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MissTica
3 years ago
7

What is it called when the system and all the components except for the real-time clock are powered down

Computers and Technology
1 answer:
Colt1911 [192]3 years ago
8 0

Answer:

The correct answer to the following question will be "Mechanical off mode".

Explanation:

  • The device will appear to be either on or off to the user. No other states are measurable. The device does, however, support various power states that suit the power states specified in the ACPI specification. Such states also differ, like hybrid sleep and rapid startup. This subject discusses certain states and how they will be represented.
  • The system is completely off, and no power is used. Only after a full reboot will the machine return to working condition.
  • It is called mechanical off mode when the machine and all parts, except for the actual-time clock, are down powered.

Therefore, Mechanical off mode is the right answer.

You might be interested in
Write a program that produces a Caesar cipher of a given message string. A Caesar cipher is formed by rotating each letter of a
PilotLPTM [1.2K]

Answer:

Here is the JAVA program that produces Caeser cipher o given message string:

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

public class Main {  

   public static void main(String args[]) {  //start of main function

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

       System.out.println("Enter the message : ");  //prompts user to enter a plaintext (message)

       String message = input.nextLine();  //reads the input message

       System.out.println("Enter the amount by which by which to rotate each letter : ");  //prompts user to enter value of shift to rotate each character according to shift

       int rotate = input.nextInt();  // reads the amount to rotate from user

       String encoded_m = "";  // to store the cipher text

       char letter;  // to store the character

       for(int i=0; i < message.length();i++)  {  // iterates through the message string until the length of the message string is reached

           letter = message.charAt(i);  // method charAt() returns the character at index i in a message and stores it to letter variable

           if(letter >= 'a' && letter <= 'z')  {  //if letter is between small a and z

            letter = (char) (letter + rotate);  //shift/rotate the letter

            if(letter > 'z') {  //if letter is greater than lower case z

               letter = (char) (letter+'a'-'z'-1); }  // re-rotate to starting position  

            encoded_m = encoded_m + letter;}  //compute the cipher text by adding the letter to the the encoded message

           else if(letter >= 'A' && letter <= 'Z') {  //if letter is between capital A and Z

            letter = (char) (letter + rotate);  //shift letter

            if(letter > 'Z') {  //if letter is greater than upper case Z

                letter = (char) (letter+'A'-'Z'-1);}  // re-rotate to starting position  

            encoded_m = encoded_m + letter;}  //computes encoded message

           else {  

         encoded_m = encoded_m + letter;  } }  //computes encoded message

System.out.println("Encoded message : " + encoded_m.toUpperCase());  }} //displays the cipher text (encoded message) in upper case letters

Explanation:

The program prompts the user to enter a message. This is a plaintext. Next the program prompts the user to enter an amount by which to rotate each letter. This is basically the value of shift. Next the program has a for loop that iterates through each character of the message string. At each iteration it uses charAt() which returns the character of message string at i-th index. This character is checked by if condition which checks if the character/letter is an upper or lowercase letter. Next the statement letter = (char) (letter + rotate);   is used to shift the letter up to the value of rotate and store it in letter variable. This letter is then added to the variable encoded_m. At each iteration the same procedure is repeated. After the loop breaks, the statement     System.out.println("Encoded message : " + encoded_m.toUpperCase()); displays the entire cipher text stored in encoded_m in uppercase letters on the output screen.

The logic of the program is explained here with an example in the attached document.

8 0
3 years ago
Explain what happens if you try to open a file for reading that does not exist.
ZanzabumX [31]

Answer:

Exception is thrown and the file is created with 0 length.

Explanation:

While opening a file there is an error occur which shows the file does not exist that means an  exception is thrown.

And this error can be occur when the size of the file is very very low means the file is a size of 0 length. So to avoid this error we have to exceed its length from the zero length.

8 0
3 years ago
Two categories of payroll deductions are required deductions and ___ deductions.
alukav5142 [94]

Answer:

1.)

- C.) Optional

2.)

- D.) Short-term Notes Payable

3.)

- A.) Payroll Sinking Funds

4.)

- A.) A Formal Timekeeping System Is Used

(I'm possibly wrong on the last question, if so then my apologies and I wish you the best of luck.)

3 0
2 years ago
Read 2 more answers
What is Brainly?<br><br> A.Yes<br> B.No
mr Goodwill [35]

Answer: C: Yesn't

Explanation:

3 0
3 years ago
Read 2 more answers
Professor Midas drives an automobile from Newark to Reno along Interstate 80. His car’s gas tank, when full, holds enough gas
zmey [24]

Answer:

The GREEDY Algorithm

Explanation:

Based on the situation given in question, the Greedy algorithm shall give the optimal solution to professor

Suppose that the cities are at locations0 =x0< x1< . . . < x

We shall use the induction method to prove that G is the optimal solution valid for numbers less than n

We assume another solution Z which we initially consider to be optimum as well, based on that when Z fills the tank, it fills it to full level

Let us state the values in case of n intervals. Given below, we say that g1 is the first stop and z1 is also the first stop.

This can be written as ;

G=g1, g2, . . . , gk

Z=z1, z2, . . . , zk’

Here k’ <= k and k < n

Let I be an idex where for the first time gi is not equal to zi

Considering t= maxi Zi

Z′=g1, z2, z3, . . . , zk′

Now since z2, z3, . . . , zk′ should be an optimal stopping pattern for the problem otherwise we have chosen Z, with smaller gas filling (not feasible)

Using induction hypothesis we conclude thatg2, . . . , gk is an optimal stopping pattern, which is based on greedy algorithm

7 0
3 years ago
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