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ryzh [129]
4 years ago
5

You interact with a program through which of the following, thereby controlling how you enter data and instructions and how info

rmation is displayed on the screen?
A. operating system
B. information
C. user interface
D.storage
Computers and Technology
2 answers:
kaheart [24]4 years ago
4 0

Answer:C)User interface

Explanation: User interface(UI) is the interaction phase of the user or programmer with the computer system and network. This phase helps the user to connect with the operating system function and let them process it according to the requirement. Example-monitor , keyboard, mouse etc.

Other options that are given are incorrect because operating system is the system that is part of the interface , information is the data and storage is for the storing of data,.Thus the correct option is option(C)

stiks02 [169]4 years ago
3 0

Answer:

Option C. User Interface

Explanation:

User interface is a platform which allows the user to interact or communicate with a program in a device. It also provides a platform to the user to establish a communication or interaction between a application, website or a program.

The emphasis on the development of a good user interface is on the priority list of many companies  and businesses as a result of increasing dependence on the technology.

the UI also allows to control the way of entry of instructions and data and also the way information is displayed on the screen.

You might be interested in
Difference between customized and packaged software​
docker41 [41]

Packaged software is a compilation of programs which are grouped together in order to provide publicly with different tools in the same group.

Custom software is a specific program that are advanced for a goal in a department or in a company.

5 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 process of determining that a user has permission to perform a particular action on a computer is called
Doss [256]
I believe that would be authorization.
5 0
4 years ago
3. For “Incident Energy Analysis” What body parts are involved in the distance
krek1111 [17]

The body part that is close to the arc flash boundary is the energized conductors or circuit parts.

<h3>What is the distance of an arc flash?</h3>

The working distance is known to be the distance that exist between a person and the center of an arc flash.

Note that The body part that is close to the arc flash boundary is the energized conductors or circuit parts.

Learn more about Energy from

brainly.com/question/13881533

#SPJ1

7 0
2 years ago
Suppose that TCP's current estimated values for the round trip time (estimatedRTT) and deviation in the RTT (DevRTT) are 400 mse
Masja [62]

Answer:

617.5 msecs, 580.89 msecs, 582.36 msecs

Explanation:

Lets look at the following formulas:

EstimatedRTT = α * SampleRTT+(1- α) * EstimatedRTT

DevRTT = β * | SampleRTT- EstimatedRTT|+(1- β)* DevRTT

TimeoutInterval = EstimatedRTT +4* DevRTT

Given values are:

EstimatedRTT = 400 msec

DevRTT = 25 msec

α = 0.125

β = 0.25

After the first RTT estimate: (210)

EstimatedRTT = 0.125 * 210 + (1 - 0.125) * 400 = 376.25 msecs

DevRTT = 0.25 * |210 - 376.25| + (1 - 0.25) * 25 = 60.3125 msecs

TimeoutInterval = 376.25 + 4 * 60.3125 = 617.5 msecs

After the second RTT estimate: (400)

EstimatedRTT = 0.125 * 400 + (1 - 0.125) * 376.25 = 379.218 msecs

DevRTT = 0.25 * |400 - 379.218| + (1 - 0.25) * 60.3125 = 50.42 msecs

TimeoutInterval = 379.218 + 4 * 50.42 = 580.89 msecs

After the third RTT estimate: (310)

EstimatedRTT = 0.125 * 310 + (1 - 0.125) * 379.218 = 370.56 msecs

DevRTT = 0.25 * |310 - 370.56| + (1 - 0.25) * 50.42 = 52.95 msecs

TimeoutInterval = 370.56 + 4 * 52.95 = 582.36 msecs

5 0
3 years ago
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