The elements of an integer-valued array can be initialized so that a[i] == i in a recursive fashion as follows: An array of size

Answer:

boolean rsvp;

int selection;

String option1;

String option2;

(a) if(rsvp){

System.out.println("Attending");

} else {

System.out.println("Not attending");

}

(b) if (selection == 1){

System.out.println("Beef");

} else if (selection == 2){

System.out.println("Chicken");

} else if (selection == 3) {

System.out.println("Pasta");

} else {

System.out.println("Fish");

}

(c) switch(selection){

case 1:

if(rsvp){

option1 = "Thanks for attending. You will be served beef.";

} else{

option1 = "Sorry you can't make it";

}

break;

case 2:

if(rsvp){

option1 = "Thanks for attending. You will be served chicken.";

} else{

option1 = "Sorry you can't make it";

}

break;

case 3:

if(rsvp){

option1 = "Thanks for attending. You will be served pasta.";

} else{

option1 = "Sorry you can't make it";

}

break;

default:

if(rsvp){

option1 = "Thanks for attending. You will be served fish.";

} else{

option1 = "Sorry you can't make it";

}

break;

}

(d) if (option1.equals(option2)){

System.out.println("True");

} else {

System.out.println("False");

}

Explanation:

This code is written in Java.

a) It uses an if statement to print "attending" if rsvp evaluates to true and "not attending" if rsvp is false.

b) An if-else structure determines which food choice is printed depending on user input. If selection equals 1, beef is printed; if it is 2, chicken appears; if it is 3, pasta is displayed; any other value results in fish being printed.

c) The switch statement checks the value of selection and uses an inner if statement to assign "Thanks for attending. You will be served beef." to option1 if rsvp is true. This assignment varies with the selection's value. When rsvp is false, option1 gets the message "Sorry you can't make it."

d) An if statement evaluates whether option1 and option2 are equivalent and outputs true if they match; otherwise, it indicates false.

Answer:

The following C++ program is displayed below. No output is produced as per the requirements stated in the question.

#include <iostream>

using namespace std;

int main() {    

   // declaration and initialization of integer variables

   int k, memberID = 12, nMembers=5;

   bool isAMember;    

   // declaration and initialization of integer array

   int currentMembers[] = {12, 34, 56, 78, 90};    

   for(k=0; k<nMembers; k++)

   {

       if(memberID == currentMembers[k])

       {

           // if the member is found in the array, the loop exits using the break statement

           isAMember = true;

           break;

       }

       else

           isAMember = false;

   }    

   return 0;

}

Explanation:

The program starts with the declaration and initialization of integer variables, followed by the definition of an array that holds the identifiers of all members.

The Boolean variable is declared, although it isn't initialized at this point.

int k, memberID = 12, nMembers=5;

bool isAMember;

int currentMembers[] = {12, 34, 56, 78, 90};

Subsequently, the array containing the members' IDs is searched for the specific member ID. This action is performed using a for loop along with an if-else condition within that loop.

If the member ID exists in the array, the variable isAMember is set to true; otherwise, it is marked as false.

If isAMember is set to true, the break statement allows for the loop to be exited early.

for(k=0; k<nMembers; k++)

   {

       if(memberID == currentMembers[k])

       {

           isAMember = true;

           break;

       }

       else

           isAMember = false;

 }

The break statement is utilized since any other IDs in the array won't match the given member ID and thus, the variable isAMember would incorrectly be initialized to false even if the member ID is present in the array. Therefore, it's crucial to terminate the loop as soon as the matching member ID is located.

This program can be tested by utilizing different member ID values and varying the size of the array.

Answer:

Step 1: Create a table reflecting the three potential scenarios and two possible outcomes. There should be a total of 8 distinct rules for the three varying conditions, structured similarly to the discount table shown here in the example: attached is the discount table

Step 2: The rules can now be made simpler by taking the following aspects into account:

a) When a consumer completes the survey form AND opts into the newsletter, according to Rules 1 and 2, they qualify for a discount if their order exceeds $100. This leads to two distinct rules being formulated while the third condition (order quantity) holds significance.

b) If the buyer fills out the survey form OR subscribes to the newsletter, as indicated by Laws 3, 4, 5, and 6, they will benefit from free shipping, regardless of the order amount. As a result, this situation can be divided into two individual rules, where at least one requirement is satisfied, but not both.

c) When a customer fails to meet any requirements, corresponding to Rules 7 and 8, the order value will not qualify for either free shipping or discount. This can be seen as a single law. The linked simplification table illustrates this.

Answer:

Below is the explanation for the C code.

Explanation:

#include <stdio.h>

#include <stdbool.h>

int main(void) {

int userNum;

bool isPositive;

bool isEven;

scanf("%d", &userNum);

isPositive = (userNum > 0);

isEven = ((userNum % 2) == 0);

if(isPositive && isEven){

  printf("Positive even number");

}

else if(isPositive &&!isEven){

  printf("Positive number");

}

else{

  printf("Not a positive number");

}

printf("\n");

return 0;

}