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zzz [600]
3 years ago
12

The code selection above is taken from the Color Sleuth activity you just completed. This selection would count as an abstractio

n on the Create PT, but to earn Row 8 you will need to describe how this abstraction manages complexity. Write response 2d describing the abstraction above. Use the writing prompt and scoring guidelines as a guide.
Computers and Technology
1 answer:
Vinvika [58]3 years ago
4 0

Answer:

b

Explanation:

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You have to display or connect to network shares on remote computers. Which command-line utility will you use to accomplish the
Xelga [282]

Answer:

net use X: \\SERVER\Share

Explanation:

1. Connect securely to the remote computer or ensure you have access to it on the network

2. then follow the step below:

Where X: is the drive letter you wish to map the share to, and \\SERVER\Share is the UNC path to the share. This should make the share visible in My Computer and the command line as well like all other shares mapped through the GUI.

In order to later disconnect the share, you would use

net use X: /delete

8 0
3 years ago
Chat messages are most likely to be found where on a computer? firewall nic internet history ram
miv72 [106K]
<span>The answer is RAM.  Chat messages are most likely to be found in the RAM of the computer.  It is Random Access Memory.  </span><span>RAM is considered volatile memory, which means that the stored information is lost when there is no more power.</span>
3 0
3 years ago
The following checksum formula is widely used by banks and credit card companies to validate legal account numbers: d0 + f(d1) +
Aleksandr [31]

Answer:

Here is the JAVA program:

import java.util.Scanner; //to import Scanner class

public class ISBN

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

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

//prompts the user to enter 10 digit integer

   System.out.println("Enter the digits of an ISBN as integer: ");    

   String number = s.next(); // reads the number from the user

   int sum = 0; // stores the sum of the digits

   for (int i = 2; i <= number.length(); i++) {

//loop starts and continues till the end of the number is reached by i

          sum += (i * number.charAt(i - 1) ); }

/*this statement multiplies each digit of the number with i and adds the value of sum to the product result and stores in the sum variable*/

          int remainder = (sum % 11);  // take mod of sum by 11 to get checksum  

   if (remainder == 10)

/*if remainder is equal to 10 adds X at the end of given isbn number as checksum value */

  { System.out.println("The ISBN number is " + number + "X"); }

  else

// displays input number with the checksum value computed

 {System.out.println("The ISBN number is " + number + remainder); }  }  }  

Explanation:

This program takes a 10-digit integer as a command line argument and uses Scanner class to accept input from the user.

The for loop has a variable i that starts from 2 and the loop terminates when the value of i exceeds 10 and this loop multiplies each digit of the input number with the i and this product is added and stored in variable sum. charAt() function is used to return a char value at i-1.

This is done in the following way: suppose d represents each digit:

sum=d1 * 1 + d2 * 2 + d3 * 3 + d4 * 4 + d5 * 5 + d6 * 6 + d7 * 7 + d8 * 8 + d9 * 9

Next the mod operator is used to get the remainder by dividing the value of sum with 11 in order to find the checksum and stores the result in remainder variable.

If the value of remainder is equal to 10 then use X for 10 and the output will be the 10 digits and the 11th digit checksum (last digit) is X.

If the value of remainder is not equal to 10, then it prints a valid 11-digit number with the given integer as its first 10 digits and the checksum computed by sum % 11 as the last digit.  

8 0
3 years ago
Write a loop that reads c-strings from standard input where the c-string is either "land", "air", or "water". The loop terminate
Anvisha [2.4K]

Answer:

I am writing a C++ code.        

#include <iostream> // for input output functions

using namespace std; // identifies objects like cin cout

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

string c_string; // stores the string entered by user from land water or air

int land=0; // contains the number of times land string is read in

int air=0; //contains the number of times air string is read in

int water = 0; //contains the number of times water string is read in

cin>>c_string; //reads the string entered by user from air water or land

/*while loop continues to execute until user enters xxxxx or maximum length of a string exceeds 8 */

while(c_string!= "xxxxx" && c_string.length()<=8) {

if(c_string=="land"){ // if string entered by user is land

land = land + 1;} //counts and increments each occurrence of land string by 1

else if(c_string=="air"){// if string entered by user is air

air = air + 1;}// counts and increments each occurrence of string air by 1

else if(c_string=="water"){// if string entered by user is water

water = water + 1;}//counts and increments each occurrence of air by 1

cin>>c_string;}

/* keeps reading the string entered by user from land air or water, until the loop breaks after the user enters xxxxx or user enters a string whose length is greater than 8 */

//prints the number of times land, air and water are read in

cout << "land:"<<land;

cout << endl<<"air:"<<air;

cout << endl<< "water:"<<water; }                      

                                                                                     

Explanation:

Everything is well explained in the comments above.

The program prompts the user to input strings. These strings are either land air or water. The while loop continues to read the input strings until user enters xxxxx or the string entered by user exceeds the length 8. Both these terminating conditions are added in the while loop. After the loop terminates, the number of times land, air and water strings are read is displayed on the output screen. Any other string entered by user other than these 3 is ignored. The program along with the output is attached.

3 0
3 years ago
Suppose you have a 16-bit machine with a page size of 16B. Assume that any unsigned 16-bit integer can be a memory address. Also
irinina [24]

Answer:

2^11

Explanation:

Physical Memory Size = 32 KB = 32 x 2^10 B

Virtual Address space = 216 B

Page size is always equal to frame size.

Page size = 16 B. Therefore, Frame size = 16 B

If there is a restriction, the number of bits is calculated like this:  

number of page entries = 2^[log2(physical memory size) - log2(n bit machine)]

where

physical memory size = 32KB  which is the restriction

n bit machine = frame size = 16

Hence, we have page entries = 2^[log2(32*2^10) - log2(16)] = 2ˆ[15 - 4 ] = 2ˆ11

7 0
3 years ago
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