(a) What is the difference between a compare validator and a range validator?

(b) When would you choose to use one versus the other?

Answer:

There are no limitation, policies or regulations that limit these tools for use on privately owned machines or home networks. In most businesses networks, intranets or internets the use of them is illegal if used for malicious intent. Penetration testing teams sign rules of engagement before using these tools.

Explanation:

There are no limitation, limitations, policies, or regulations in their use on local machines, privately owned machines or home networks, or small business networks.

In most businesses networks, intranets or internets the use of them is often and mostly illegal especially in a situation where they are been used for malicious intent.

Penetration testing teams would often or always make sure that they sign rules of engagement before using these tools.

Answer:

Check the explanation

Explanation:

solution a:

def Rotate(string) :

n = len(string)

temp = string + string

for i in range(n) :

for j in range(n) :

print(temp[i + j], end = "")

print()

string = ("abcde")

Rotate(string)

solution b:

nums = [3,9,5,8,2,4,7]

res = list(filter(lambda n : n < 5, nums))

print(res)

Answer:

Check Explanation.

Explanation:

Immediately there is a connection between your laptop and the wireless hotspot, data will be transferred via the Physical layer that will be used for encoding of the network and the modulation of the network and that layer is called the OSI MODEL.

Another Important layer is the APPLICATION LAYER in which its main aim and objectives is for support that is to say it is used in software supporting.

The next layer is the PRESENTATION LAYER which is used during the process of sending the e-mail. This layer is a very important layer because it helps to back up your data and it is also been used to make connections of sessions between servers of different computers

Answer:

Pay raise is an example of the rewards & consequences characteristic of an organization.

Answer:

See explaination

Explanation:

An Academic Database deals with the information pertaining to the records of students of an institute. The various fields which can be associated with a student are Name, Unique Identification Number, Marks in various subjects, Grades, Courses Taken and so on. Most of the fields mentioned above have some or other form of validation that is required for the DB to be consistent and follow the basic properties of a database.

One field where range validation can be used is MARKS. In the marks field, the range of marks will be say 0 to 100 and anything below 0 or above 100 must be reported to the database administrator. Another field where we can apply a range validation is AGE in which the range of age allowed could greater than say 10, if it is a college and maximum age could be say 60. Thus, the range check on the AGE field is 10 to 60.

One field where choice validation can be used is GENDER. In the gender field, there could be multiple choices like Male, Female and Others. Thus, out of the available choices we have to select only the available choice and only one choice must be selected where we can apply choice validation. Another field is COURSE, where we can have a list of courses a student can opt for and out of the available courses can select one.

Thus, there are multiple fields where we can apply various types of validation and it is important to explore the area for which the DB has been created understanding all the scenarios and attributes of the problems that are associated with that area.

Answer:

see explaination

Explanation:

a)Worst Case-time complexity=O(n)

def intersection_list(lst1, lst2):

lst3 = [value for value in lst1 if value in lst2]

return lst3

lst1 = []

lst2 = []

n1 = int(input("Enter number of elements for list1 : "))

for i in range(0, n1):

ele = int(input())

lst1.append(ele) # adding the element

n2 = int(input("Enter number of elements for list2 : "))

for i in range(0, n2):

ele = int(input())

lst2.append(ele) # adding the element

print(intersection_list(lst1, lst2))

b)Average case-time complexity=O(min(len(lst1), len(lst2))

def intersection_list(lst1, lst2):

return list(set(lst1) & set(lst2))

lst1 = []

lst2 = []

n1 = int(input("Enter number of elements for list1 : "))

for i in range(0, n1):

ele = int(input())

lst1.append(ele)

n2 = int(input("Enter number of elements for list2 : "))

for i in range(0, n2):

ele = int(input())

lst2.append(ele)

print(intersection_list(lst1, lst2))

Answer:

Explanation:

Please find attachment for the Python program

Answer:

Hewo, Here are some ways in which apps earn money :-

hope it helps!

Answer:

The Integrated video may not be working properly well or functioning well.

Explanation:

Solution:

In this case, it is known that the computer was working fine a night before the problem.

If the video cable is connected to a wrong video port, then the computer was unable to work previous night.the integrated video cord has not seated because most times it occurs, due to the heat sink in the card. after restoring the video cable to the video card port, it works.

If the video was disabled in  BIOS/UEFI, it does not show video, even after restoring the cable.

Answer:

Check the explanation

Explanation:

PSEUDO CODE:

for(int i=0,j=0;i<phone.length,j<company.length;i++,j++)

{

print(phone[i]," - ",company[j]);

}

this for loop will print each phone number associated with the company names.

Answer:

The software can't locate the device

Explanation:

The device wasn't properly setup in order for the software to locate it.

Here is the complete question.

In this lab, you use the flowchart and pseudocode found in the figures below to add code to a partially created C++ program. When completed, college admissions officers should be able to use the C++ program to determine whether to accept or reject a student, based on his or her test score and class rank.

start input testScore,

classRank if testScore >= 90 then if classRank >= 25 then output "Accept"

else output "Reject" endif else if testScore >= 80

then if classRank >= 50 then output "Accept" else output "Reject" endif

else if testScore >= 70

then if classRank >= 75 then output "Accept"

else output "Reject"

endif else output "Reject"

endif

endif

endif

stop

Study the pseudocode in picture above. Write the interactive input statements to retrieve: A student’s test score (testScore) A student's class rank (classRank) The rest of the program is written for you. Execute the program by clicking "Run Code." Enter 87 for the test score and 60 for the class rank. Execute the program by entering 60 for the test score and 87 for the class rank.

[comment]: <> (3. Write the statements to convert the string representation of a student’s test score and class rank to the integer data type (testScore and classRank, respectively).)

Function: This program determines if a student will be admitted or rejected. Input: Interactive Output: Accept or Reject

*/ #include using namespace std; int main() { // Declare variables

// Prompt for and get user input

// Test using admission requirements and print Accept or Reject

if(testScore >= 90)

{ if(classRank >= 25)

{ cout << "Accept" << endl; }

else

cout << "Reject" << endl; }

else { if(testScore >= 80)

{ if(classRank >= 50)

cout << "Accept" << endl;

else cout << "Reject" << endl; }

else { if(testScore >= 70)

{ if(classRank >=75) cout << "Accept" << endl;

else cout << "Reject" << endl; }

else cout << "Reject" << endl; } } } //End of main() function

Answer:

Explanation:

The objective here is to use the flowchart and pseudocode found in the figures below to add code to a partially created C++ program. When completed, college admissions officers should be able to use the C++ program to determine whether to accept or reject a student, based on his or her test score and class rank.

PROGRAM:

#include<iostream>

using namespace std;

int main(){

// Declare variables

int testScore, classRank;

// Prompt for and get user input

cout<<"Enter test score: ";

cin>>testScore;

cout<<"Enter class rank: ";

cin>>classRank;

// Test using admission requirements and print Accept or Reject

if(testScore >= 90)  

{ if(classRank >= 25)

{ cout << "Accept" << endl; }

else

cout << "Reject" << endl;

}

else { if(testScore >= 80)

{ if(classRank >= 50)

cout << "Accept" << endl;

else cout << "Reject" << endl; }

else { if(testScore >= 70)

{ if(classRank >=75) cout << "Accept" << endl;

else cout << "Reject" << endl;

}

else cout << "Reject" << endl; }

}

return 0;

} //End of main() function

OUTPUT:

See the attached file below:

Answer:

And user documentation consists of instructions and information to users who will interact with the system

Explanation:

user documentation will not allow user to prepare overall documentation because it is already prepared so not A.

user documentation does not need implementation, So not B.

user documentation may not be a printed one . So not C

And user documentation consists of instructions and information to users who will interact with the system, so D is the right option

Answer:

See explaination

Explanation:

#include <iostream>

#include <map>

#include <string>

#include <algorithm>

#include <cstdlib>

#include <iomanip>

using namespace std;

int main()

{

map<int, int> histogram;

int a=0;

while(a>=0)

{

cout << " enter value of a" ;

cin >>a;

histogram[a]++;

}

map<int,int>::iterator it;

for ( it=histogram.begin() ; it != histogram.end(); it++ )

{

cout << (*it).first << " occurs " << setw(3) << (*it).second << (((*it).second == 1) ? " time." : " times.") <<endl;

}

return 0;

}

Explanation:

The most reliable hard drives are those whose most common size is 3.5 inches, their advantages are their storage capacity and their speed and a disadvantage is that they usually make more noise.

For greater speed, it is ideal to opt for two smaller hard disks, since large disks are slower, but are partitioned so that there are no problems with file loss.

For a person who needs to use content creation programs, it is ideal to opt for a hard drive that has reliability.

Answer: Provided in the explanation segment

Explanation:

The following code provides answers to the given problem.

import java.util.*;

public class untitled{

public static void main(String[] args)

{int tens,fives,ones,quarters,dimes,nickles,pennies,amt;

double amount;

Scanner in = new Scanner(System.in);

System.out.println("What do you need to find the change for?");

amount=in.nextDouble();

System.out.println("Your change is");

amt=(int)(amount*100);

tens=amt/1000;

amt%=1000;

fives=amt/500;

amt%=500;

ones=amt/100;

amt%=100;

quarters=amt/25;

amt%=25;

dimes=amt/10;

amt%=10;

nickles=amt/5;

pennies=amt%5;

System.out.println(tens +" ten dollar bills");

System.out.println(fives +" five dollar bills");

System.out.println(ones +" one dollar bills");

System.out.println(quarters +" quarters");

System.out.println(dimes +" dimes");

System.out.println(nickles +" nickles");

System.out.println(pennies +" pennies");

}

}

cheers i hope this helped !!

Python Code with Explanation:

# create a function named func to implement the required logic

def func():

# get the input from the user and store it in a variable named number

   number = int(input("Please enter an integer between 1 to 10 inclusive\n"))

# if the input number is equal or greater than 1 and equal to 10 or less then the input is correct    

   if number>=1 and number<=10:

# print correct input

       return print("Correct input")

# else the input is wrong

   else:

# print wrong input

       return print("Wrong input")

# call the function func

func()

Output:

Test 1:

Please enter an integer between 1 to 10 inclusive

4

Correct input

Test 2:

Please enter an integer between 1 to 10 inclusive

0

Wrong input

Test 3:

Please enter an integer between 1 to 10 inclusive

11

Wrong input

Answer:

dont know if it works!

Explanation:

input=(input("enter number")

if input > 1

print("correct input")

if input <10

print("correct input")

else

print("wrong input")

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

Answer:

99.6

Explanation:

The cloud provider  provides the certain modules of the cloud service like infrastructure as a service , software as a service ,etc to other companies .The cloud provider implements a new SaaS service that includes a cloud platform also the cloud provider intends to disclose the Service Level Agreement that is score based on its approximate accessibility during the next 12 months.

The Cloud providers predict that, depending on past cloud system data, it is a 25 % risk that a physical server holding the cloud platform will fail therefore such a failure will require 2 days to recover the cloud service so 99.6 is the Cloud storage accessibility ranking, to be released in the SLA.

=======================================================

//Class header definition

public class DisArray {

   //First method with int array as parameter

   public static void convert2D(int[] oneD) {

       //1. First calculate the number of columns

       //a. get the length of the one dimensional array

       int arraylength = oneD.length;

       //b. find the square root of the length and typecast it into a float

       float squareroot = (float) Math.sqrt(arraylength);

       //c. round off the result and save in a variable called row

       int row = Math.round(squareroot);

       //2. Secondly, calculate the number of columns

       //a. if the square of the number of rows is greater than or equal to the

       //length of  the one dimensional array,

       //then to minimize padding, the number of

       //columns  is the same as the number of rows.

       //b. otherwise, the number of columns in one more than the

       // number of rows

       int col = ((row * row) >= arraylength) ? row : row + 1;

       //3. Create a 2D int array with the number of rows and cols

       int [ ][ ] twoD = new int [row][col];

       //4. Place the elements in the one dimensional array into

       //the two dimensional array.

       //a. First create a variable counter to control the cycle through the one

       // dimensional array.

       int counter = 0;

       //b. Create two for loops to loop through the rows and columns of the

       // two  dimensional array.

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

           for (int j = 0; j < col; j++) {

               //if counter is less then the length of the one dimensional array,

               //then  copy the element at that position into the two dimensional

               //array.  And also increment counter by one.

               if (counter < oneD.length) {

                   twoD[i][j] = oneD[counter];

                   counter++;

              }

              //Otherwise, just pad the array with zeros

               else {

                   twoD[i][j] = 0;

               }

           }

       }

       //You might want to create another pair of loop to print the elements

       //in the  populated two dimensional array as follows

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

           for (int j = 0; j < twoD[i].length; j++) {

               System.out.print(twoD[i][j] + " ");

           }

           System.out.println("");

       }

   }     //End of first method

   //Second method with String array as parameter

   public static void convert2D(String[] oneD) {

       //1. First calculate the number of columns

       //a. get the length of the one dimensional array

       int arraylength = oneD.length;

       //b. find the square root of the length and typecast it into a float

       float squareroot = (float) Math.sqrt(arraylength);

       //c. round off the result and save in a variable called row

       int row = Math.round(squareroot);

       //2. Secondly, calculate the number of columns

       //a. if the square of the number of rows is greater than or equal to the length of  

       //the one dimensional array, then to minimize padding, the number of

       //columns  is the same as the number of rows.

       //b. otherwise, the number of columns in one more than the

       //number of rows.

       int col = (row * row >= arraylength) ? row : row + 1;

       //3. Create a 2D String array with the number of rows and cols

       String[][] twoD = new String[row][col];

       //4. Place the elements in the one dimensional array into the two

       // dimensional array.

       //a. First create a variable counter to control the cycle through the one

       // dimensional array.

       int counter = 0;

       //b. Create two for loops to loop through the rows and columns of the

       //two  dimensional array.

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

           for (int j = 0; j < col; j++) {

               //if counter is less then the length of the one dimensional array,

               //then  copy the element at that position into the two dimensional

               //array.  And also increment counter by one.

               if (counter < oneD.length) {

                   twoD[i][j] = oneD[counter];

                   counter++;

               }

               //Otherwise, just pad the array with null values

               else {

                   twoD[i][j] = null;

               }

           }

       }

       //You might want to create another pair of loop to print the elements  

       //in the populated two dimensional array as follows:

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

           for (int j = 0; j < twoD[i].length; j++) {

               System.out.print(twoD[i][j] + " ");

           }

           System.out.println("");

       }

   }    // End of the second method

   //Create the main method

   public static void main(String[] args) {

       //1. Create an arbitrary one dimensional int array

       int[] x = {23, 3, 4, 3, 2, 4, 3, 3, 5, 6, 5, 3, 5, 5, 6, 3};

       //2. Create an arbitrary two dimensional String array

       String[] names = {"abc", "john", "dow", "doe", "xyz"};

       //Call the respective methods

       convert2D(x);

       System.out.println("");

       convert2D(names);

   }         // End of the main method

}            // End of class definition

=========================================================

==========================================================

23 3 4 3  

2 4 3 3  

5 6 5 3  

5 5 6 3  

abc john dow  

doe xyz null

==========================================================

The above code has been written in Java and it contains comments explaining each line of the code. Please go through the comments. The actual executable lines of code are written in bold-face to distinguish them from the comments.

Answer:

See explaination

Explanation:

clc;

clear all;

close all;

x=input(' choose abc or fgh or mno:','s');

to=input('enter intial fluid temperature in celcius:');

t1=input('enter time in minutes:');

ta=input('enter ambient temperature in celcius:');

abc=1;

fgh=2;

mno=3;

if x==1

k=0.01;

elseif x==2

k=0.03;

else

k=0.02;

end

t=ta+((to-ta)*exp((-k)*t1));

X = sprintf('%s has temp %f degrees celcius after %d minutes.',x,t,t1);

disp(X);

Answer:

The program in cpp is given below.

#include <iostream>

using namespace std;

int main()

{

   //variables declared to hold sum, count and average of movie ratings

   int sum=0;

   int count=0;

   int rating;

   double avg;

   //audience is prompted to enter ratings

   do{

       std::cout << "Enter the rating (0 to 4): ";

       cin>>rating;

       if(rating>4)

           std::cout << "Invalid rating. Please enter correct rating again." << std::endl;

       if(rating>=0 && rating<=4)

       {

           sum=sum+rating;

           count++;

       }

       if(rating<0)

           break;

   }while(rating>=0);

   //average rating is computed

   avg=sum/count;

   std::cout << "The average rating for the movie is " << avg << std::endl;  

   return 0;

}

OUTPUT

Enter the rating (0 to 4): 1

Enter the rating (0 to 4): 2

Enter the rating (0 to 4): 0

Enter the rating (0 to 4): 5

Invalid rating. Please enter correct rating again.

Enter the rating (0 to 4): 1

Enter the rating (0 to 4): -1

The average rating for the movie is 1

Explanation:

1. Variables are declared to hold the rating, count of rating, sum of all ratings and average of all the user entered ratings.

2. The variables for sum and count are initialized to 0.

3. Inside do-while loop, user is prompted to enter the rating for the movie.

4. In case the user enters a rating which is out of range, the user is prompted again to enter a rating anywhere from 0 to 4.

5. The loop executes until a negative number is entered.

6. Inside the loop, the running total sum of all the valid ratings is computed and the count variable is incremented by 1 for each valid rating.

7. Outside the loop, the average is computed by dividing the total rating by count of ratings.

8. The average rating is then displayed to the console.

9. The program is written in cpp due to simplicity since all the code is written inside main().

Answer:

Check the explanation

Explanation:

#include <iostream>

using namespace std;

void CoordTransform(int x, int y, int& xValNew,int& yValNew){

  xValNew = (x+1)*2;

  yValNew = (y+1)*2;

}

int main() {

  int xValNew = 0;

  int yValNew = 0;

  CoordTransform(3, 4, xValNew, yValNew);

  cout << "(3, 4) becomes " << "(" << xValNew << ", " << yValNew << ")" << endl;

  return 0;

}

Answer:

public class RectangleArea {

   public static void main(String[] args) {

       Rectangle myRectangle = new Rectangle(15.8, 7.9);

       double calArea = myRectangle.calcArea(15.8, 7.9);

       System.out.println("The area is "+ calArea);

   }

}

class Rectangle{

   private double length;

   private double width;

   //Defined Constructor

   public Rectangle(double length, double width) {

       this.length = length;

       this.width = width;

   }

   public double calcArea(double len, double wi){

       return len* wi;

   }

}

Explanation:

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Answer:

Explanation:

Program:

#include<iostream>

#include <bits/stdc++.h>

using namespace std;

//check for operator

bool isOperator(char c)

{

switch(c)

{

case '+': case '-': case '/': case '*': case '^':

return true;

}

return false;

}

//Converter class

class Converter

{

private:

string str;

public:

//constructor

Converter(string s):str(s){}

//convert from infix to postfix expression

string toPostFix(string str)

{

stack <char> as;

int i, pre1, pre2;

string result="";

as.push('(');

str = str + ")";

for (i = 0; i < str.size(); i++)

{

char ch = str[i];

if(ch==' ') continue;

if (ch == '(')

as.push(ch);

else if (ch == ')')

{

while (as.size() != 0 && as.top() != '('){

result = result + as.top() + " ";

as.pop();

}

as.pop();

}

else if(isOperator(ch))

{

while (as.size() != 0 && as.top() != '(')

{

pre1 = precedence(ch);

pre2 = precedence(as.top());

if (pre2 >= pre1){

result = result + as.top() + " ";

as.pop();

}

else break;

}

as.push(ch);

}

else

{

result = result + ch;

}

}

while(as.size() != 0 && as.top() != '(') {

result += as.top() + " ";

as.pop();

}

return result;

}

//return the precedence of an operator

int precedence(char ch)

{

int choice = 0;

switch (ch) {

case '+':

choice = 0;

break;

case '-':

choice = 0;

break;

case '*':

choice = 1;

break;

case '/':

choice = 1;

break;

case '^':

choice = 2;

default:

choice = -999;

}

return choice;

}

};

//Node class

class Node

{

public:

string element;

Node *leftChild;

Node *rightChild;

//constructors

Node (string s):element(s),leftChild(nullptr),rightChild(nullptr) {}

Node (string s, Node* l, Node* r):element(s),leftChild(l),rightChild(r) {}

};

//ExpressionTree class

class ExpressionTree

{

public:

//expression tree construction

Node* covert(string postfix)

{

stack <Node*> stk;

Node *t = nullptr;

for(int i=0; i<postfix.size(); i++)

{

if(postfix[i]==' ') continue;

string s(1, postfix[i]);

t = new Node(s);

if(!isOperator(postfix[i]))

{

stk.push(t);

}

else

{

Node *r = nullptr, *l = nullptr;

if(!stk.empty()){

r = stk.top();

stk.pop();

}

if(!stk.empty()){

l = stk.top();

stk.pop();

}

t->leftChild = l;

t->rightChild = r;

stk.push(t);

}

}

return stk.top();

}

//inorder traversal

void infix(Node *root)

{

if(root!=nullptr)

{

cout<< "(";

infix(root->leftChild);

cout<<root->element;

infix(root->rightChild);

cout<<")";

}

}

//postorder traversal

void postfix(Node *root)

{

if(root!=nullptr)

{

postfix(root->leftChild);

postfix(root->rightChild);

cout << root->element << " ";

}

}

//preorder traversal

void prefix(Node *root)

{

if(root!=nullptr)

{

cout<< root->element << " ";

prefix(root->leftChild);

prefix(root->rightChild);

}

}

};

//main method

int main()

{

string infix;

cout<<"Enter the expression: ";

cin >> infix;

Converter conv(infix);

string postfix = conv.toPostFix(infix);

cout<<"Postfix Expression: " << postfix<<endl;

if(postfix == "")

{

cout<<"Invalid expression";

return 1;

}

ExpressionTree etree;

Node *root = etree.covert(postfix);

cout<<"Infix: ";

etree.infix(root);

cout<<endl;

cout<<"Prefix: ";

etree.prefix(root);

cout<<endl;

cout<< "Postfix: ";

etree.postfix(root);

cout<<endl;

return 0;

}

Answer:

Following are the step to this question:

i) search "intitle passwd" in google search engine.

ii) Open first link of the page.

iii) use commands to access data.  

Explanation:

Google Hacking operates as follows of web engines like google and yahoo to find web pages and sensitive information, that are insecure. It builds on the idea that web pages index a large number of public pages and files and make their discovery easy to create the correct query.

Answer:

Check the explanation

Explanation:

CPI means Clock cycle per Instruction

given Clock rate 600 MHz then clock time is Cー 1.67nSec clockrate 600M

Execution time is given by following Formula.

Execution Time(CPU time) = CPI*Instruction Count * clock time = [tex]\frac{CPI*Instruction Count}{ClockRate}[/tex]

a)

for system A CPU time is 1.3 * 100, 000 600 106

= 216.67 micro sec.

b)

for system B CPU time is [tex]=\frac{2.5*100,000}{750*10^6}[/tex]

= 333.33 micro sec

c) Since the system B is slower than system A, So the system A executes the given program in less time

Hence take CPU execution time of system B as CPU time of System A.

therefore

216.67 micro = =[tex]\frac{2.5*Instruction}{750*10^6}[/tex]

Instructions = 216.67*750/2.5

= 65001

hence 65001 instruction are needed for executing program By system B. to complete the program as fast as system A

The number of instructions that the program would need to have when compiled for Computer B is; 65000 instructions

Formula for Execution time is;

Execution time = (CPI × Instruction Count)/Clock Time

We are given;

CPI for computer A = 1.3

Instruction Count = 100000

Clock time = 600 MHz = 600 × 10⁶ Hz

Thus;

Execution time = (1.3 * 100000)/(600 × 10⁶)

Execution time(CPU Time) = 216.67 * 10⁻⁶ second

For CPU B;

CPU Time = (2.5 * 100000)/(750 × 10⁶)

CPU Time = 333.33 * 10⁻⁶ seconds

Thus, instructions for computer B for the two computers to have same execution time is;

216.67 * 10⁻⁶ = (2.5 * I)/(750 × 10⁶)

I = (216.67 * 10⁻⁶ * 750 × 10⁶)/2.5

I = 65000 instructions

Read more about programming instructions at; https://brainly.com/question/15683939

Answer:

The formula used in this question is called the probability of combinations or combination formula.

Explanation:

Solution

Given that:

Formula applied is stated as follows:

nCr = no of ways to choose r objects from n objects

  = n!/(r!*(n-r)!)

The Data given :

Menu A : 5 appetizers and 3 main dishes

Menu B : 3 appetizers and 4 main dishes

Total appetizers - 6

Total main dishes - 5

Now,

Part A :

Total ways = No of ways to select menu A + no of ways to select menu B

          = (no of ways to select appetizers in A)*(no of ways to select main dish in A) + (no of ways to select appetizers in B)*(no of ways to select main dish in B)

          = 6C5*5C3 + 6C3*5C4

          = 6*10 + 20*5

          = 160

Part B :

Since, we can select the same number of appetizers/main dish again so the number of ways to select appetizers/main dishes will be = (total appetizers/main dishes)^(no of appetizers/main dishes to be selected)

Total ways = No of ways to select menu A + no of ways to select menu B  

          = (no of ways to select appetizers in A)*(no of ways to select main dish in A) + (no of ways to select appetizers in B)*(no of ways to select main dish in B)

          = (6^5)*(5^3) + (6^3)*(5^4)

          = 7776*125 + 216*625

          = 1107000

Part C :

No of ways to select same appetizers and main dish for all the options

= No of ways to select menu A + no of ways to select menu B  

= (no of ways to select appetizers in A)*(no of ways to select main dish in A) + (no of ways to select appetizers in B)*(no of ways to select main dish in B)

=(6*5) + (6*5)

= 60

Total ways = Part B - (same appetizers and main dish selected)      

= 1107000 - 60

= 1106940