How long would it take to transfer a 600-MB database from disk to memory over a DMA channel with a bandwidth of 2.5 GB/s?

Answer:

The program to this question as follows:

Program:

#include <iostream> //defining header file

using namespace std;

int main() //defining main method

{

int n; //defining integer variables

cout<<"Enter number: "; //message

cin>>n; //input value by user

while (n< 100) //loop for check condition

{

cout<<n<<" ";//print value

n=n*2; //calculate value

}

return 0;

}

Output:

Enter number: 8

8 16 32 64  

Explanation:

In the above C++ language code, a header file is included, then defining the main method, inside the method an integer variable n is defined, which is used for user input for calculating their double number.  

In the next step, The while loop is declared, and the loop variable n range is defined, that its value is less than 100, inside the loop the variable n is used to calculate, its double number, and print function "cout" to print its value.

Answer:

A good deal of restraint must be observed by the language designer when revising programming languages. The greatest danger is that the revision process of programming language will continually add new features, so that the programming language continue to grow more and more complex. Compounding the problem during language revision is the reluctance to remove obsolete features because of existing software. to remove obsolete features.

Comparing the evolution of Fortran as a case study:

Fortran has evolved since it started. The evolved version of Fortran include:

Fortran I: no data-typing statements in the Fortran I language. Variables whose names began with I, J, K, L, M, and N were implicitly integer type, and all others were implicitly floating-point.

Fortran II: fixed many of the bugs in the Fortran I compilation system and added some significant features to the language, the most important being the independent compilation of subroutines.

Fortran IV: Fortran IV was an improvement over Fortran II in many ways. Among its most important additions were explicit type declarations for variables, a logical If construct, and the capability of passing subprograms as parameters to other subprograms.

Fortran 77: Fortran 77 retained most of the features of Fortran IV and added character string handling, logical loop control statements, and an If with an optional Else clause.

Fortran 90: two simple syntactic changes that altered the appearance of both programs and the literature describing the language. First, the required fixed format of code, which required the use of specific character positions for specific parts of statements, was dropped. The second change was that the official spelling of FORTRAN became Fortran.This change was accompanied by the change in convention of using all uppercase letters for keywords and identifiers in Fortran programs. The new convention was that only the first letter of keywords and identifiers would be uppercase.

Fortran 95: continued the evolution of the language, but only a few changes were made. A new iteration construct, Forall, was added to ease the task of parallelizing Fortran programs.

Fortran 2003: added support for object-oriented programming, parameterized derived types, procedure pointers, and interoperability with the C programming language.

Fortran 2008: added support for blocks to define local scopes, co-arrays, which provide a parallel execution model, and the DO CONCURRENT construct, to specify loops without inter-dependencies.

Final answer:

Programming language evolution, like human language, incorporates new features for modern needs while aiming to maintain backward compatibility. Fortran exemplifies this balance with incremental changes over decades, accommodating modern programming paradigms while preserving older codes' functionality.

Explanation:

The evolution of programming languages, much like human languages, involves changes to adapt to new concepts and technologies, whilst also maintaining backward compatibility. Just as human languages adopt new words and grammar for clarity and efficiency, programming languages evolve to include new features and paradigms, improve performance, and fix bugs. However, changes must be carefully managed to avoid breaking existing code, which can be costly and time-consuming to update. Comparing this to the evolution of Fortran, which has seen numerous revisions since its inception in 1957, we notice that each new standard carefully introduces improvements while retaining the ability to compile older code, thus maintaining a balance between progress and stability.

With programming languages like LISP and Fortran, we have seen that their growth comes with a coevolution with technological advancement. Organizations might place restrictions on how rapidly and drastically programming languages can change in order to ensure reliability, maintainability, and backwards compatibility. Much like Fortran, which has evolved incrementally over decades, it has managed to stay relevant in scientific computing by adding features such as object-oriented programming, while still keeping a large proportion of legacy code functional.

Understanding the historical development of programming languages alongside their restrictions and adaptability is crucial. Despite personal preferences for 'pure' versions of languages, most modern languages undergo systematic changes and maintain a level of complexity that is essential for contemporary needs. Therefore, languages like Fortran achieve a delicate balance by embracing modern concepts without discarding the underlying principles that have been proven effective over the years.

Answer:

Option A i.e., Detective.

Explanation:

When an organization, in a very bad area, recently started a new office. The manager installs a CCTV device for 24-hour surveillance of the area and entrance. So, the following scenario is about the detective control because If users start understanding that their actions are registered and tracked through authenticating into the computer to execute a function.

Answer:

Spyware

Explanation:

Internet is a type of computer network that allow device communicate with each other world wide. So, it is not the correct option.

Worm: this is a standalone malware computer program that replicates itself in order to spread to other computers. It spreads copies of itself from computer to computer. A worm can replicate itself without any human interaction, and it does not need to attach itself to a software program in order to cause damage. The major feature of a worm is ability to replicate on it own. So, it is not the correct option.

Bot: this is a software application that runs automated tasks. So, it is not the correct option.

Middleware: this is software that lies between an operating system and the applications running on it, enabling communication and data management. It provides services to software applications beyond those available from the operating system. So, it is not the correct option.

Spyware: this is the correct answer. Spyware is similar to trojan horse because it hides itself in a system and a user may not know that it exist on the system. Spyware is a form of malware that hides on your device, monitors your activity, and steals sensitive information without knowledge of the user.

Answer:

Spyware

Explanation:

Computer network systems are intercommunication of connected devices, to share resources within and to other remote networks. Network security policies and technical control system technologies are used by a network or system administrator to prevent unauthorized and unauthenticated access to user account in the network by cyber attackers.

Cyber attackers are individuals or group of individuals with the traditional hacking skills to exploit network and system vulnerabilities, to steal information. Malwares are used by hackers to covertly access networks. The spyware is a monitoring Trojan horse malware, used to monitor the activities of a user, steal information for fraudulent purposes.

Answer:

freeware

Explanation:

a freeware is a software that is available free of charge but is not distributed with the source code.

Answer:

int sumNumber(int n){

if(n < 0)

return 0;

int sum = 0;

int i;

for(i = 0; i <= n; i++){

sum = sum + i;

}

return sum;

}

Explanation:

I am going to write a C code for this.

A for loop and an initialized variable is sufficient to solve this problem.

int sumNumber(int n){

if(n < 0)

return 0;

int sum = 0;

int i;

for(i = 0; i <= n; i++){

sum = sum + i;

}

return sum;

}

Answer:

Answer explained below

Explanation:

Joel is the sole shareholder of Manatee Corp. C Corporation has separate legal identity and limited liability for the owner but the double taxation is its demerit as its revenue is taxed at the company level and again as shareholder dividends.

Therefore Joel should purchase the warehouse and lease it to Manatee to take the tax advantage and reduce his overall tax liability. Leasing assets to the corporation is legal also and corporation will pay rent to the owner and owner has to show it as rental income.

By this way the tax liability will be lower for the corporation. Joel will get deduction on the items like acquisition interest, depreciation, repairs and maintenance, insurance and administrative costs etc. This way the owner of the corporation can avoid double taxation of any gain from the asset’s appreciation and sale.

As there are limited liability for the owner in C Corporation so having more asset with owner and less with corporation is good strategic decision of a volatile industry.

Answer:

The complete program is given below with step by step comments for explanation.

Explanation:

So we need to write a program that can calculate the average daily temperature and humidity. We can create a class named Weather_Station to do the job.

public class Weather_Station

{

public static void main(String[] args)

{

// we are told that the data is for 10 days and 24 hours so they are fixed and cannot be changed

final int Days = 10;

final int Hours = 24;

// Then we define a three-dimensional array named station for storing days, hours and temperature/humidity

double[][][] station = new double[Days][Hours][2];

// Then we read input from a file

Scanner input = new Scanner(System.in);

// Since there is a lot of data, using a loop to extract the data would be a good idea so we run a loop for 10*24 times to get the day, hour, temperature and humidity values.

for (int k = 0; k < Days*Hours; k++)

{

int day = input.nextInt();

int hour = input.nextInt();

double temp = input.nextDouble();

double humidity = input.nextDouble();

station[day - 1][hour - 1][0] = temp;

station[day - 1][hour - 1][1] = humidity;

}

// Since we want to get the average temperature and humidity of each day for 10 days so we run a loop for days times (thats 10)

for (int i = 0; i < Days; i++)

{

double Total_temp = 0;

double Total_humidity = 0;

// Then we run another loop to add up all the temperatures and humidity values for each hour in a day and store them in Total_temp and Total_humidity respectively.

for (int j = 0; j < Hours; j++)

{

Total_temp += station[i][j][0];

Total_humidity += station[i][j][1];

}

}

// The average is simply the total number of temperature values divided by the number of hours, same goes for humidity.

double Avg_temp=Total_temp / Hours;

double Avg_humidity=Total_humidity / Hours;

// Then we print the values of any day's average temperature and humidity

System.out.println("For the Day of " + i + "Average Temperature is " + Avg_temp);

System.out.println("For the Day of " + i + "Average Humidity is " + Avg_humidity);

}

}

Final answer:

This response provides a Java program that simulates the collection of temperature and humidity data from a meteorology station and calculates the average daily temperature and humidity over 10 days. It makes use of a 3-dimensional array to store and process the hourly data for each day.

Explanation:

Java Program to Calculate Average Daily Temperature and Humidity

In order to calculate the average daily temperature and humidity using a Java program, we will first simulate the data collection from a meteorology station, which records temperature and humidity hourly for 10 days. The program will iterate through this data to calculate the averages.

The following Java program does just that, and it can be expanded or adapted for different datasets or time periods:
public class WeatherStation {
   public static void main(String[] args) {
       // Randomly generate data for temperature and humidity
       int[][][] data = new int[10][24][2];
       for (int day = 0; day < data.length; day++) {
           for (int hour = 0; hour < data[day].length; hour++) {
               data[day][hour][0] = (int)(Math.random() * 30) + 10; // Temperature
               data[day][hour][1] = (int)(Math.random() * 100); // Humidity
           }
       }
       
       // Calculate average temperature and humidity for each day
       for (int day = 0; day < data.length; day++) {
           double dailyTempTotal = 0, dailyHumidityTotal = 0;
           for (int hour = 0; hour < data[day].length; hour++) {
               dailyTempTotal += data[day][hour][0];
               dailyHumidityTotal += data[day][hour][1];
           }
           double dailyTempAverage = dailyTempTotal / data[day].length;
           double dailyHumidityAverage = dailyHumidityTotal / data[day].length;
           System.out.println("Day " + (day + 1) + ": Average Temperature = " + dailyTempAverage
               + ", Average Humidity = " + dailyHumidityAverage);
       }
   }
}

This simple program creates a 3-dimensional array to store 10 days of hourly temperature and humidity readings, then calculates the average values for each day, providing crucial insights into the climate and environmental conditions of the region.

To create an application named TestSoccerPlayer, you will need to define a class called SoccerPlayer with the properties Name, JerseyNum, Goals, and Assists. In your TestSoccerPlayer application, you can create an instance of the SoccerPlayer class and display its properties.

To create an application named TestSoccerPlayer, you will need to define a class called SoccerPlayer with the properties Name, JerseyNum, Goals, and Assists. Here is an example:

In your TestSoccerPlayer application, you can create an instance of the SoccerPlayer class and display its properties. Here is an example:

Final answer:

The subject of this question is Computers and Technology and it is suitable for High School level students. To create the TestSoccerPlayer application, you need to instantiate and display a SoccerPlayer object. The SoccerPlayer class contains properties such as Name (a string), JerseyNum (an integer), Goals (number of goals scored, an integer), and Assists (number of assists, an integer).

Explanation:

The subject of this question is Computers and Technology and it is suitable for High School level students.

To create the TestSoccerPlayer application, you need to instantiate and display a SoccerPlayer object. The SoccerPlayer class contains properties such as Name (a string), JerseyNum (an integer), Goals (number of goals scored, an integer), and Assists (number of assists, an integer).

Here's an example of how you can create the application in Java:

public class TestSoccerPlayer {

   public static void main(String[] args) {

       SoccerPlayer player = new SoccerPlayer();

       player.setName("John Doe");

       player.setJerseyNum(10);

       player.setGoals(5);

       player.setAssists(3);

       System.out.println("Player Name: " + player.getName());

       System.out.println("Jersey Number: " + player.getJerseyNum());

       System.out.println("Goals: " + player.getGoals());

       System.out.println("Assists: " + player.getAssists());

   }

}

Remember to replace the placeholders with appropriate values for the player's name, jersey number, goals, and assists.

Answer:

You must attach the database files

Explanation:

Answer:

#include <iostream>

bool validateDate(int, int, int);

using namespace std;

int main()

{

int year;

int month;

int day;

const int MIN_YEAR = 0, MIN_MONTH = 1, MAX_MONTH = 12, MIN_DAY = 1, MAX_DAY = 31;

bool validDate = true;

   cout<<"Enter the year"<<endl;

   cin>>year;

   cout<<"Enter the month"<<endl;

   cin>>month;

   cout<<"Enter the day"<<endl;

   cin>>day;

if(year <= MIN_YEAR) // invalid year

    validDate = false;

else if (month < MIN_MONTH || month > MAX_MONTH) // invalid month

    validDate = false;

else if (day < MIN_DAY || day > MAX_DAY) // invalid day

    validDate = false;

if(validDate == true)

{

    // Output statement

    cout<<month<<"/"<<day<<"/"<<year<<"is a valid date"<<endl;

}

else

{

    // Output statement

    cout<<month<<"/"<<day<<"/"<<year<<" is an invalid date"<<endl;

}

}

Explanation:

To fix the 'else' without a previous 'if' error in your code, you need to ensure that you have an 'if' statement before the 'else' keyword. Update your code to include conditional statements to check the validity of the entered date. Once you make the necessary changes, recompile and run your code to see the desired output.

The error you are encountering, 'else' without a previous 'if', typically occurs when there is a missing or unmatched 'if' statement before the 'else' keyword in your code. In the given instructions and code, you need to add conditional statements to check the validity of the entered date. Here is an example of how you can modify your code:

This modified code checks if the entered year, month, and day are within the valid ranges. If the date is valid, it prints a message saying so; otherwise, it prints a message indicating an invalid date. Make sure to recompile and run the updated code to see the desired output. This should resolve the error you were encountering.

Answer and explanation:

When defining a rectangle class that implements rectangleinterface and we implement the comparable interface such that rectangle objects can be compared by their perimeter, our class signature should be the following:

Public class, rectangle implements, rectangleInterface, comparable

This Python program calculates compound interest based on user-input principal, annual interest rate, compounding frequency, and the number of years, providing the resulting amount in the account.

Here's a simple Python program that takes user input for principal amount, annual interest rate, compounding frequency, and the number of years, and then calculates and displays the amount of money in the account after the specified number of years:

python

def compound_interest():

   # Get user input

   principal = float(input("Enter the principal amount: "))

   interest_rate = float(input("Enter the annual interest rate (%): "))

   compounding_frequency = int(input("Enter the compounding frequency per year: "))

   years = int(input("Enter the number of years: "))

   # Convert interest rate to decimal

   interest_rate_decimal = interest_rate / 100

   # Calculate compound interest

   amount = principal * (1 + interest_rate_decimal / compounding_frequency) ** (compounding_frequency * years)

   # Display the result

   print(f"The amount in the account after {years} years will be: ${round(amount, 2)}")

# Call the function

compound_interest()

This program uses the compound interest formula you provided and prompts the user for the necessary input to perform the calculation. It then displays the calculated amount after the specified number of years.

Answer:

World Wide Web

Explanation:

The world wide web shortened as www is a convenient way of refering to the internet. The Internet is a massive (globally) inter connection of computers which are communicating with each other and sharing information. The architecture of the word wide web (i.e the internet) is analogous to the client sever computer network architecture, because in a client/server network a powerful computer is designated as server and several work stations (clients)  are connected to this server, the clients request and retreive information stored in the servers. This is comparable to the www Whereby information stored in multiple server locations globally are requested and retrieved using the uniform resource locator (URL).

Answer:

The missing code for the above problem is as follows:

Explanation:

Missing code :

    cin>>firstNum>>secondNum; // for the first case.

    if(firstNum==0||secondNum==0) // for the second case.

    cout<<"Both the inputed numbers must be nonzero."; // for the second case.

    else if (firstNum>secondNum) // for the third case

    cout <<firstNum/secondNum; // for the third case.

    else if(secondNum>firstNum) // for the fourth case.

     cout <<secondNum/firstNum; // for the fourth case.

     else // for the fourth case.

     cout<<firstNum*secondNum; // for the forth case.

Output:

Code Explanation:

Answer:

B. Answer == "No"

Explanation:

The test condition from the code segment is If Answer == "No". The if-statement checks to see if the user input (Answer) equals "No", if it is then the code display "Try Again" and request the user to enter another input (Answer).

The general pattern for writing an if-statement is:

if (condition) then expression

The condition represent the test condition.

Answer:

The code is given below with appropriate comments for better understanding

Explanation:

#include <bits/stdc++.h>

using namespace std;

enum Digits {

zero,

one

} ;

struct CellType

{

  Digits bit;

  CellType* next;

};

class Minimum

{

public:

  struct CellType* head,*temp; // head to point MSB

  Minimum(string s)

  {

      int sz = s.size(); // binary size as per question it indicates n.

      for(int i=0;i<sz;i++)

      {

          if(s[i] == '0') // if the bit is zero , we add zero at the end of stream

              addAtEnd(zero);

          else // if the bit is one , we one zero at the end of stream

              addAtEnd(one);

      }

      addOne();

      showlist();

  }

  CellType* create(Digits x) // to create a node of CellType*

  {

      CellType* t = new CellType();

      t->bit = x;

      t->next = NULL;

      return t;

  }

  void addAtEnd(Digits x)

  {

      if(head == NULL) // if list is empty , then that will be the only node

      {

          CellType* t;

          t = create(x);

          head = temp = t;

      }

      else

      { // other wise we add the node at end indicated by the temp variable

          CellType* t ;

          t = create(x);

          temp->next = t;

          temp=temp->next;

      }

  }

  void showlist()

  {

      // this is just a normla method to show the list

      CellType* t = head;

      while(t!=NULL)

      {

          cout<<t->bit;

          t=t->next;

      }

  }

  void addOne()

  {

      /*

          here since we need to add from the end and it is a singly linked list we take a stack

          and store in last in ,first out format .

          Then we keep on changing all ones to zeroes until we find a zero int he list,

          The moment a zero is found it should be changed to one.

          If everything is one in the sequence , then we add a new zero digit node at the beginning of the list.

      */

      stack<CellType*> st;

      CellType* t = head;

      while(t!=NULL)

      {

          st.push(t);

          t=t->next;

      }

      while(st.size()>0 && (st.top())->bit == one )

      {

          CellType* f = st.top();

          f->bit = zero ;

          st.pop();

      }

      if(st.size())

      {

          CellType* f = st.top();

          f->bit = one ;

      }

      else

      {

          t = create(one);

          t->next = head;

          head = t;

      }

  }

};

int main()

{

  /*

      Here i am taking an integer as input and then converting it to binary using a string varaible s

      if you want to directly take the binary stream as input , remove the comment from "cin>>s" line.

  */

  long long int n,k;

  cin>>n;

  string s;

  k = n;

  while(k>0)

  {

      s = s + (char)(k%2 +48);

      k=k/2;

  }

  reverse(s.begin(),s.end());

  //cin>>s;

  Minimum* g = new Minimum(s);

  return 0;

}

Answer:

Option 1 is the correct answer for the above question

Explanation:

In the provided code of 'SomeClass', the variable 'VALUE1' is a class constant because it is the only one declared as public, static, and final, indicating a constant value that cannot be changed.

In Java, a class constant is typically defined with the keywords public, static, and final. A class constant is a variable with a constant value that cannot be changed. In the given class SomeClass, the variable VALUE1 is declared with these keywords: public static final. Therefore, VALUE1 is a class constant. It is important to understand that class constants are useful for defining values that should not change throughout the execution of a program. Since it is declared with the final keyword, it must be assigned a value only once, and it cannot be modified afterwards.

Answer:

2UEHDGXHXNX

Explanation:

SSUDUDUDIDJE

Answer:

Rescue disk

Explanation:

The computer system is an electronic device with a circuitry of interconnected hardware devices like processors, input devices, output devices and storage devices. It's hardware is controlled by a kernel software provided by a system software called the operating system.

The operating system is stored in the storage device (HDD or SSD) and is loaded to run on the system during the boot process.

A common practice is to create a restore point in a partition of the primary storage or in an external storage disk, when a software from an untrusted vendor is to be installed. This partition or storage is called a rescue disk.

A rescue disk is used to recover the normal operation of the computer system software by re-installing the operating system restore point.

Answer:

voice recognition

Explanation:

Voice recognition is also known as speech recognition. It is a computer software programs or in some cases a hardware devices used as a biometric security measure for decoding human voice and translating it into digital patterns. One popular use of voice recognition is for operating a device, perform commands and other functions. Voice recognition are recorded and examined for tone and pitch.

Answer:

A. the ability to quickly restore service following a service failure and provide compensation.

Explanation:

Systems like servers are bound to fail at some point. This is because, it either has running parts or electronic components that could overheat or get shorted due to external or internal malfunction.

The ability of system to quickly restore services and operation after it experiences a total system failure is called system recovery.

Answer:

Following are the program in Python langauge

person_name = input() # Read the person name by the user

person_age=0  #declared a vaiable person_age  

person_age = int(input()) # read person_age by the user

person_age=person_age+5  # add 5 to person_age

print('In 5 years',person_name,'will be',person_age) # display the output

Output:

  Amy

   4

   In 5 years Amy will be 9

Explanation :

Following is the description of code:

Answer:

The most capable version of Microsoft hypervisor you can install on this machine, provided the computer meets all the other requirements is

(b) Client Hyper-V.

Explanation:

A hypervisor is a software that allows the running of one or more virtual machines on a host machine. Client hyper-v is a type 1 example of a hypervisor for Microsoft 8 and Microsoft 10 and can be run on a computer's hardware.

Client hyper-v uses only hardware assisted virtualization like AMD, AMD-V which allows virtual machines to perform well.

Answer:

Client Hyper-V

Explanation:

Virtualization is a concept in computer technology where a computer is configured to run multiple operating system known as guest operating system on a host operating system.

The hyper-v manager is a software tool used by the host operating system to allow and manager virtual machines that run these operating system. The client Hyper-V is a type-1 Hyper-v or hypervisor that allows multiple operating system run inside a virtual machine.

AMD-V or advanced micro Dynamics virtualization technology is a extension is AMD processor that repeats tasks done by softwares, increasing the resource use and virtual machine performance.

Answer:

SELECT

NOW() AS 'today_unformatted',

DATE_FORMAT(NOW(), '%d-%b-%Y') AS 'today_formatted';

Explanation:

%d represents date.

%b represents month.

%Y represents year.

To ensure a loop executes as long as the user inputs a non-negative number, you can utilize a while loop in your code. The loop should continue if the condition is true, where the input number is greater than or equal to zero. Here's a sample code snippet:

number = int(input('Enter a number: '))

while number >= 0:

   print('Body')

   number = int(input('Enter a number: '))

print('Done')

This loop begins by prompting the user to enter a number. The while condition checks if the entered number is non-negative (i.e., greater than or equal to zero). If the condition is met, the loop 'Body' is executed, followed by another prompt for the user to input a number. This process repeats until the user inputs a negative number, which causes the while loop to end, and the program prints 'Done' after exiting the loop. Using a break statement isn't necessary here because the loop condition naturally terminates when a negative number is entered.

Answer:

The given question is incomplete as it does not contains the data or calculation. The complete question is given below in the attached image. The solution to the question is also attached and all the steps are explained in it.

I hope it will help you!

Answer:

Following are the code in the Java Programming Language.

//print first element of the array

System.out.println(runTimes[0]);

//Print second element of the array

System.out.println(runTimes[1]);

//print third element of the array

System.out.println(runTimes[2]);

Explanation:

Following are the description of the Program:

In the above program, There is the integer type array variable that is 'runTimes' and its index value is 5. Then, initialize values in the array one by one which means firstly, they insert in index 0, then in index 1, and so on. Finally, we have to print the first three elements of the array. So, we set the System.out.println() function three times to print array by passing array name and first three index value with it.

Print statements are used to display statements and expressions in a program.

The three print statements are:

The above statements are straightforward; however, we need to take not of the following highlights.

The question says, the print statement should print a new line, after printing each array element.

This can be achieved using System.out.println() or System.out.print('\n').

In this program, we used the former.

Also, array elements start its indices from 0.

This means that, the first element of the array is at index 0, the second is at index 1, and so on.

Read more about similar programs at:

https://brainly.com/question/7182610

Answer: Leveraged creativity

Explanation:

Improving functionality and performance of any technology through new and unique methods so that it can lift above the competing organization and technologies. Experts and guides try innovating the technology through various attempts which is caused by maximizing advantage through creativity.

Answer:

import java.util.Scanner;

public class Sales

{

public static void main(String[] args) {

    int sum;

    Scanner input = new Scanner(System.in);

    System.out.print("Enter number of salespeople: ");

    int salespeople = input.nextInt();

    int[] sales = new int[salespeople];

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

    {

     System.out.print("Enter sales for salesperson " + (i+1) + ": ");

     sales[i] = input.nextInt();

    }

    System.out.println("\nSalesperson   Sales");

    System.out.println("--------------------");

    sum = 0;

       int maxSale = sales[0];

       int minSale  = sales[0];

       int minId=1;

       int maxId=1;

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

    {

               System.out.println("     " + (i+1) + "         " + sales[i]);

 sum += sales[i];

  if(maxSale < sales[i])

           {

               maxSale = sales[i];

               maxId = i+1;

           }

           if(minSale > sales[i])

           {

               minSale = sales[i];

               minId = i+1;

           }

   }

System.out.println("Total sales: " + sum);

       System.out.println("The average sale is: " + sum / salespeople );

       System.out.println("Salesperson "+ maxId + " had the highest sale with "+ maxSale + ".");

       System.out.println("Salesperson "+ minId + " had the lowest sale with "+ minSale + ".");

       int amount = 0;

       int counter = 0;

       System.out.print("Enter an amount: ");

       amount = input.nextInt();

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

    {

     if(sales[i] > amount) {

         counter++;

         System.out.println("Salesperson "+ (i+1) + " exceeded given amount. (Number of sales: " + sales[i] +")");

     }

    }

       System.out.println("Number of salespeople exceeded given amount is: "+ counter);

}

}

Explanation:

- Ask the user for the number of salesperson

- According to entered number, create an array called sales to hold sale number for each person

- In the first for loop, assign the given sales numbers

-  In the second for loop, print all the sales people with sales, calculate the total sales. Also, find the minimum and maximum sales numbers with associated individuals.

- Print total, average, highest, and lowest sales.

- Ask user enter an amount

- In the third for loop, find the sales people who exceeded this amount.

- Print number of sales people who exceeded the amount.

This detailed explanation modifies Sales.java to include computing and printing the average, maximum, and minimum sales, and then prompts the user for a threshold to find salespeople exceeding it.

To modify the Sales.java program as per the requirements, start by reading the sales data and modify it according to the task.

Here is an example implementation:

import java.util.Scanner;

public class Sales {

   public static void main(String[] args) {

       Scanner scan = new Scanner(System.in);

       System.out.print("Enter the number of salespeople: ");

       int numSalespeople = scan.nextInt();

       int[] sales = new int[numSalespeople];

       int sum = 0, maxSale = Integer.MIN_VALUE, minSale = Integer.MAX_VALUE;

       int maxId = -1, minId = -1;

       for (int i = 0; i < numofSalespeople; i++) {

           System.out.print("Put sales for the salesperson " + (i + 1) + ": ");

           sales[i] = scan.nextInt();

           sum += sales[i];

           if (sales[i] > maxSale) {

               maxSale = sales[i];

               maxId = i + 1;

           }

           if (sales[i] < minSale) {

               minSale = sales[i];

               minId = i + 1;

           }

       }

       double average = (double) sum / numofSalespeople;

       System.out.println("\nSalesperson" + " Sales");

       System.out.println("------------------");

       for (int i = 0; i < numofSalespeople; i++) {

           System.out.println(" " + (i + 1) + " " + sales[i]);

       }

       System.out.println("\nTotal sales: " + sum);

       System.out.println("Average sales: " + average);

       System.out.println("Salesperson " + maxId + " had the highest sale with $" + maxSale);

       System.out.println("Salesperson " + minId + " had the lowest sale with $" + minSale);

       System.out.print("\nEnter a sales amount to compare: ");

       int threshold = scan.nextInt();

       int count = 0;

       for (int i = 0; i < numofSalespeople; i++) {

           if (sales[i] > threshold) {

               System.out.println("Salesperson " + (i + 1) + " exceeded with sales of $" + sales[i]);

               count++;

           }

       }

       System.out.println("Total salespeople who surpass the amount: " + count);

   }

}