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Ivan
3 years ago
12

Multiple layers provide multiple road blocks for a ________.

Computers and Technology
1 answer:
Hatshy [7]3 years ago
4 0

Individual or other words user and attackers

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What is the main purpose of a graphic organizer?
timama [110]
B. To organize information using shapes.
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3 years ago
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The process of starting or restarting a computer or mobile device is called __________.
Neko [114]
Hi!

When ever we start or restart any device we may hold - this process is called <em>booting. </em>

Hopefully, this helps! =)
3 0
3 years ago
HELP!!!
aleksley [76]

Answer:

The detail answer of this question is given in explanation section.

The simple answer is option B

Explanation:

Let took at each option:

A) an html tag

An html tag is used to start and end html document. It does not have anything to do with rendering.

B) a doctype declaration

The document type declaration is necessary because it tell the browser which version of html should be rendered.

C)Body tag tell the browser. it is the visible area of website.

D) A hear tag is used to clear meta data about website.

3 0
2 years ago
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Create an array of numbers filled by the random number generator. (value = (int)(Math.random() * 100 + 1);) Print the array and
Y_Kistochka [10]

Answer:

Explanation:

the following is the code to run this (JAVA)

MeanStandardDev.java

import java.util.Random;

import java.util.Scanner;

public class MeanStandardDev {

public static void main(String[] args) {

// Declaring variables

int N;

double lower, upper, min, max, mean, stdDev;

/*

* Creating an Scanner class object which is used to get the inputs

* entered by the user

*/

Scanner sc = new Scanner(System.in);

// Getting the input entered by the user

System.out.print(" How many Random Numbers you want to generate :");

N = sc.nextInt();

System.out.print("Enter the Lower Limit in the Range :");

lower = sc.nextDouble();

System.out.print("Enter the Upper Limit in the Range :");

upper = sc.nextDouble();

// Creating Random class object

Random rand = new Random();

double nos[] = new double[N];

// this loop generates and populates 10 random numbers into an array

for (int i = 0; i < nos.length; i++) {

nos[i] = lower + (upper - lower) * rand.nextDouble();

}

//calling the methods

min = findMinimum(nos);

max = findMaximum(nos);

mean = calMean(nos);

stdDev = calStandardDev(nos, mean);

//Displaying the output

System.out.printf("The Minimum Number is :%.1f\n",min);

System.out.printf("The Maximum Number is :%.1f\n",max);

System.out.printf("The Mean is :%.2f\n",mean);

System.out.printf("The Standard Deviation is :%.2f\n",stdDev);

}

//This method will calculate the standard deviation

private static double calStandardDev(double[] nos, double mean) {

//Declaring local variables

double standard_deviation=0.0,variance=0.0,sum_of_squares=0.0;

/* This loop Calculating the sum of

* square of eeach element in the array

*/

for(int i=0;i<nos.length;i++)

{

/* Calculating the sum of square of

* each element in the array    

*/

sum_of_squares+=Math.pow((nos[i]-mean),2);

}

//calculating the variance of an array

variance=((double)sum_of_squares/(nos.length-1));

//calculating the standard deviation of an array

standard_deviation=Math.sqrt(variance);

return standard_deviation;

}

//This method will calculate the mean

private static double calMean(double[] nos) {

double mean = 0.0, tot = 0.0;

// This for loop will find the minimum and maximum of an array

for (int i = 0; i < nos.length; i++) {

// Calculating the sum of all the elements in the array

tot += nos[i];

}

mean = tot / nos.length;

return mean;

}

//This method will find the Minimum element in the array

private static double findMinimum(double[] nos) {

double min = nos[0];

// This for loop will find the minimum and maximum of an array

for (int i = 0; i < nos.length; i++) {

// Finding minimum element

if (nos[i] < min)

min = nos[i];

}

return min;

}

//This method will find the Maximum element in the array

private static double findMaximum(double[] nos) {

double max = nos[0];

// This for loop will find the minimum and maximum of an array

for (int i = 0; i < nos.length; i++) {

// Finding minimum element

if (nos[i] > max)

max = nos[i];

}

return max;

}

}

the OUTPUT should give;

How many Random Numbers you want to generate :10

Enter the Lower Limit in the Range :1.0

Enter the Upper Limit in the Range :10.0

The Minimum Number is :1.1

The Maximum Number is :9.9

The Mean is :6.30

The Standard Deviation is :2.98

cheers i hope this helps!!!

4 0
3 years ago
List some of the ways that healthcare information technology is making healthcare more of a self-service industry?
harkovskaia [24]

Health information technology (HIT)) is "the utilization of data preparing including both PC equipment and programming that manages the capacity, recovery, sharing, and utilization of medicinal services data.

<u>Explanation:</u>

  • Improving personal satisfaction is one of the primary advantages of incorporating new advancements into medication.
  • Restorative advancements like negligibly intrusive medical procedures, better-observing frameworks, and progressively happy with checking gear are enabling patients to invest less energy in recuperation and additional time getting a charge out of a solid life.
  • An essential advantage of offering self-administration checkouts is that clients need them, and effective retailers give what their clients need.
  • The principal innovation is advantageous to people for a few reasons. At the therapeutic level, innovation can help treat increasingly wiped out individuals and subsequently spare numerous lives and battle exceptionally hurtful infections and microscopic organisms.
  • Indeed, three explicit reasons that innovation is acceptable is that it spares lives by improving prescription, keeps us associated with one another, and gives instruction and stimulation.
  • One motivation behind why innovation is acceptable is that it has spared numerous lives.
  • Innovation additionally can be used to improve instructing and learning and help our students be successful.
3 0
3 years ago
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