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Darina [25.2K]
3 years ago
14

Arctic Gear LLC. recently adopted a new ERP​ system, but quickly found that it​ wasn't a perfect fit with their existing procedu

res. These differences between the​ organization's requirements and the​ application's capabilities are known as​ ________.
a. employee resistance.
b. transition problems.
c. requirements gaps.
d. lack of collaborative management.
Computers and Technology
1 answer:
inysia [295]3 years ago
4 0

Answer:

The answer is "Option c".

Explanation:

GAP analyses are worked to specify the business operations and function. It  helps developed the ERP framework across the analysis process, and the variance in business, which uses the existing resource, that should be defined, reported and allowed, and wrong choices can be described as follows:

  • In option a, It described its action or a failure to alter, that's why it is wrong.
  • In option b, This technique fails the system, that's why it is wrong.
  • In option d, It combines the potential for social application, that's why it is wrong.
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Program 7 - Circle You write ALL the code, then run it - Produce the correct output. Turn in code and screen print of successful
fgiga [73]

Answer:

Here is the Circle class:

public class Circle {   // class definition

private int radius;    // private data member of type int of class Circle named radius that stores the value of radius

public Circle(int r) {   // parameterized constructor of class Circle that takes radius as argument

 radius = r;  }    // sets value of radius as r

public int getRadius() {   // accessor method to get the value of radius

 return radius;  }   // returns the current value of radius

public int Diameter() {   // method to compute diameter of a circle

 return radius * 2;  }   // multiply radius value by 2 to compute diameter of Circle

 

public double Area() {   // method to compute area of a circle

 return Math.PI  * Math.pow(radius, 2);   }   //the formula of area is

π x radius² where value of π is get using Math.PI

 

 public double Circumference() {   // // method to compute circumference of a circle

 return 2* Math.PI * radius;   }   }  //the formula of circumference is

2 x π x radius where value of π is get using Math.PI

Explanation:

Here is the Main class:

import java.util.Scanner;  //to accept input from user

public class Main {  //definition of Main class

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

   

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

    System.out.println("Enter radius: ");  //prompts user to enter radius

    int r = scanner.nextInt();  //reads the value of radius from user

 Circle c = new Circle(r);  // calls Constructor of Circle passing r as argument to it using the object c of class Circle

 

    if(c.getRadius()<=0){  //calls getRadius method to get current value of radius using objec and checks if this value (input value of r ) is less than or equal to 0

        System.out.println("Error!");   }  //when above if condition evaluates to true then print this Error! message

    else {  //if the value of radius is greater than 0

System.out.println("the radius of this Circle is: " +c.getRadius());  //calls getRadius method to return current value of r (input value by user)

System.out.println("the diameter of this Circle  is: " + c.Diameter());  //calls Diameter method to compute the diameter of Circle and display the result on output screen

System.out.printf("the circumference of this Circle  is: %.2f", c.Circumference());  //calls Circumference method to compute the Circumference of Circle and display the result on output screen

System.out.println();  //prints a line

System.out.printf("the Area of this Circle  is: %.2f", c.Area()); } }  }

//calls Area method to compute the Area of Circle and display the result on output screen

The program and its output is attached.

7 0
3 years ago
A finance company wants to upgrade its accounting software to a higher version. This version change requires a change in data fo
Alina [70]

Answer:

The answer to this question is option b which is data re-engineering.

Explanation:

In computer science, data re-engineering is a part of the software development life cycle(SDLC). In the SDLC the data re-engineering is a technique that provides the facility to increase the size of the data formats, design,data-view, etc. It is also known as the software development process in this process, there are seven-stage for software development. If we want to upgrade the software to use the data re-engineering technique so we used the software development process. We use only this process to develop the software because there is no other process to development. So the correct answer is data re-engineering

4 0
2 years ago
Read 2 more answers
The _________ in an internet are responsible for receiving and forwarding packets through the interconnected set of networks and
raketka [301]

Answer:

routers

Explanation:

<h2><em><u>Fill in the blanks</u></em></h2>

The<u> routers </u> in an internet are responsible for receiving and forwarding packets through the interconnected set of networks and making routing decisions based on knowledge of the topology and traffic/delay conditions of the internet.

3 0
3 years ago
MODS ONLY answer this, I have something for you. I'm reporting this to make it easier
aliya0001 [1]

Answer:

whyyyyy

have  a good day :)

Explanation:

7 0
3 years ago
let m be a positive integer with n bit binary representation an-1 an-2 ... a1a0 with an-1=1 what are the smallest and largest va
Ahat [919]

Answer:

Explanation:

From the given information:

a_{n-1} , a_{n-2}...a_o in binary is:

a_{n-1}\times 2^{n-1}  + a_{n-2}}\times 2^{n-2}+ ...+a_o

So, the largest number posses all a_{n-1} , a_{n-2}...a_o  nonzero, however, the smallest number has a_{n-2} , a_{n-3}...a_o all zero.

∴

The largest = 11111. . .1 in n times and the smallest = 1000. . .0 in n -1 times

i.e.

(11111111...1)_2 = ( 1 \times 2^{n-1} + 1\times 2^{n-2} + ... + 1 )_{10}

= \dfrac{1(2^n-1)}{2-1}

\mathbf{=2^n -1}

(1000...0)_2 = (1 \times 2^{n-1} + 0 \times 2^{n-2} + 0 \times 2^{n-3} + ... + 0)_{10}

\mathbf {= 2 ^{n-1}}

Hence, the smallest value is \mathbf{2^{n-1}} and the largest value is \mathbf{2^{n}-1}

3 0
2 years ago
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