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OverLord2011 [107]
3 years ago
14

Which of the following binary (base-2) numbers is LARGEST?

Computers and Technology
1 answer:
Lemur [1.5K]3 years ago
7 0

Answer:

A. 11000000

Explanation:

A. 11000000 is 192

B. 01111111 is 127

C. 00000001 is 1

D. 10111111 is 191

Therefore, making A the largest.

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Network issues; screen resolution and volume issues
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Business intelligence is gained through industrial espionage and the gathering of this information often includes illegal or une
Nataly [62]

Answer:

A) True

Explanation:

Industrial espionage utilizes both illegal and unethical methods in gathering information about a corporate organization in order to get business intelligence. This involves stealing intellectual property and trade secrets to use them for a competitive advantage.  because Information about company's products, services, finances, sales, etc can be toold for  economic warfare

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Which is a common problem caused by the widespread use of computers?
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Hacking would be a widespread problem.
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4 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
George enters the types of gases and the amount of gases emitted in two columns of an Excel sheet. Based on this data he creates
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Answer:

data source

Explanation:

The main aim of a data source is for the gathering of all necessary information that is needed to access a data. Since he has used the information to create a pie chart, this means that some data were used for the creation of this pie chart. Hence the information used for the creation of the pie chart is the data source for the information illustrated on the pie chart.

3 0
3 years ago
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