Let's revisit our lucky_number function. We want to change it, so that instead of printing the message, it returns the message.

The answer is b. Explanation: The general population started accessing the internet following the advent of the World Wide Web.

Answer:

Here is the C++ code:

#include <iostream> //for utilizing input and output functions

using namespace std; //for defining objects cin cout

int main(){ //beginning of the main function

     

   double risingLevel; //declares a variable of type double to store the rising level

   cin>>risingLevel; //captures the risingLevel input from the user

   

   cout<<"Level: "<<risingLevel<<endl; //outputs the rising level

   cout << "Years: 5, Rise: " << risingLevel * 5<<endl; //outputs the increase in ocean level over 5 years

   cout << "Years: 10, Rise: " << risingLevel * 10<<endl; //outputs the increase in ocean level over 10 years

   cout << "Years: 50, Rise: " << risingLevel * 50<<endl; //outputs the increase in ocean level over 50 years

}

Explanation:

The functioning of the program is as follows:

Supposing the user inputs a rising level of 2.1, then,

risingLevel = 2.1

At that point, the first print statement (cout):

cout<<"Level: "<<risingLevel<<endl; outputs the value of risingLevel to the display. Therefore, this line shows:

Level: 2.1

Next, the program proceeds to the following statement:

   cout << "Years: 5, Rise: " << risingLevel * 5<<endl;

which calculates the increase in ocean levels for a duration of 5 years using the equation:

risingLevel * 5 = 2.1 * 5 = 10.5

It subsequently shows the calculated result on the display. So, this line shows:

Years: 5, Rise: 10.5 Then, the control shifts to the next statement:

   cout << "Years: 10, Rise: " << risingLevel * 10<<endl;

that computes the rise in ocean levels over a span of 10 years using the equation:

risingLevel * 10 = 2.1 * 10 = 21

It then showcases the calculated figure on the output screen. Thus, the output is:

Years: 10, Rise: 21

Next, the program control progresses to this statement:

   cout << "Years: 50, Rise: " << risingLevel * 50<<endl;

which determines the rise in ocean levels for a duration of 50 years using the equation:

risingLevel * 50 = 2.1 * 50 = 105

This then displays the computed result on the output screen, leading to the output:

Years: 50, Rise: 105 Consequently, the complete output from the program will be:

Level: 2.1 Years: 5, Rise: 10.5

Years: 10, Rise: 21

Years: 50, Rise: 105

Answer: Sole Proprietorship. This question seeks to identify the business type mentioned in the scenario. In a Sole Proprietorship, the owner effectively is the business, managing all elements from operations to finance. Typically, this is the initial business form in entrepreneurship. The description fits Doug well; he operates the farming business independently, taking care of all its functions. Thus, it qualifies as a sole proprietorship.

Answer:

The following is the C program:

#include<stdio.h>

#include<conio.h>

struct employee{

char empname[50]; int empcode, day, mon, yr;

} employees[30];

int main(){

int total;

printf("Numbers of Employees: "); scanf("%d",&total);

for(int kt=0; kt<total; kt++){

printf("Employee Code: "); scanf("%d",&employees[kt].empcode);

printf("Name: "); scanf("%s", employees[kt].empname);

printf("Date of Joining [dd mm yyyy]: "); scanf("%d%d%d",&employees[kt].day,&employees[kt].mon,&employees[kt].yr); }

int year;

printf("\nCurrent Date [Year only]: "); scanf("%d", &year);

printf("Code\t\t\t Name\t\t\t Date of Joining\n");

for(int kt=0; kt<total; kt++)

if((year - employees[kt].yr) >= 15)

printf("%d\t\t\t %s\t\t\t %d/%d/%d\n", employees[kt].empcode, employees[kt].empname, employees[kt].day, employees[kt].mon, employees[kt].yr);

}

Explanation:

Refer to the attached document for clarification, including comments that elucidate certain lines

Here are the programs. I have written C++ and Python scripts:

a)

C++

#include<iostream>  

using namespace std;  

int factorial(int num)  {  

   if (num == 0)  

       return 1;  

   return num * factorial(num - 1);  }    

int main()  {  

   int integer;

   cout<<"Enter a non negative integer: ";

   cin>>integer;

   cout<< "Factorial of "<< integer<<" is "<< factorial(integer)<< endl;  }

Python:

def factorial(num):  

   if num == 0:  

       return 1

   return num * factorial(num-1)  

integer = int(input("Enter a non negative integer: "))  

print("Factorial of", integer, "is", factorial(integer))

b)

C++

#include <iostream>  

using namespace std;

double factorial(int number) {  

if (number == 0)  

 return 1;  

return number * factorial(number - 1); }  

 

double estimate_e(int num){

    double e = 1;

    for(int i = 1; i < num; i++)

     e = e + 1/factorial(i);

     cout<<"e: "<< e; }  

 

int main(){

int term;

cout<<"Enter a term to evaluate: ";

cin>>term;

estimate_e(term);}

Python:

def factorial(number):  

   if number == 0:  

       return 1

   return number * factorial(number-1)  

def estimate_e(term):

   if not term:

       return 0

   else:

       return (1 / factorial(term-1)) + estimate_e(term-1)

number = int(input("Enter how many terms to evaluate "))

print("e: ", estimate_e(number))

c)

C++

#include <iostream>

using namespace std;

int main(){

   float terms, sumSeries, series;

   int i, number;

   cout << " Input the value of x: ";

   cin >> number;

   cout << " Input number of terms: ";

   cin >> terms;

   sumSeries = 1;

   series = 1;

   for (i = 1; i < terms; i++)      {

       series = series * number / (float)i;

       sumSeries = sumSeries + series;     }

   cout << " The sum  is: " << sumSeries << endl;  }  

Python    

def ePowerx(number,terms):

   sumSeries = 1

   series =1

   for x in range(1,terms):

       series = series * number / x;

       sumSeries = sumSeries + series;

   return sumSeries    

num = int(input("Enter a number: "))

term=int(input("Enter a number: "))

print("e^x: ",ePowerx(num,term))

Explanation:

a)

The program includes a factorial method that takes a number as an argument and calculates its factorial using recursion. For instance, if number = 3

The base case occurs at  if (number == 0)

and the recursion is handled with return number * factorial(number - 1);  

With number = 3 not equaling zero, the function calls itself recursively to get the factorial of 3

return 3* factorial(3- 1);

3 * factorial(2)

3* [2* factorial(2- 1) ]

3 * 2* [ factorial(1)]

3 * 2 * [1* factorial(1- 1) ]

3 * 2 * 1* [factorial(0)]

At this point at factorial(0), the base condition is satisfied as number==0, so factorial(0) returns 1

The resulting output is:

3 * 2 * 1* 1

yielding 6

So, the final program output will be

Factorial of 3 is 6

b)

The estimate_e method takes a number, termed as num, which signifies the term to estimate the mathematical constant e

The for loop extends through each term. For example, if num is set to 3

Then the core statement:

e = e + 1/factorial(i);  

The preceding calculation works as:

e = 1 + 1/1! +1/2!

Since the term count is 3

Initially, e is set to 1

i is initialized at 1

Inserting this into the calculation gives us:

e = 1 + 1/factorial(1)

The factorial function computes and returns 1, as the factorial of 1 is 1. Thus,

e = 1 + 1/1

This results in e = 2

Proceeding to the next iteration, where i = 2 and e = 2, we calculate e = 2 + 1/factorial(2)

Thus, e = 2 + 1/2 results in e = 2.5

Following to the next iteration with i = 3, we have e = 3 + 1/factorial(3)

This yields e = 3 + 1/6 resulting in approximately e = 3.16666

Therefore, the output is:

e: 3.16666

c)

This program calculates the sum of a series based on the formula:

e^x = 1 + x/1! + x^2/2! + x^3/3! +...

The for loop iterates according to the number set for terms. Assuming x is 2, and the number of terms is set to 3, the series would read:

e^2 = 1 + 2/1! + 2^2/2!

In this setup: number = 2 and terms = 3

Initial values for series and sumSeries are both 1

Starting with i equal to 1, the update statement series = series * number / (float)i; applies as follows:series = 1 * 2 /1 results in series = 2

Then, for sumSeries, we have sumSeries = sumSeries + series; Outputs sumSeries as 1 + 2, yielding 3

Continuing to the next iteration: i=2, with series = 2 and sumSeries = 3, we recalculate as series = 2 * 2/2 imposing series = 2 again. Thus, we find: sumSeries = 3 + 2 giving a final sumSeries value of 5

After the loop concludes, the result shows the value of sumSeries, leading finally to the output value of 5