Write a program that allows the user to play a guessing game. The game will choose a "secret number", a positive integer less than 10000. The user has 10 tries to guess the number.

Answer:

Here is the Python program:

import random  #to generate random numbers

low = 1  #lowest range of a guess

high = 100  #highest range of a guess

attempts=0   # no of attempts

def guess_number(low, high, attempts):  #function for guessing a number

   print("Guess a number between 1 and 100")      

#prompts user to enter an integer between 1 to 100 as a guess

   number = random.randint(low, high)  

#return randoms digits from low to high

   for _ in range(attempts):  #loop to make guesses

       guess = input("Enter a number: ")   #prompts user to enter a guess

       num = int(guess)  #reads the input guess

       if num == number:  # if num is equal to the guess number

           print('Congratulations! You guessed the right number!')    

           command= input("Do you want to play another game? ")

#asks user if he wants to play another game

           if command=='Y':  #if user enters Y  

               guess_number(low,high,attempts)  #starts game again

           elif command=='N':  #if user types N

               exit() #exits the program

       elif num < number:   #if user entered a no less than random no

           print('Too low, Try Higher')  # program prompts user to enter higher no

       elif num > number: #if user entered a no greater than random no

           print('Too high, Try Lower') # program prompts user to enter lower no

       else:  #if user enters anything else

           print("You must enter a valid integer.")  

#if user guessed wrong number till the given attempts

   print("You failed to guess the number correctly in {attempts} attempts.".format(attempts=attempts))  

   exit() # program exits

Explanation:

The program has a function guess_number which takes three parameters, high and low are the range set for the user to enter the number in this range and attempts is the number of tries given to the user to guess the number.

random.randint() generates integers in a given range, here the range is specified i.e from 1 to 100. So it generates random number from this range.

For loop keeps asking the user to make attempts in guessing a number and the number of tries the user is given is specified in the attempts variable.

If the number entered by the user is equal to the random number then this message is displayed: Congratulations! You guessed the right number! Then the user is asked if he wants to play one more time and if he types Y the game starts again but if he types N then the exit() function exits the game.

If the number entered as guess is higher than the random number then this message is displayed Too high, Try Lower and if the number entered as guess is lower than the random number then this message is displayed Too low, Try Higher.

If the user fails to guess the number in given attempts then the following message is displayed and program ends. You failed to guess the number correctly in {attempts} attempts.

You can call this function in main by passing value to this function to see the output on the screen.

guess_number(1, 100, 5)

The output is shown in screenshot attached.

Answer:

Explanation:

AVL trees are used frequently for quick searching as searching takes O(Log n) because tree is balanced. Where as insertion and deletions are comparatively more tedious and slower as at every insertion and deletion, it requires re-balancing. Hence, AVL trees are preferred for application, which are search intensive.

The attached diagramm further ilustrate this

Answer:

A right grants the authority to perform an action on a system. A permission grants access to a resource.

Explanation:

In System administration there are different rights and permissions are given to the user as per the requirement to perform some actions and accessing some resources.

Rights are given to the users may be at admin level to perform some action or make some changes in the system. The users who have rights of the system can add, edit or delete data from the system as per requirement of the company.

Many peoples granted by permissions by providing them some passwords to access the resources of the systems. These users did not have the right to change, add, edit or delete any data. They just can use the information.

A right grants the authority to perform an action on a system. A permission grants access to a resource.

In the context of system security and access control:

- A right is a privilege granted to a user or process that allows them to perform a specific action on a system. For example, a right might allow a user to shut down the computer or to modify system settings.  

- A permission is a rule that grants access to a specific resource, such as a file, directory, or network resource. Permissions determine what actions (such as read, write, or execute) can be performed on that resource.

Thus, a right defines the actions a user can perform, while a permission defines what resources a user can access and what they can do with those resources.

Answer:

The answer is A. Cloud computing

Explanation:

Cloud computing is the process of using the internet to store, mange and process data rather than using a local server or a personal computer.

The term  cloud computing is generally used to describe data centers available to many users over the Internet.  

Therefore the company's approach to computing resources can be said to be cloud computing because it uses the internet  for data storage, data backup, and software development  

the function and loop will be

def factorial(x):

   total = 1

   if x != 1 and x != 0:

       for i in range(x,1,-1):

           total *= i

   return total

The program is an illustration of loops

Loops are program statements that are used to perform repeated operations

The program in python, where comments are used to explain each line is as follows:

#This gets input for N

N = int(input())

#This initializes factorial to 1

factorial = 1

#This opens the while loop

while N >1:

   #This calculates the factorial

   factorial*=N

   N-=1

#This prints the factorial

print(factorial)

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Answer:

Following are the variables which is visible in when the last module is called:

Explanation:

Missing information :

Flash and RAM are two types of memory used in microcontrollers like PIC, with differences such as Flash being non-volatile and RAM being volatile. Configuration words are usually stored in Flash memory, and special linear data memory can be used for larger structures. Even while accessing data RAM, the Bank Select Register should typically be set to the correct bank, but some common areas are accessible from all banks.

1. Flash and RAM are two types of memories used in microcontrollers like PIC, with significant differences. Flash Memory, being non-volatile, retains its information even when power is turned off, generally used for storing the program. On the other side, RAM (Random Access Memory) is volatile and loses its information when power is removed, typically used for temporary storage during program execution.

2+3. The specific number of RAM and Flash locations, as well as the bit size at each location, depends on the specific PIC microcontroller model. In general, one could read these details from the datasheet for the specific part.

4. Configuration words are usually associated with the Flash memory, as they are used to set up certain parameters for the chip that remain constant during operation.

5. Linear Data Memory is important for larger data structures such as arrays or buffers that require more than 80 bytes. Having this extra memory space allows more complex calculations and data manipulations.

6. Lastly, when accessing data RAM with direct addressing, the correct memory bank that includes our desired memory location should be selected into the BSR (Bank Select Register). However, there are ‘common’ areas in data RAM accessible from all banks, typically which include Special Function Registers (SFRs) and some general purpose registers.

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Answer:

C++.

#include <iostream>

using namespace std;

////////////////////////////////////////////////////////////////

int main() {

   int weekly_hours = 0;

   int hourly_rate;

   float gross_pay = 0;

   cout<<"Enter weekly hours worked: ";

   cin>>weekly_hours;

   cout<<"Enter hourly rate: ";

   cin>>hourly_rate;

   cout<<endl;

   ////////////////////////////////////////////////

   if (weekly_hours > 40) {

       gross_pay = (weekly_hours*hourly_rate) + ((weekly_hours*hourly_rate)*0.5);

   }

   else

       gross_pay = weekly_hours*hourly_rate;

   cout<<"Weekly gross pay: $"<<gross_pay;

   ////////////////////////////////////////////////

   return 0;

}

Answer:

The solution code is written in Python:

Explanation:

Based on the information given in the question, we presume the cost per 500 miles is $1.10 per pound ($2.20 / 2 pound).

Create a variable COST_PER_500MI to hold the value of cost per 500 miles (Line 1). Next, prompt user input the weight and assign it to variable weight (Line 3). Calculate the shipping charge and display it using print function (Line 4-5).

Final answer:

A program to calculate shipping charges can be created by converting the weight of a package to ounces and using the provided rate structure. For a 2-pound package, the cost to ship would be $6.88, following the given rate system where the first 3 ounces cost $1.95 and each additional ounce costs $0.17. The per 500-mile charges are not prorated.

Explanation:

The given scenario requires the creation of a simple program that calculates shipping charges based on the weight of a package and the distance it needs to be shipped. According to the details, there's a flat rate for the first 3 ounces, and then an additional, per ounce charge after that. If a 2-pound package is shipped, we can calculate the cost based on the conversion of pounds to ounces, and then apply the pricing rules provided. Since there are 16 ounces in one pound, a 2-pound package would be equivalent to 32 ounces. The cost would then be computed as follows:

Total Cost = Base cost for first 3 ounces + (Number of additional ounces x Cost per additional ounce)

For a 32 ounce package, this would be:

Total Cost = $1.95 + (29 x $0.17) = $1.95 + $4.93 = $6.88

There is no need to prorate the charge per 500 miles as the example provided indicates a flat charge is applied per 500-mile increment. Handling user input correctly in a program is also essential to ensure the right calculations are made based on what the user enters for the package weight and distance.

Answer:

#include<bits/stdc++.h>

using namespace std;

int main(){

  // Defining Variables

  int no_of_weeks;

  int total_cases = 0;

  //Declaring Vector of Pair of Integer and string

  std::vector<pair<int,string>> data;

  // Taking Input for the Number of Weeks

  cout<<"Enter No. of Weeks\n";

  cin >> no_of_weeks;

  // Running the Loop for no_of_weeks times

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

      int A,B,C;

      // Taking Input for different types of flus

      cout<<"Enter No. of Cases of Flu A, B, C for week" << i + 1 << " seperated by space : \n";

      cin >> A >> B >>C;

      // Adding all the cases in a week

      int cases_in_a_week = A + B + C;

      // Updating total cases

      total_cases += cases_in_a_week;

      // Declaring the level variable

      string level;

      // Updating the level of the week corresponding to each case

      if(cases_in_a_week < 500) level = "Low";

      else if(cases_in_a_week >= 500 && cases_in_a_week < 2000) level = "Moderate";

      else level = "Widespread";

      // Storing the Week's information by using a vector of pairs

      // in which pair's first is the number of cases which is of type int

      // while the second is the level of the flu which is of the type string

      data.push_back(make_pair(cases_in_a_week,level));

  }

  // Linking the stdoutput to the flu_report.txt file

  // this also creates the file with the same name if it doesn't exists

  freopen("flu_report.txt", "w", stdout);

  // Printing the respective output data with Bar Chart of stars for each level

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

      //printing the week no. and number of cases

      cout<<i+1<<" "<<data[i].first<<" "<<data[i].second<<" |";

      //calculating the number of stars

      int stars = data[i].first/250;

      //printing the stars of the bar chart

      for(int j = 0; j < stars ; j++) cout<<"*";

      cout<<endl;

  }

  //printing the total number of cases

  cout<<total_cases;

}

Explanation:

Answer:

C code explained below

Explanation:

I have provided the proper commented code below.

I hope that you find the answer helpful.

CODE:

-------------------------------------------------------------------------------------------------------------

#include<bits/stdc++.h>

using namespace std;

int main(){

  // Defining Variables

  int no_of_weeks;

  int total_cases = 0;

  //Declaring Vector of Pair of Integer and string

  std::vector<pair<int,string>> data;

  // Taking Input for the Number of Weeks

  cout<<"Enter No. of Weeks\n";

  cin >> no_of_weeks;

  // Running the Loop for no_of_weeks times

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

      int A,B,C;

      // Taking Input for different types of flus

      cout<<"Enter No. of Cases of Flu A, B, C for week" << i + 1 << " seperated by space : \n";

      cin >> A >> B >>C;

      // Adding all the cases in a week

      int cases_in_a_week = A + B + C;

      // Updating total cases

      total_cases += cases_in_a_week;

      // Declaring the level variable

      string level;

      // Updating the level of the week corresponding to each case

      if(cases_in_a_week < 500) level = "Low";

      else if(cases_in_a_week >= 500 && cases_in_a_week < 2000) level = "Moderate";

      else level = "Widespread";

      // Storing the Week's information by using a vector of pairs

      // in which pair's first is the number of cases which is of type int

      // while the second is the level of the flu which is of the type string

      data.push_back(make_pair(cases_in_a_week,level));

  }

  // Linking the stdoutput to the flu_report.txt file

  // this also creates the file with the same name if it doesn't exists

  freopen("flu_report.txt", "w", stdout);

  // Printing the respective output data with Bar Chart of stars for each level

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

      //printing the week no. and number of cases

      cout<<i+1<<" "<<data[i].first<<" "<<data[i].second<<" |";

      //calculating the number of stars

      int stars = data[i].first/250;

      //printing the stars of the bar chart

      for(int j = 0; j < stars ; j++) cout<<"*";

      cout<<endl;

  }

  //printing the total number of cases

  cout<<total_cases;

}

Answer and Explanation:

So, it's the right answer.

Answer:

The set is a object of the class 'H' and the setY method is a member of G class which is not called by the object of H class.

Explanation:

Answer:

The correct answer to the following question will be "B9:B18-ROUND(0.1*B9:B18,0) +C9:C18".

Explanation:

The ROUND function of Excel seems to be an adaptive feature in excel that calculates the first round amount of a specific number with either the numerical value or digits to be given as just a statement.

For rounding rolling quantities to a defined degree of accuracy you can use this function named ROUND.

The description of the above statement is as follow :

The formula will be used in that manner B9 along with colon B18 after that subtract (-) and using the round function (In this we pass the range i.e B9:B18,0) Plus C9 :C18.

Answer:see the picture attached

Explanation:

Answer:

#include<stdio.h>

int sumDigitCube(int n);

int main(){

   int n, i;

   printf("Enter number: ");

   scanf("%d", &n);

printf("The Armstrong numbers are:" );

   for(i=1; i<=n; i++){

       if(sumDigitCube(i)==i){

           printf(" %d", i);

       }

   }

   printf("\n");

   return 0;

}

int sumDigitCube(int n){

   int s = 0;

   int digit;

   while(n>0){

       digit = n%10;

       s += digit * digit *digit;

       n = n/10;

   }

   return s;

}

Explanation:

Answer:

The bandwidth is 3200 Kbps

Explanation:

Given:

The length of the point to point link = 50 km = 50×10³ = 50000 m

Speed of transmission = 2 × 10⁸ m/s

Therefore the propagation delay is the ratio of the length of the link to speed of transmission

Therefore,  [tex]t_{prop} = \frac{distance}{speed}[/tex]

[tex]t_{prop} = \frac{50000}{2*10^{8} }=2.5*10^{-4}[/tex]

Therefore the propagation delay in 25 ms

100 byte = (100 × 8) bits

Transmission delay = [tex]\frac{(100*8)bits }{(x)bits/sec}[/tex]

If propagation delay is equal to transmission delay;

[tex]\frac{(100*8)bits }{(x)bits/sec}= \frac{50000}{2*10^{8} }[/tex]

[tex]x=\frac{100*8*2*10^{8} }{50000}[/tex]

[tex]x=\frac{100*8}{2.5*10 ^-4}=3200000[/tex]

x = 3200 Kbps

Answer:

The complete question is here:

Consider a point-to-point link 2 km in length. At what bandwidth would propagation delay (at a speed of 2 x 108 m/s) equal transmission delay for

(a) 100-byte packets?

(b) 512-byte packets?

Explanation:

Given:

Length of the link = 50 km

Propagation delay = distance/speed = d/s

To find:

At what speed would propagation delay (at a speed 2 x 10^8 meters/second) equal transmit delay for 100 byte packets?

Solution:

Propagation Delay = t prop

                                 = 50 * 10^3 / 2 * 10^8

                                 = 50000 / 200000000

                                 = 0.00025

                                 = 25 ms

(a) When propagation delay is equal to transmission delay for 100 packets:

Transmission Delay = packet length / data rate = L / R

So when,

Transmission Delay = Propagation Delay

100 * 8  / bits/sec = 50 * 10^3  / 2 * 10^8

Let y denotes the data rate in bit / sec

speed = bandwidth = y bits/sec

         100 * 8  / y bits/sec = 50 * 10^3  / 2 * 10^8

                        y bit/sec   =  100 * 8  / 25 * 10^ -5

                                         = 800 / 0.00025

                                     y  = 3200000 bit/sec

                                      y = 3200 Kbps

(b) 512-byte packets

512 * 8  / y bits/sec = 50 * 10^3  / 2 * 10^8

                        y bit/sec   =  512 * 8  / 25 * 10^ -5

                                         = 4096 / 0.00025

                                     y  = 16384000 bit/sec

                                      y = 16384 Kbps

Answer:

Data encapsulation

Explanation:

When we send data to someone else across another network, this data travels down a stack of layers of the TCP/IP model. Each layer of the TCP/IP layer encapsulates data or hides the data by adding headers and sometimes trailers. This packaging is what is known as data encapsulation. The data starts at the application layer and trickles down beneath. As it moves down the layers, it is encoded or compressed to a standard format and sometimes encrypted for security purposes.

Answer:

Attached is the solution. Hope it helps:

a) The class of context-free languages is not closed under intersection, as shown by the example of[tex]A = [ a^m b^n c^n | m, n \geq 0 ][/tex] and [tex]B = [a^n b^n c^m | m, n \geq 0 ][/tex]

Given that;

We have seen in class that the sets of both regular and context-free languages are closed under the union, concatenation, and star operations.

(a) We're given two languages A and B:

[tex]A = [ a^m b^n c^n | m, n \geq 0 ][/tex]

[tex]B = [a^n b^n c^m | m, n \geq 0 ][/tex]

We want to show that the intersection of A and B is not context-free.

To do this, we can use the fact that the language [tex]C = [a^n b^n c^n | n \geq 0 ][/tex] is not context-free.

Assume for contradiction that the intersection of A and B, which we'll call D, is context-free.

Since context-free languages are closed under intersection, we'd have D = A ∩ B is also a context-free language.

Now, notice that D is a subset of C (if a string belongs to D, it must belong to C as well).

However, C is not context-free, as given.

This contradicts our assumption that D is context-free.

Therefore, we can conclude that the class of context-free languages is not closed under intersection, as shown by the example of[tex]A = [ a^m b^n c^n | m, n \geq 0 ][/tex] and [tex]B = [a^n b^n c^m | m, n \geq 0 ][/tex]

(b) Using the result from part (a), we can now show that the set of context-free languages is not closed under complementation.

Let's consider the complement of a context-free language.

If the complement of a context-free language were always context-free, we could take the complement of C and obtain a context-free language.

However, as established earlier, C is not context-free.

Hence, we can conclude that the set of context-free languages is not closed under complementation.

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Answer:

See attached file for detailed code.

Explanation:

See attached file for explanation.

Answer:

The program in Java will be:

// Java program to demonstrate working parseInt()  

public class GFG  

{  

   public static void main(String args[])  

   {  

       int decimalExample = Integer.parseInt("20");  

       int signedPositiveExample = Integer.parseInt("+20");  

       int signedNegativeExample = Integer.parseInt("-20");  

       int radixExample = Integer.parseInt("20",16);  

       int stringExample = Integer.parseInt("geeks",29);  

       // Uncomment the following code to check  

       // NumberFormatException  

       //   String invalidArguments = "";  

       //   int emptyString = Integer.parseInt(invalidArguments);  

       //   int outOfRangeOfInteger = Integer.parseInt("geeksforgeeks",29);  

       //   int domainOfNumberSystem = Integer.parseInt("geeks",28);  

       System.out.println(decimalExample);  

       System.out.println(signedPositiveExample);  

       System.out.println(signedNegativeExample);  

       System.out.println(radixExample);  

       System.out.println(stringExample);  

   }  

}

Answer:

canadian_capitals = {"Alberta":"Edmonton", "British Columbia":"Victoria", "Manitoba":"Winnipeg"}

mexica_capitals ={"Aguascalientes":"Aguascalientes", "Baja California":"Mexicali", "Baja California Sur":"La Paz"}

us_capitals ={"Alabama":"Montgomery", "Alaska":"Juneau", "Arizona":"Phoenix"}

nafta_capitals = {}

nafta_capitals.update(canadian_capitals)

nafta_capitals.update(mexica_capitals)

nafta_capitals.update(us_capitals)

print(nafta_capitals)

Explanation:

The code is written in python.

canadian_capitals = {"Alberta":"Edmonton", "British Columbia":"Victoria", "Manitoba":"Winnipeg"}   I created the first dictionary with the variable name called canadian_capital.  Notice the province of Canada is mapped to to their respective capitals.

mexica_capitals ={"Aguascalientes":"Aguascalientes", "Baja California":"Mexicali", "Baja California Sur":"La Paz"}   I created the second dictionary with the variable name called mexican_capital. The states are also mapped to their respective capitals.

us_capitals ={"Alabama":"Montgomery", "Alaska":"Juneau", "Arizona":"Phoenix"}   I created the third dictionary with the variable name called us_capital.  The states are also mapped to their respective capitals.

nafta_capitals = {}  This is an empty dictionary with the variable name nafta_capitals to combine the 3 dictionaries.

nafta_capitals.update(canadian_capitals)  I updated the empty dictionary with the canadian_capitals dictionary.

nafta_capitals.update(mexica_capitals)  I also added the mexica_capitals dictionary to the nafta_capitals dictionary.

nafta_capitals.update(us_capitals)  I also added the us_capitals dictionary to the nafta_capitals dictionary.

print(nafta_capitals)   The whole 3  dictionaries have been combined to form the nafta_capitals dictionary. The print function displays the already combine dictionaries.

Answer:

See attached file for detailed code.

Explanation:

See attached file.

Answer:

The answers are A)176 B)44 C)56.

Explanation:

According to the information given in the question about the architecture that is used and its communication times;

For option A: The goal is to retrieve one cache block which consists of 16 words. Doing this one word at a time over a period of 16 main memory accesses and the miss penalty for this configuration is 10*16 = 160 bus clock cycles for data transfer and 1*16 = 16 bus clock cycles for memory transfer time which comes up to 176.

For option B: The goal is to retrieve one cache block which consists of 16 words. Doing this four words at a time over a period of 4 main memory accesses and the miss penalty for this configuration is 10*4 = 40 bus clock cycles for data transfer and 1*4 = 4 bus clock cycles for memory transfer which comes up to 44.

For option C: The goal is to retrieve one cache block which consists of 16 words. Doing this 4 words at a time over a period of 4 main memory accesses and the miss penalty for this configuration is 10*4 = 40 bus clock cycles for data transfer and 1*16 = 16 bus clock cycles for memory transfer since words of the block are transferred not 4 but 1 at a time which comes up to 56.

I hope this answer helps.

Answer:

public class Rectangle {

   private double width;

   private double heigth;

   public Rectangle(double width, double heigth) {

       this.width = width;

       this.heigth = heigth;

   }

   public double getWidth() {

       return width;

   }

   public void setWidth(double width) {

       this.width = width;

   }

   public double getHeigth() {

       return heigth;

   }

   public void setHeigth(double heigth) {

       this.heigth = heigth;

   }

   public double perimeter(double width, double heigth){

       double peri = 2*(width+heigth);

       return peri;

   }

   public double area(double width, double heigth){

       double area = width*heigth;

       return  area;

   }

}

class RectangleTest{

   public static void main(String[] args) {

       //Creating two Rectangle objects

       Rectangle rectangle1 = new Rectangle(4,40);

       Rectangle rectangle2 = new Rectangle(3.5, 35.7);

       //Calling methods on the first Rectangel objects

       System.out.println("The Height of Rectangle 1 is: "+rectangle1.getHeigth());

       System.out.println("The Width of Rectangle 1 is: "+rectangle1.getWidth());

       System.out.println("The Perimeter of Rectangle 1 is: "+rectangle1.perimeter(4,40));

       System.out.println("The Area of Rectangle 1 is: "+rectangle1.area(4,40));

       // Second Rectangle object

       System.out.println("The Height of Rectangle 2 is: "+rectangle2.getHeigth());

       System.out.println("The Width of Rectangle 2 is: "+rectangle2.getWidth());

       System.out.println("The Perimeter of Rectangle 2 is: "+rectangle2.perimeter(4,40));

       System.out.println("The Area of Rectangle 2 is: "+rectangle2.area(4,40));

   }

}

Explanation:

Answer:

void printAttitude (int pint){

if (pint == 1){

System.out.println("disagree");

}

if (pint == 2){

System.out.println("no opinion");

}

if (pint == 3){

System.out.println("agree");

}

}

Explanation:

Answer:

public class num4 {

   public static void main(String[] args) {

       int para =2;

       printAttitude(para);

   }

   static void printAttitude( int para){

       if(para==1){

           System.out.println("disagree");

       }

       else if (para == 2){

           System.out.println("no option");

       }

       else if (para==3){

           System.out.println("agree");

       }

   }

}

Explanation:

Using Java Programming language:

Three if and else if statements are created to handle the conditions as given in the question (1,2,3) The output of disagree, no option and agree respectively is displayed using Java's output method System.out.println When these conditions are true

No else statement is provided to handle situations when the parameter value is neither (1,2,3), as a result, then program will do nothing and simply exits.

Ootions: TRUE OR FALSE

Answer: FALSE

Explanation: According to (Steer,2015) there are many platforms for learning and teaching purposes in corporate Organisations, they include PINTEREST,WIKI,GOOGLE+, LINKEDIN,TWITTER etc, The social media platforms are available for effective and efficient Communication and has been the main driving force for Businesses the world over.

These social media platforms have aided the growth and expansion of product marketing strategies and enhanced the over all business efficiency.

Answer:

case 0:

           System.out.println("Rock");

           break;

        case 1:

           System.out.println("Paper");

           break;

        case 2:

           System.out.println("Scissors");

           break;

        default:

           System.out.println("Unknown");

           break;

     }

Explanation:

In this exercise we have to use the knowledge of computational language in python to describe a code, like this:

The code can be found in the attached image.

To make it easier the code can be found below as:

case 0:

          System.out.println("Rock");

          break;

case 1:

          System.out.println("Paper");

          break;

case 2:

          System.out.println("Scissors");

          break;

       default:

          System.out.println("Unknown");

          break;

    }

See more about python at brainly.com/question/26104476

The answer & explanation for this question is given in the attachment below.

The SQL query selects vendor contacts' first names followed by last initials and phone numbers without area codes from the Vendors table. It filters vendors with the 559 area code and sorts results alphabetically by first name and last name.

Below is the SQL SELECT statement that fulfills your requirements:

```sql

SELECT

   CONCAT(SUBSTRING_INDEX(ContactName, ' ', 1), ' ', LEFT(SUBSTRING_INDEX(ContactName, ' ', -1), 1), '.') AS Contact,

   SUBSTRING(VendorPhone, 5) AS Phone

FROM

   Vendors

WHERE

   VendorPhone LIKE '559%'

ORDER BY

   SUBSTRING_INDEX(ContactName, ' ', 1),

   SUBSTRING_INDEX(ContactName, ' ', -1);

```

This query selects the ContactName column from the Vendors table, concatenates the first name and the first letter of the last name, followed by a period, to create the Contact column.

It then extracts the phone number without the area code from the VendorPhone column and stores it in the Phone column.

Finally, it filters the results to include only vendors with the 559 area code and sorts the result set by first name and last name.

Answer:

C

Explanation:

W50.1XXA, Y92.322, Y93.66

Cheers

Answer:

// Program is written in Java Programming Language

// Comments are used for explanatory purpose

// Program starts here

public class RandomOddEve {

/** Main Method */

public static void main(String[] args) {

int[] nums = new int[100]; // Declare an array of 100 integers

// Store the counts of 100 random numbers

for (int i = 1; i <= 100; i++) {

nums[(int)(Math.random() * 10)]++;

}

int odd = 0, even = 0; // declare even and odd variables to 0, respectively

// Both variables will serve a counters

// Check for odd and even numbers

for(int I = 0; I<100; I++)

{

if (nums[I]%2 == 0) {// Even number.

even++;

}

else // Odd number.

{

odd++;

}

}

//.Print Results

System.out.print("Odd number = "+odd);

System.out.print("Even number = "+even);

}

Answer:

Complete python code along with comments to explain the whole code is given below.

Python Code with Explanation:

# import random module to use randint function

import random

# counter variables to store the number of even and odd numbers

even = 0

odd = 0

# list to store randomly generated numbers

random_num=[]

# A for loop is used to generate 100 random numbers

for i in range(100):

# generate a random number from 0 to 100 and append it to the list

   random_num.append(random.randint(0,100))

# check if ith number in the list is divisible by 2 then it is even

   if random_num[i] % 2==0:

# add one to the even counter

        even+=1

# if it is not even number then it must be an odd number

   else:

# add one to the odd counter

        odd+=1

# finally print the count number of even and odd numbers

print("Total Even Random Numbers are: ",even)

print("Total Odd Random Numbers are: ",odd)

Output:

Total Even Random Numbers are: 60

Total Odd Random Numbers are: 40

Total Even Random Numbers are: 54

Total Odd Random Numbers are: 46