Which one of the following characteristics or skills of a ScrumMaster are closelyaligned with coaching?Select one:

Response:

Procedure:

1. Define and set the variable one_dog_year to 7.

2.Request the user to input the dog's age.

   2.1 Capture the entered dog age and store it in variable "dog_age".

3.Generate a variable "human_age" for the corresponding human age.

   3.1 Compute the equivalent human age using the formula "human_age=one_dog_year*dog_age".

4.Display the calculated human age that corresponds to the dog's age.

7. Conclude the program.

// following is the algorithm executed in c++

#include <bits/stdc++.h>

using namespace std;

int main()

{

   // set one dog year

   int one_dog_year=7;

   int dog_age;

   int equi_h_age;

   cout<<"Enter the dog age:";

   // get the dog age

   cin>>dog_age;

   //determine the corresponding human age

   equi_h_age=dog_age*one_dog_year;

   cout<<"equivalent human age is: "<<equi_h_age<<endl;

return 0;

}

Result:

Enter the dog age:2

equivalent human age is: 14

Answer:

C3H8

Explanation:

Step 1:

Relevant details from the question are noted below:

Mass of the alkane is 1.17g

Volume (V) = 674 mL

Temperature (T) = 28°C

Pressure (P) = 741 mmHg.

Gas constant (R) = 0.08206 atm.L/Kmol

Step 2:

Convert to the appropriate units.

For Volume:

1000mL = 1L

So, 674mL = 674/1000 = 0.674L

For Temperature:

Temperature in Kelvin = Temperature in Celsius + 273

Temperature in Celsius = 28°C

Temperature in Kelvin = 28°C + 273 = 301K

For Pressure:

760mmHg = 1atm

So, 741 mmHg = 741/760 = 0.975atm

Step 3:

Calculate the moles of the alkane.

The moles of the alkane can be found using the ideal gas equation, presented below:

Volume (V) = 0.674L

Temperature (T) = 301K

Pressure (P) = 0.975atm

Gas constant (R) = 0.08206 atm.L/Kmol

Number of moles (n) =?

PV = nRT

n = PV /RT

n = (0.975 x 0.674)/(0.08206x301)

n = 0.0266 moles

Step 4:

Calculate the alkane’s molar mass.

Mass of alkane = 1.17g

Number of moles = 0.0266moles

Molar Mass =?

Number of moles = Mass/Molar Mass

Molar Mass = Mass/Number of moles

Molar Mass of alkane = 1.17/0.0266 = 44g/mol

Step 5:

Determine the molecular formula of the alkane.

This is based on:

The general formula for alkanes is CnH2n+2

To find the molecular formula, we start with n = 1, 2, 3, etc., until we find the molar mass of 44.

When n = 1

CnH2n+2 = CH4 = 12 + (4x1) = 16g/mol

When n = 2

CnH2n+2 = C2H6 = (12x2) + (6x1) = 30g/mol

When n = 3

CnH2n+2 = C3H8 = (12x3) + (8x1) = 44g/mol

It can be observed that for n equal to 3, the molar mass is 44g/mol.

Thus, the molecular formula of the alkane is C3H8.

Answer

Continue on your current heading and maintain your speed unless the vessel required to give way does not act.

Explanation

When two vessels approach each other, the one positioned on the right side is designated as the stand-on vessel, which means the other vessel, the give-way vessel, must yield. In this circumstance, you should keep your course and speed unless the give-way vessel fails to do so. If that happens, you must take evasive maneuvers to avoid collision by steering clear, never turning toward or crossing in front of the other vessel.

Answer:

Multimedia has become an essential and vital part of various sectors such as healthcare, education, entertainment, and marketing.

Explanation:

In the past, children primarily learned using textbooks, but now they engage with smart boards and other interactive digital devices that provide animated content.

The role of multimedia is crucial in entertainment, where there has been a significant increase in attendance for multimedia events.

Similarly, in the medical field, students can learn more effectively with the use of multimedia tools. Looking ahead, multimedia is expected to draw interest across all disciplines and its future seems promising.

Answer:

// here's the Java code.

import java.util.*;

// definition of the class

class BookstoreCredit

{

/* method that prints a message including the name and grade point */

   public static void fun(String name,double grade)

   {

// multiplying grade by 10

       grade=grade*10;

// display the message

       System.out.println(name+"\'s average grade is $"+grade);

   }

// main method for the class

public static void main (String[] args) throws java.lang.Exception

{

   try{

    // scanner instance to read input

       Scanner s=new Scanner(System.in);

        // declaring variables

      String s_name;

      double grade;

      System.out.print("Please enter the name:");

      s_name=s.nextLine();

      System.out.print("Please enter the grade point:");

      grade=s.nextDouble();

   }catch(Exception ex){

       return;}

}

}

Explanation:

Define a string variable named s_name and a double variable named grade. Obtain the name and grade from the user, then invoke the fun() method with these parameters, which will multiply the grade by 10 and output a message that includes the student's name along with their average grade.

Output:

Please enter the name:john

Please enter the grade point:3.2

john's average grade is $32.0