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Likurg_2 [28]
3 years ago
12

Lori Redford, who has been a member of the Project Management group, was recently promoted to manager of the team. She has been

added as a member of the Managers group. Several days after being promoted, Lori needs to have performance reviews with the team she manages but she cannot access the performance management system. As a member of the Managers group, she should have the Allow permission to access this system. What is most likely preventing her from accessing this system
Computers and Technology
1 answer:
Marat540 [252]3 years ago
3 0

Answer:

she is still a member of the Project Management group

Explanation:

The most likely reason for this is that she is still a member of the Project Management group. This would prevent her from accessing the performance management system. The performance management system most likely allows access to the Managers Group but denies access to the Project Management Group. Therefore, since the deny access overrides the allow access, it would ultimately cause Lori to be unable to access the system. In order to fix this, she would simply need to leave the Project Management Group.

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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;

/*

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* entered by the user

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Scanner sc = new Scanner(System.in);

// Getting the input entered by the user

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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;

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private static double calMean(double[] nos) {

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

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

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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

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return min;

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double max = nos[0];

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

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

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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!!!

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Explanation:

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num2 = int(input())

If both are positive, the sum is calculated and printed

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<em>    print(num1+num2)</em>

If both are negative, the products is calculated and printed

<em>elif num1 <0 and num2 < 0:</em>

<em>    print(num1*num2)</em>

If only one of them is positive

else:

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