All categories

3 years ago

How long is the latency time of a typical hard-disk drive spinning at 360 revolutions per second?

Computers and Technology

1 answer:

HACTEHA [7]3 years ago

8 0

Solution:

The latency time of a typical hard-disk drive spinning at 360 revolutions per second is 5400.

Latency is a measure of delay. In a network, latency measures the time it takes for some data to get to its destination across the network. It is usually measured as a round trip delay - the time taken for information to get to its destination and back again. The round trip delay is an important measure because a computer that uses a TCP/IP network sends a limited amount of data to its destination and then waits for an acknowledgement to come back before sending any more. Thus, the round trip delay has a key impact on the performance of the network.

Thus the required answer is 5400

You might be interested in

A network on the internet has a subnet mask of 255.255.240.0. what is the maximum number of hosts it can handle

Lera25 [3.4K]

It is a class B network, so for a class B network, the upper 16 bits form the network address and the lower 16 bits are subnet and host fields. Of the lower 16 bits, most significant 4 bits are 1111. This leaves 12 bits for the host number. So, 4096(2^12) host address exists. First and last address are special so the maximum number of address is 4096-2=4094.

7 0

4 years ago

What does a coder do on a daily basis?

Gnoma [55]

medical charts and assinging codes

8 0

3 years ago

For this lab, you will write a Java program to prompt the user to enter two integers. Your program will display a series of arit

Mekhanik [1.2K]

Answer:

Following are the program to this question:

import java.util.*;//for user-input value

public class Project01//Decalring a class Project01

{

public static void main(String asq[])//main method

{

int ax1,ax2;//Decalring integer variables

Scanner oscr=new Scanner(System.in);//creating Scanner class Object

System.out.print("Enter the First number: ");//print message

ax1=oscr.nextInt();//input first number

System.out.print("Enter the Second number: ");//print message

ax2=oscr.nextInt();//input second number

System.out.println(ax1+ "+"+ax2+"= "+(ax1+ax2));//use print method to perform the arithmetic operation

System.out.println(ax1+"-"+ax2+"= "+(ax1-ax2));//use print method to perform the arithmetic operation

System.out.println(ax1+"*"+ax2+"= "+(ax1*ax2));//use print method to perform the arithmetic operation

System.out.println(ax1+"/"+ax2+"= "+(ax1/ax2));///use print method to perform the arithmetic operation

System.out.println(ax1+"%"+ax2+"= "+(ax1%ax2));//use print method to perform the arithmetic operation

System.out.println("The average of your two numbers is: "+(ax1+ax2)/2);//calculating average

}

Output:

Please find the attached file.

Explanation:

In the program, inside the class "Project01" and the main method two integer variable "ax1,ax2" is declared that uses the scanner class concept for input the value from the user-end, and in the next step, it uses the print method with the arithmetic operation to perform and prints its calculated value.

7 0

3 years ago

An object's state is defined by the object's

Eva8 [605]

Answer:

The right answer is option 2: instance variables and their values

Explanation:

Let us define what an object is.

An object is a blueprint of a real-world identity. The instances are the reference to the object. Multiple instances of an object type can be made.

The instance variables and their values help us to determine the state of the object.

Hence,

The right answer is option 2: instance variables and their values

6 0

3 years ago

g Write a program that prompts the user for an integer n between 1 and 100. If the number is outside the range, it prints an err

grin007 [14]

Answer:

The cpp program is given below.

#include<iostream>

#include<iomanip>

using namespace std;

int main() {

// variables declared

int n;

int sum=0;

float avg;

{

// user input taken for number

cout<< "Enter a number between 1 and 100 (inclusive): ";

cin>>n;

if(n<1 || n>100)

cout<<" Number is out of range. Enter valid number."<<endl;

}while(n<1 || n>100);

cout<<" "<<endl;

// printing even numbers between num and 50

for(int num=1; num<=n; num++)

{

sum = sum + num;

}

avg = sum/n;

// displaying sum and average

cout<<"Sum of numbers between 1 and "<<n<<" is "<<sum<<endl;

cout<<"Average of numbers between 1 and "<<n<<" is ";

printf("%.2f", avg);

return 0;

}

OUTPUT

Enter a number between 1 and 100 (inclusive): 123

Number is out of range. Enter valid number.

Enter a number between 1 and 100 (inclusive): 56

Sum of numbers between 1 and 56 is 1596

Average of numbers between 1 and 56 is 28.00

Explanation:

The program is explained below.

1. Two integer variables are declared to hold the number, n, and to hold the sum of numbers from 1 to n, sum. The variable sum is initialized to 0.

2. One float variable, avg, is declared to hold average of numbers from 1 to n.

3. User input is taken for n inside do-while loop. The loop executes till user enters value between 1 and 100. Otherwise, error message is printed.

4. The for loop executes over variable num, which runs from 1 to user-entered value of n.

5. Inside for loop, all the values of num are added to sum.

sum = sum + num;

6. Outside for loop, average is computed and stored in avg.

avg = sum/n;

7. The average is printed with two numbers after decimal using the following code.

printf("%.2f", avg);

8. The program ends with return statement.

9. All the code is written inside main() and no classes are involved.

3 0

3 years ago