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tiny-mole [99]
3 years ago
10

You open a folder Properties box to encrypt the folder, click Advanced, and discover that Encrypt contents to secure data is dim

med. What is the most likely problem?a. Encryption has not been enabled. Use the Computer Management console to enable it.
b. You are not using an edition of Windows that supports encryption.
c. Most likely a virus has attacked the system and is disabling encryption.
d. Encryption applies only to files, not folders
Computers and Technology
1 answer:
yaroslaw [1]3 years ago
5 0

Most likely a virus has attacked the system and is disabling encryption is the most likely problem.

c. Most likely a virus has attacked the system and is disabling encryption.

<u>Explanation:</u>

The end users can protect the folder by enabling encrypt the folder or files. But in windows encrypting the file or folder is not available. The computer management console is used to open all other services such as risk management, performance management extra.

In some other operating encryption of folder or files is allowed.  Third-party software is available to protect folder or files once the third party used any files possible to share outside the world.

As far as my knowledge there is encrypting technology is available in the windows operating system.

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A small company with 100 computers has hired you to install a local area network. All of the users perform functions like email,
charle [14.2K]

Answer:

Answer to the following question is as follows;

Explanation:

Windows 7 for 100 users and Windows 8.1 for 25 users are the best options since they both enable networking and active directory, and Windows 8.1 also offers Windows server capabilities.

Winxp would therefore work for $100, but it is unsupported and has no updates.

If you wish to go with open source, you can choose Ubuntu 16 or 18 or Linux.

8 0
3 years ago
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
what are some of the challenges that could arise from setting up a file management system on a computer
seropon [69]
A virus maybe or ransomware
7 0
3 years ago
Jim, the IT director, is able to complete IT management task very well but is usually two weeks late in submitting results compa
oee [108]

Answer:Effective but not efficient

Explanation:

Jim is effective because he was able to complete the IT tasks well but he is not efficient because he didn't submit the result on time because being efficient includes management of time.

5 0
3 years ago
True or false :Beyond fulfilling legal obligations, a diverse workplace offers many other benefits as well.
KengaRu [80]
The answer to your question is true.
7 0
3 years ago
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