Write a program that calculates an adult's fat-burning heart rate, which is 70% of 220 minus the person's age. Complete fat_burn

In summary, the skills listed are referred to as Technical capabilities. These capabilities encompass the Knowledge and skillset required to execute specific tasks. Meanwhile, Soft skills are equally vital, as they pertain to the capability of effectively engaging and communicating with others.

Answer:

#include <iostream>

using namespace std;

class Digits

{

   public:

   int num;

   int read()       //method to read num from user

   {

       cout<<"Enter number(>0)\n";

       cin>>num;

       return num;

   }

   int digit_count(int num)  //method to count number of digits of num

   {

       int count=0;

       while(num>0)    //loop till num>0

       {

           num/=10;

           count++;   //counter which counts number of digits

       }

       return count;

   }

   int countDigits(int num)   //method to return remainder

   {

       int c=digit_count(num); //calls method inside method

       return num%c;  

   }

};

int main()

{

   Digits d;    //object of class Digits is created

   int number=d.read();   //num is read from user

   cout<<"\nRemainder is: "<<d.countDigits(number);  //used to find remainder

   return 0;

}

Output:

Enter number(>0)

343

Remainder is: 1

Explanation:

The program has a logical error that needs rectification. A correctly structured program calculates the remainder when a number is divided by the count of its digits. A class named Digits is created, consisting of the public variable 'num' and methods for reading input, counting digits, and calculating the remainder.

• read() - This function asks the user to enter the value for 'num' and returns it.
• digit_count() - This function accepts an integer and counts how many digits it has, incrementing a counter until 'num' is less than or equal to 0. It ultimately returns the digit count.
• countDigits() - This function takes an integer and delivers the remainder from dividing that number by its digit count. The digit count is computed using the 'digit_count()' method.

Finally, in the main function, a Digits object is instantiated, and its methods are utilized to produce an output.

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:

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