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zepelin [54]
2 years ago
5

Which method allows a computer to react accordingly when it requests data from a server and the server takes too long to respond

?
(A) response timeout
(B) encapsulation
(C) flow control
(D) access method
Computers and Technology
1 answer:
Rina8888 [55]2 years ago
6 0

Answer:

A. Request timeout.

Explanation:

The end devices like the computer systems in a network seeks to share resources with one another and/ or request resources from central server.

With this, there are two ways computers in a network can communicate. They are peer to peer network communication and client-server network communication.

The client-server communication requires a dedicated central server where computers in the network require data. Peer to peer describes a network where computers serve as both client and server to each other.

Request timeout is a message sent to a source when the time to live period (TTL) of a packet expires.

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If the object instance is created in a user program, then the object instance can access both the public and private members of
rewona [7]

Answer:

False

Explanation:

The private member of a class is not accessible by using the Dot notation ,however the private member are those which are not accessible inside the class they are accessible outside the class  .The public member are accessible inside the class so they are accessible by using the dot operator .

<u>Following are the example is given below in C++ Language </u>

#include<iostream>   // header file

using namespace std;  

class Rectangle

{    

   private:  

       double r; // private member  

   public:      

       double  area()  

       {     return 3.14*r*r;  

       }        

};  

int main()  

{    

  Rectangle r1;// creating the object  

   r1.r = 3.5;  

 double t= r1.area(); // calling

cout<<" Area is:"<<t;  

   return 0;  

}  

Output:

compile time error is generated

<u>The correct program to access the private member of class is given below </u>

#include<iostream>   // header file

using namespace std;  

class Rectangle

{    

   private:  

       double r; // private member  

   public:      

       double  area()  

       {    

r1=r;

double t2=3.14*r2*r2;

return(t2); // return the value  

       }        

};  

int main()  

{    

  Rectangle r1;// creating the object  

   r1.r = 1.5;  

 double t= r1.area(); // calling

cout<<" Area is:"<<t;  

   return 0;  

}  

Therefore the given statement is False

5 0
2 years ago
Get user input as a boolean and store it in the variable tallEnough. Also, get user input as a boolean and store it in the varia
frozen [14]

Answer:

I will code in JAVA.

import java.util.Scanner;

public class Main {

public static void main(String[] args) {

 

boolean tallEnough;

boolean oldEnough;

Scanner input = new Scanner(System.in);

tallEnough = input.nextBoolean();<em> //wait the input for tallEnough</em>

oldEnough = input.nextBoolean(); <em>//wait the input for OldEnough</em>

   if(tallEnough && oldEnough){

   System.out.print(true);

   } else {

   System.out.print(false);

   }

}

}

Explanation:

First, to accept user inputs you have to import the class Scanner. Then declare both variables before allowing the user to set input values for both boolean variables.

In the if-else statement checks if both variables are true, then prints true. Another case prints always false.

8 0
1 year ago
Create a program named Auction that allows a user to enter an amount bid on an online auction item. Include three overloaded met
Olin [163]

Answer:

Explanation:

The following code is written in Java and creates the overloaded methods as requested. It has been tested and is working without bugs. The test cases are shown in the first red square within the main method and the output results are shown in the bottom red square...

class Auction {

   public static void main(String[] args) {

       bid(10);

       bid(10.00);

       bid("10 dollars");

       bid("$10");

       bid("150 bills");

   }

   public static void bid(int bid) {

       if(bid >= 10) {

           System.out.println("Bid Accepted");

       } else {

           System.out.println("Bid not high enough");

       }

   }

   public static void bid(double bid) {

       if(bid >= 10) {

           System.out.println("Bid Accepted");

       } else {

           System.out.println("Bid not high enough");

       }

   }

   public static void bid(String bid) {

       if (bid.charAt(0) == '$') {

           if (Integer.parseInt(bid.substring(1, bid.length())) > 0) {

               System.out.println("Bid Accepted");

               return;

           } else {

               System.out.println("Bid not in correct Format");

               return;

           }

       }

       int dollarStartingPoint = 0;

       for (int x = 0; x < bid.length(); x++) {

           if (bid.charAt(x) == 'd') {

               if (bid.substring(x, x + 7).equals("dollars")) {

                   dollarStartingPoint = x;

               } else {

                   break;

               }

           }

       }

       if (dollarStartingPoint > 1) {

           if (Integer.parseInt(bid.substring(0, dollarStartingPoint-1)) > 0) {

               System.out.println("Bid Accepted");

               return;

           } else {

               System.out.println("Bid not in correct format");

               return;

           }

       } else {

           System.out.println("Bid not in correct format");

           return;

       }

   }

}

3 0
1 year ago
Write a python 3 function named words_in_both that takes two strings as parameters and returns a set of only those words that ap
s344n2d4d5 [400]

Answer:

def words_in_both(a, b):

 a1 = set(a.lower().split())

 b1 = set(b.lower().split())

 return a1.intersection(b1)

common_words = words_in_both("She is a jack of all trades", 'Jack was tallest of all')

print(common_words)

Explanation:

Output:

{'all', 'of', 'jack'}

5 0
2 years ago
Read 2 more answers
a.Write a Python function sum_1k(M) that returns the sum푠푠= ∑1푘푘푀푀푘푘=1b.Compute s for M = 3 by hand and write
stealth61 [152]

Answer:

def sum_1k(M):

   s = 0

   for k in range(1, M+1):

       s = s + 1.0/k

   return s

def test_sum_1k():

   expected_value = 1.0+1.0/2+1.0/3

   computed_value = sum_1k(3)

   if expected_value == computed_value:

       print("Test is successful")

   else:

       print("Test is NOT successful")

test_sum_1k()

Explanation:

It seems the hidden part is a summation (sigma) notation that goes from 1 to M with 1/k.

- Inside the <em>sum_1k(M)</em>, iterate from 1 to M and calculate-return the sum of the expression.

- Inside the <em>test_sum_1k(),</em> calculate the <em>expected_value,</em> refers to the value that is calculated by hand and <em>computed_value,</em> refers to the value that is the result of the <em>sum_1k(3). </em>Then, compare the values and print the appropriate message

- Call the <em>test_sum_1k()</em> to see the result

8 0
1 year ago
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