Web-Based Game that serves multiple platforms
Recommendations
Analyze the characteristics of and techniques specific to various systems architectures and make a recommendation to The Gaming Room. Specifically, address the following:
1. Operating Platform: Recommend an appropriate operating platform that will allow The Gaming Room to expand Draw It or Lose It to other computing environments.>
2. Operating Systems Architectures:
3. Storage Management:
4. Memory Management:
5. Distributed Systems and Networks:
6. Security: Security is a must-have for the client. Explain how to protect user information on and between various platforms. Consider the user protection and security capabilities of the recommended operating platform.>

Answer:

A triangle on the Clock input.

Answer:

PMT function. PMT, one of the financial functions, calculates the payment for a loan based on constant payments and a constant interest rate. Use the Excel Formula Coach to figure out a monthly loan payment.

Explanation:

It’s a click-to-call ad

Answer:

Vulnerability assessment

Explanation:

Cyber security can be defined as preventive practice of protecting computers, software programs, electronic devices, networks, servers and data from potential theft, attack, damage, or unauthorized access by using a body of technology, frameworks, processes and network engineers.

Some examples of cyber attacks are phishing, zero-day exploits, denial of service, man in the middle, cryptojacking, malware, SQL injection, spoofing etc.

A vulnerability assessment is a systematic and methodical evaluation used by network security experts to determine the exposure of network assets such as routers, switches, computer systems, etc., to attackers, forces of nature, or any other entity that is capable of causing harm i.e potential harms. Thus, vulnerability assessment simply analyzes and evaluate the network assets of an individual or business entity in order to gather information on how exposed these assets are to hackers, potential attackers, or other entities that poses a threat.

Answer: Web services

Explanation:

Web service refers to a piece of software which uses a XML messaging system that's standardized and makes itself available over the internet.

Web services provides standardized web protocols which are vital for communication and exchange of data messaging throughout the internet.

They are are designed according to a set of constraints which help in shaping the service architecture in order to emulate the properties of the World Wide Web.

Answer:

The correct approach is "REST Services".

Explanation:

It merely offers accessibility to commodities as well as authorization to REST customers but also alters capabilities or sources.

Answer:

The three basic types of control structures are sequential, selection and iteration. They can be combined in any way to solve a specified problem. Sequential is the default control structure, statements are executed line by line in the order in which they appear.

Answer:

PLS programming language supporting.

Explanation:

When she manages a software development project, Candace uses a program called PLS programming language supporting.

Software development project is the project that is undertaken by the team of experts that developed through different format of coding language. The software development has specific time, budget, and resources.

Thus, it is PLS programming language

For more details about Software development project, click here:

https://brainly.com/question/14228553

#SPJ2

Answer:

The program in C++ is as follows:

#include <iostream>

#include <iomanip>

using namespace std;

int* sortArray(int temp [], int n){

   int tmp;

   for(int i = 0;i < n-1;i++){

 for (int j = i + 1;j < n;j++){

  if (temp[i] > temp[j]){

   tmp  = temp[i];

   temp[i] = temp[j];

   temp[j] = tmp;   }  } }

return temp;

}

float average(int temp [], int n){

   float sum = 0;

   for(int i = 0; i<n;i++){

       sum+=temp[i];    }

   return sum/n;

}

int main(){

   int n;

   cout<<"Number of temperatures: ";    cin>>n;

   while(n>10){

       cout<<"Number of temperatures: ";    cin>>n;    }

   int temp[n];

   for(int i = 0; i<n;i++){

       cout<<"Input temperature "<<i+1<<": ";

       cin>>temp[i];    }

   int *sorted = sortArray(temp, n);

   cout<<"The sorted temperature in ascending order: ";

   for( int i = 0; i < n; i++ ) {  cout << *(sorted + i) << " ";   }

   cout<<endl;

   float ave = average(temp,n);

   cout<<"Average Temperature: "<<setprecision(2)<<ave<<endl;

   cout<<"Highest Temperature: "<<*(sorted + n - 1)<<endl;

   cout<<"Lowest Temperature: "<<*(sorted + 0)<<endl;

   return 0;

}

Explanation:

See attachment for complete source file with explanation

Answer:

The Operating System allocates resources when a program need them. When the program terminates, the resources are de-allocated, and allocated to other programs that need them

Answer:

This can be done by a greedy solution. The algorithm does the following:  

Put song 1 on CD1.

For song 1, if there's space left on the present CD, then put the song on the present CD. If not, use a replacement CD.

If there are not any CDs left, output "no solution".

Explanation:  

The main thing is prove the correctness, do that by the "greedy stays ahead argument". For the primary song, the greedy solution is perfect trivially.  

Now, let the optimal solution match the greedy solution upto song i. Then, if the present CD has space and optimal solution puts song (i+1) on an equivalent CD, then the greedy and optimal match, hence greedy is not any worse than the optimal.Else, optimal puts song (i + 1) on subsequent CD. Consider an answer during which only song (i + 1) is moved to the present CD and zip else is modified. Clearly this is often another valid solution and no worse than the optimal, hence greedy is not any worse than the optimal solution during this case either. This proves the correctness of the algorithm. As for every song, there are constantly many operations to try to do, the complexity is O(n).

Solution :

ENTITY 1               CARDINALITY RATIO                                     ENTITY 2

1. Student                        1 to many                                     Social security card

                      A student may have more than one          

                     social security card (legally with the

                      same unique social security number),

                     and every social security number belongs

                   to a unique.

2. Student                   Many to Many                                    Teacher

                    Generally students are taught by many

                    teachers and a teacher teaches many students.

3.ClassRoom               Many to Many                                   Wall

                      Do not forget that the wall is usually

                      shared by adjacent rooms.

4. Country                 1 to 1                                             Current President

                       Assuming a normal country under normal

                       circumstances having one president at a

                      time.

5. Course               Many to Many                               TextBook

                        A course may have many textbooks and

                       text book may be prescribed for different

                      courses.

6. Item                             Many to Many                           Order

                        Assuming the same item can appear

                         in different orders.

7. Student                       Many to Many                          Class

                        One student may take several classes.

                        Every class usually has several students.

8. Class                                 Many to 1                             Instructor

                       Assuming that every class has a unique

                        instructor. In case instructors were allowed

                        to team teach, this will be many-many.

9. Instructor                  1 to 1                                             Office

                       Assuming every instructor has only one

                      office and it is not shared. In case of offices

                     shared by 2 instructors, the relationship

                    will be 2-1. Conversely, if any instructor has a joint

                    appointment and offices in both departments,

                    then the relationship will be 1-2. In a very general

                  case, it may be many-many.

10. E-bay Auction item             1-Many                           E-bay bid

                        1 item has many bids and a bid is unique

                        to an item.

Answer:

see the code snippet below writing in Kotlin Language

Explanation:

fun main(args: Array<String>) {

   sumOfNumbers()

}

fun sumOfNumbers(): Int{

   var firstNum:Int

   var secondNum:Int

   println("Enter the value of first +ve Number")

   firstNum= Integer.valueOf(readLine())

   println("Enter the value of second +ve Number")

   secondNum= Integer.valueOf(readLine())

   var sum:Int= firstNum+secondNum

  println("The sum of $firstNum and $secondNum is $sum")

   return sum

}

Answer:

Hence the answer is top and peek.

Explanation:

In a stack, we insert and delete an item from the top of the stack. So we use variables top, peek to stay track of the position.

The insertion operation is understood as push and deletion operation is understood as entering stack.

Hence the answer is top and peek.

Answer:

line of code :

newString=myString+myString.charAt(0)+myString.charAt(myString.length()-1);

Explanation:

In Java, the + is used for concatenation

method

charAt() will return the character from the given index from string

myString.charAt(0) will return E

myString.charAt(myString.length()-1)

Answer:

Encapsulation.

Explanation:

Information hiding involves using private fields within classes and it's a feature found in all object-oriented languages such as Java, Python, C#, Ruby etc.

In object-oriented programming language, a process known as encapsulation is used for the restrictions of the internal data of a software program from the outside code, therefore preventing an unauthorized direct access to the codes. This is achieved through the use of classes.

In Computer programming, class members can be defined as the members of a class that typically represents or indicates the behavior and data contained in a class.

Basically, the members of a class are declared in a class, as well as all classes in its inheritance hierarchy except for destructors and constructors.

In a class, member variables are mainly known as its attributes while its member function are seldomly referred to as its methods or behaviors.

One of the main benefits and importance of using classes is that classes helps to protect and safely guard their member variables and methods by controlling access from other objects.

Therefore, the class members which should be declared as public are methods and occasionally final attributes because a public access modifier can be accessed from anywhere such as within the current or external assembly, as there are no restrictions on its access.

Answer:

Here the code is given in java as follows,

Answer:

Explanation:

#include <stdarg.h>  

main()  

{  

display("Hello", 4, 12, 13, 14, 44);  

}  

display(char *s,...)  

{  

va_list ptr;  

va_start(ptr, s);  

show(s,ptr);  

}  

show(char *t, va_list ptr1)  

{  

int a, n, i;  

a=va_arg(ptr1, int);  

for(i=0; i<a; i++)  

 {  

n=va_arg(ptr1, int);  

printf("\n%d", n);  

}  

}

Answer:

LAB means laboratory

IEEE means institute of electrical and electronic engineer

Answer:

10

Explanation:

An enqueue operation is a function that adds an element(value) to a queue array. A dequeue operations removes an element from a queue array. Queue arrays follow a first-in-first-out approach, so elements that are first stored in the queue are removed/accessed first(enqueue operations add elements at the rear of the queue array).

The following operations leave 10 elements in the queue of array size 12 after its done:

10 enqueue operations= adds 10 elements

5 dequeue operations= removes 5 elements( 5 elements left in queue)

6 enqueue operations= adds 6 elements(11 elements in queue)

10 dequeue operations= removes 10 elements(1 element left in queue)

8 enqueue operations= adds 8 elements(9 elements in queue)

2 dequeue operations= removes 2 elements(7 elements left in queue)

3 enqueue operations= adds 3 elements(10 elements in queue)

Therefore there are 10 elements in the queue after enqueue and dequeue operations.

Answer:

3) If number is zero, number is displayed;

Explanation:

The code snippet created is supposed to take any input number from the user (positive or negative) and print it to the console if it is less than 0. In this code the IF statement does not need an else clause and will work regardless. The System.out.println() statement does not need to be inside braces since it is a simple one line statement. Therefore, the only statement in the question that is actually true would be ...

3) If number is zero, number is displayed;

Answer:

If number is zero, number is displayed

Explanation:

Given

The above code segment

Required

What is true about the code segment

Analyzing the options, we have:

(1) Wrong statement because there is no else clause

The above statement is not a requirement for an if statement to be valid; i.e. an if statement can stand alone in a program

Hence, (1) is false

(2) The print statement must be in { }

There is only one statement (i.e. the print statement) after the if clause. Since the statement is just one, then the print statement does not have to be in {} to be valid.

Hence, (2) is false

(3) number is displayed, if input is 0

In the analysis of the program, we have: number <=0

i.e. the if statement is true for only inputs less than 0 or 0.

Hence, number will be printed and (3) is false

(4) number is displayed for positive inputs

The valid inputs have been explained in (3) above;

Hence, (4) is false.

(5) is also false

Answer:

Explanation:

The code that was provided in the question contained a couple of bugs. These bugs were fixed and the new code can be seen below as well as with the solution for the number of one-dollar bills included. The program was tested and the output can be seen in the attached image below highlighted in red.

amount_to_change = int(input())

num_fives = amount_to_change // 5

#Your solution goes here

num_ones = amount_to_change % 5

print('Change for $' + str(amount_to_change))

print(num_fives, 'five dollar bill(s) and', num_ones, 'one dollar bill (s)')

Answer:

Explanation:

The code provided worked for the most part, it had some gramatical errors in when printing out the information as well as repeated values. I fixed and reorganized the printed information so that it is well formatted. The only thing that was missing from the code was the input validation to make sure that no negative values were passed. I added this within the getMonths() method so that it repeats the question until the user inputs a valid value. The program was tested and the output can be seen in the attached image below.

import java.io.*;

import java.util.*;

class Rainfall

{

   Scanner in = new Scanner(System.in);

   private int month = 12;

   private double total = 0;

   private double average;

   private double standard_deviation;

   private double largest;

   private double smallest;

   private double months[];

   public Rainfall()

   {

       months = new double[12];

   }

   public void setMonths()

   {

       for(int n=1; n <= month; n++)

       {

           int answer = 0;

           while (true) {

               System.out.println("Enter the rainfall (in inches) for month #" + n + ":" );

               answer = in.nextInt();

               if (answer > 0) {

                   months[n-1] = answer;

                   break;

               }

           }

       }

   }

   public void getTotal()

   {

       total = 0;

       for(int i = 0; i < 12; i++)

       {

           total = total + months[i];

       }

       System.out.println("The total rainfall for the year is " + total);

   }

   public void getAverage()

   {

       average = total/12;

       System.out.println("The average monthly rainfall is " + average);

   }

   public void getLargest()

   {

       double largest = 0;

       int largeind = 0;

       for(int i = 0; i < 12; i++)

       {

           if (months[i] > largest)

           {

               largest = months[i];

               largeind = i;

           }

       }

       System.out.println("The largest amount of rainfall was " + largest +

               " inches in month " + (largeind + 1));

   }

   public void getSmallest()

   {

       double smallest = Double.MAX_VALUE;

       int smallind = 0;

       for(int n = 0; n < month; n++)

       {

           if (months[n] < smallest)

           {

               smallest = months[n];

               smallind = n;

           }

       }

       System.out.println("The smallest amount of rainfall was " + smallest +

               " inches in month " + (smallind + 1));

   }

   public static void main(String[] args)

   {

       Rainfall r = new Rainfall();

       r.setMonths();

       r.getTotal();

       r.getSmallest();

       r.getLargest();

       r.getAverage();

   }

}

Answer:

The program is as follows:

[tex]mult\_table = [[1, 2, 3], [2, 4, 6], [3, 6, 9]][/tex]

count = 1

for row in mult_table:

   for item in row:

       if count < len(row):

           count+=1

           print(item,end="|")

       else:

           print(item)

   count = 1

Explanation:

This initializes the 2D list

[tex]mult\_table = [[1, 2, 3], [2, 4, 6], [3, 6, 9]][/tex]

This initializes the count of print-outs to 1

count = 1

This iterates through each rows

for row in mult_table:

This iterates through each element of the rows

   for item in row:

If the number of print-outs is less than the number of rows

       if count < len(row):

This prints the row element followed by |; then count is incremented by 1

           count+=1

           print(item,end="|")

If otherwise

       else:

This prints the row element only

           print(item)

This resets count to 1, after printing each row

   count = 1

To delete the whitespace, you have to print each element as:            print(item,end="|")

See that there is no space after "/"

Answer:

count = 1

for row in mult_table:

  for item in row:

      if count < len(row):

          count+=1

          print(item,end=" | ")

      else:

          print(item)

  count = 1

Explanation: Other answers didnt have the spacing correct

Answer:

Explanation:

The following code is written in Python. It is a function that takes two string parameters and compares the lengths of both strings. If the string in variable s1 is greater then the function returns True, otherwise the function returns False. A test case has been provided and the output can be seen in the attached image below.

def compareStrings(s1, s2):

   if len(s1) > len(s2):

       return True

   else:

       return False

print(compareStrings("Brainly", "Competition"))

Answer:

Known as software development life cycle, these steps include planning, analysis, design, development & implementation, testing and maintenance.

Answer:

E). User education and training.

Explanation:

In the context of network security, the most significant aspect of ensuring security from workstation cyberattacks would be 'education, as well as, training of the users.' If the employee users are educated well regarding the probable threats and attacks along with the necessary safety measures to be adopted and trained adequately to use the system appropriately so that the sensitive information cannot be leaked while working on the workstation and no networks could be compromised. Thus, option E is the correct answer.

Answer:

Hybrid Computer has features of both analogue and digital computer. It is fast like analogue computer and has memory and accuracy like digital computers. It can process both continuous and discrete data. So it is widely used in specialized applications where both analogue and digital data is processed.

Answer: