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Sunny_sXe [5.5K]
3 years ago
11

The basic form of backup used in magnetic tape operations is called

Computers and Technology
1 answer:
Harrizon [31]3 years ago
5 0
Answer is the son-father-grandfather concept. Good luck
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In file hashing, a file is read by a special algorithm that uses the value of the bits in the file to compute a single number ca
Zielflug [23.3K]

Answer:

Option (B) i.e., Hash is the correct option.

Explanation:

A hash value is the number that has the fixed length and the data recognition is uniquely. It is also a type of file that is read by the special type algo, it uses the value of the bits in the following computer files which is used to manipulates the single number. That's why the following option is correct.

7 0
3 years ago
Write a program that will open the file random.txt and calculate and display the following: A. The number of numbers in the file
Rama09 [41]

Answer:

Here is the C++ program:

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

#include <fstream>  //to manipulate files

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

int main(){  //start of main function

  ifstream file;   //creates an object of ifstream

   file.open("random.txt"); //open method to open random.txt file using object file of ifstream

       

    int numCount = 0;  //to store the number of all numbers in the file          

    double sum = 0;   //to store the sum of all numbers in the file

    double average = 0.0;   //to store the average of all numbers in the file

    int number ; //stores numbers in a file

                 

        if(!file){   //if file could not be opened

           cout<<"Error opening file!\n";    }   //displays this error message

       

       while(file>>number){   //reads each number from the file till the end of the file and stores into number variable

           numCount++; //adds 1 to the count of numCount each time a number is read from the file          

           sum += number;  }  //adds all the numbers and stores the result in sum variable

           average = sum/numCount;  //divides the computed sum of all numbers by the number of numbers in the file

     

       cout<<"The number of numbers in the file: "<<numCount<<endl;  //displays the number of numbers

       cout<<"The sum of all the numbers in the file: "<<sum<<endl;  //displays the sum of all numbers

       cout<<"The average of all the numbers in the file: "<< average<<endl;  //displays the average of all numbers

       file.close();     }  //closes the file    

   

Explanation:

Since the random.txt is not given to check the working of the above program, random.txt is created and some numbers are added to it:

35

48

21

56

74

93

88

109

150

16

while(file>>number) statement reads each number and stores it into number variable.

At first iteration:

35 is read and stored to number

numCount++;  becomes

numCount = numCount + 1

numCount = 1      

sum += number; this becomes:

sum = sum + number

sum = 0 + 35

sum = 35

At second iteration:

48 is read and stored to number

numCount++;  becomes

numCount = 1+ 1

numCount = 2    

sum += number; this becomes:

sum = sum + number

sum = 35 + 48

sum = 83

So at each iteration a number is read from file, the numCount increments to 1 at each iteration and the number is added to the sum.

At last iteration:

16 is read and stored to number

numCount++;  becomes

numCount = 9 + 1

numCount = 10    

sum += number; this becomes:

sum = sum + number

sum = 674 + 16

sum = 690

Now the loop breaks and the program moves to the statement:

       average = sum/numCount;  this becomes:

       average = 690/10;

       average = 69

So the entire output of the program is:

The number of numbers in the file: 10                                                                                                           The sum of all the numbers in the file: 690                                                                                                     The average of all the numbers in the file: 69

The screenshot of the program and its output is attached.

5 0
3 years ago
Suppose an IP packet is fragmented into 10 fragments, each with a 1% (independent) probability of loss. To a reasonable approxim
Marysya12 [62]

Answer:

a. 0.01

b. 0.001

c. The identification field of the packet fragment can be used to uniquely identify and collate the fragments lost in transmission.

Explanation:

The probability of losing a packet is 10% or 0.1, so the probability of losing the packet twice during transmission;

= 0.1 x 0.1 = 0.01

When any fragments have been part of the transmission, the probability of the packet is dependent on the fragments;

= 0.01 x 0.1 = 0.001

The identification field is a unique 16-bit value assigned to an IPv4 packet, when a packet is fragmented for transmission, its field is used to collate the unique fragments in the packet.

6 0
3 years ago
What are two distinctive types of unmanned aircraft systems
chubhunter [2.5K]
Probes and Drones
Hope this helps and please give brainliest!
8 0
3 years ago
You wrote a program to allow the user to guess a number. Complete the code to generate a random integer from one to 10.
just olya [345]

Answer:

randint

Explanation:

Just did my quiz and got a 100%

7 0
3 years ago
Read 2 more answers
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