them

cout<<real<<" "<<imag;}  

// computes the sum of complex numbers using operator overloading

ProblemSolution operator + (ProblemSolution const &P){  //pass by ref

          ProblemSolution sum;  // object of ProblemSolution

          sum.real = real + P.real;  // adds the real part of the  complex nos.

          sum.imag = imag + P.imag;  //adds imaginary parts of  complex nos.

//returns the resulting object

          return sum;       }  //returns the sum of complex numbers

};   //end of the class ProblemSolution

int main(){    //start of the main() function body

int real,imag;  //declare variables for real and imaginary part of complex nos

//reads values of real and imaginary part of first input complex no.1

cin>>real>>imag;  

//creates object problemSolution1 for first complex number

ProblemSolution problemSolution1(real, imag);  //creates object

//reads values of real and imaginary part of first input complex no.2

cin>>real>>imag;

//creates object problemSolution2 for second complex number

ProblemSolution problemSolution2(real,imag);

//creates object problemSolution2 to store the addition of two complex nos.

ProblemSolution problemSolution3 = problemSolution1 + problemSolution2;

problemSolution3.print();} //calls print() method to display the result of the //sum with real and imaginary part of the sum displayed with a space

Explanation:

The program is well explained in the comments mentioned with each statement of the program. The program has a class named ProblemSolution which has two data members real and imag to hold the values for the real and imaginary parts of the complex number. A default constructor ProblemSolution() which initializes an the objects for complex numbers 1 and 2 automatically when they are created.

ProblemSolution operator + (ProblemSolution const &P) is the operator overloading function. This performs the overloading of a binary operator + operating on two operands. This is used here to add two complex numbers.  In order to use a binary operator one of the operands should be passed as argument to the operator function. Here one argument const &P is passed. This is call by reference. Object sum is created to add the two complex numbers with real and imaginary parts and then the resulting object which is the sum of these two complex numbers is returned.  

In the main() method, three objects of type ProblemSolution are created and user is prompted to enter real and imaginary parts for two complex numbers. These are stored in objects problemSolution1 and problemSolution2.  Then statement ProblemSolution problemSolution3 = problemSolution1 + problemSolution2;  creates another object problemSolution3 to hold the result of the addition. When this statement is executed it invokes the operator function ProblemSolution operator + (ProblemSolution const &P). This function returns the resultant complex number (object) to main() function and print() function is called which is used to display the output of the addition.

4 0
4 years ago
Urgent will give brainliest who solves this....
Ratling [72]
THIS IS MY PERSONAL OPINION:
The world is indeed changing very fast, and the future brings new possibilities and new realities. I foresee that humans will thrive for another 50 years, until the global warming situation reaches its limits. The people in the future I think will not have any ( maybe rarely) social interactions, maybe we will go to the moon etc. but i still strongly believe that the end of humanity is decades close.
I hope this helps you :)
5 0
3 years ago
Read 2 more answers
What is computer assisted translation​
Vlad1618 [11]

Answer:

the use of software to assist a human translator in the translation process.

Explanation:

8 0
3 years ago
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