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Arte-miy333 [17]
2 years ago
5

Transparency is an important concept in policies related to the handling and use of customer data. Organizations should be trans

parent and should notify individuals of the distribution, use, collection, and maintenance of personally identifiable information (PII). Which of the following elements does not need to be included with regard to handling of customer data? a) individual participation b) purpose specification c) response controls d) data minimization
Computers and Technology
1 answer:
Tpy6a [65]2 years ago
6 0

Answer: (C) Response controls

Explanation:

  According to the given question, the response control is not the necessary element while handling the users or customer data as compared to all the other given options that is individual participation, data minimization and the purpose specification.

  • The response control is basically used for exchange the data or any information in an organization but the main disadvantage is that they share all the response and information as they not wait for the request from the users.
  • The all the three other given options expect response control are the important element for handling the customer data as it helps in maintaining the Personal information.  

 Therefore, Option (C) is correct answer.  

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saveliy_v [14]

Answer:

System software.

Explanation:

System software are the in-built functions that perform a particular pre-defined task when used in a computer program. An example of a system software is an operating system.

An operating system (OS) is a system software pre-installed on a computing device to manage or control software application, computer hardware and user processes. Some examples of an operating system are QNX, Linux, OpenVMS, MacOS, Microsoft windows, IBM, Solaris, VM etc.

4 0
2 years ago
Which of the following protocols is used by Web servers to establish communication between a web page and a web browser?
Iteru [2.4K]
HTTP or HTTPS is most likely the answer you are looking for.
6 0
3 years ago
Real numbers which, when squared, become smaller than the original number."
8_murik_8 [283]
Those would be numbers below 1.

Explanation:
.5•.5 = .25
-2•-2 = -4
Hope this helps!
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7 0
2 years ago
Write a program that prompts the user to enter positive integers (as many as the user enters) from standard input and prints out
julia-pushkina [17]

Answer:

Here is the JAVA program that prompts the user to enter positive integers from standard input and prints out the maximum and minimum values:

import java.util.*;

public class Ex0603 {

   public static void main(String[] args) { //start of main function

   Scanner in = new Scanner(System.in); //creates Scanner class object to take input from user

   int minimum = 0; //stores the minimum of the integers

   int maximum = 0; //stores the minimum of the integers

   int num; //stores the positive integer values

   System.out.println("enter a positive integer: ");//prompts user to enter a positive integer

   num = in.nextInt();//scans and reads the input integer value from user

   minimum = num;  //assigns value of num to minimum

   while(true) {//while loop keeps prompting user to enter a positive integer

       System.out.println("enter a positive integer: ");//prompts user to enter a positive integer

       num = in.nextInt();//scans and reads the input integer value from user

       if(num <=0)//if the user inputs a negative integer or 0

           break; //breaks the loop

       if (num > maximum) {//if the value of positive integer is greater than the maximum of value

           maximum = num; } //assigns that maximum value of num to maximum variable

       if (num < minimum) {//if the value of positive integer is less than the minimum of value

           minimum = num; //assigns that minimum value of num to minimum       }    }            

   System.out.println("The maximum integer is : " + maximum); //displays the maximum of the positive integers

   System.out.println("The minimum integer is : " + minimum); }}   //displays the minimum of the positive integers

Explanation:

Here is the JAVA program that that prompts the user to enter an integer N, then reads N double values, and prints their mean (average value) and sample standard deviation:

import java.util.*;  

public class Ex0603 {  

public static void main(String[] args) { //start of main function

Scanner in = new Scanner(System.in);  //creates Scanner class object to take input from user

double integer= 0;  //to store the number of input values

double sum = 0;  //to store the sum of input numbers

double mean = 0;  //to store the average of numbers

double sd = 0;  //to store the standard deviation result

double variance = 0;  //to store the variance result

double sumSquare = 0;  //to store the sum of (num-mean)^2

int n = 0;  //to store the sample size N

   System.out.println("Enter an integer: ");  //prompts user to enter an integer that is how many elements user wants to input

   integer = in.nextInt();  //reads the value of integer from user

   double size=integer; // declares a double type variable and assigns value of int type variable integer to this double type variable size

   int i=0;  //used to point each element of the num array

   double num[] = new double[(int)size];  //declares an array to hold the numbers input by user

   while(integer>0){  //loop executes until value of integer is greater than 0

   System.out.println("Enter a number: "); // prompts to enter a number

   num[i] = in.nextDouble(); //reads double values

   sum += num[i];  //adds the values stored in num array

   i++;  //increments i to point to the next element in num array

   n++;  //increments n to count the total number of elements

  integer--; //decrements integer value at each iteration

   mean = sum / n;    }  //compute the average of values

   i=0;  

    while(size>0){  //loop executes until value of size exceeds 0

       sumSquare += Math.pow(num[i]-mean,2); //takes the sum of square difference between each element of num array and value of mean

       i++; //increments i

      size--;   }  //decrements size

variance = sumSquare / (n-1);  //compute variance by dividing the result of sum of the squares of their differences from the mean by n-1

   sd = Math.sqrt(variance);  //computes standard deviation by using sqrt function which takes the sqrt of the result of variance computed above

System.out.println("Average value is: " + mean+ " and the standard deviation is " + String. format("%.2f", sd));   }} //displays the average of numbers stored in mean variable and value of standard deviation is displayed up to 2 decimal places

Screenshots of both programs and their outputs is attached.

7 0
3 years ago
The java compiler is able to find all programmer errors.
Troyanec [42]
No compiler can do that. Maybe with AI, but that's beyond the scope of a compiler.
5 0
3 years ago
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