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Develop a C program that calculates the final score and the average score for a student from his/her (1)class participation, (2)

Ghella [55]

Answer:

#include <iomanip>

#include<iostream>

using namespace std;

int main(){

char name[100];

float classp, test, assgn, exam, prctscore,ave;

cout<<"Student Name: ";

cin.getline(name,100);

cout<<"Class Participation: "; cin>>classp;

while(classp <0 || classp > 100){ cout<<"Class Participation: "; cin>>classp; }

cout<<"Test: "; cin>>test;

while(test <0 || test > 100){ cout<<"Test: "; cin>>test; }

cout<<"Assignment: "; cin>>assgn;

while(assgn <0 || assgn > 100){ cout<<"Assignment: "; cin>>assgn; }

cout<<"Examination: "; cin>>exam;

while(exam <0 || exam > 100){ cout<<"Examination: "; cin>>exam; }

cout<<"Practice Score: "; cin>>prctscore;

while(prctscore <0 || prctscore > 100){ cout<<"Practice Score: "; cin>>prctscore; }

ave = (int)(classp + test + assgn + exam + prctscore)/5;

cout <<setprecision(1)<<fixed<<"The average score is "<<ave;

return 0;}

Explanation:

The required parameters such as cin, cout, etc. implies that the program is to be written in C++ (not C).

So, I answered the program using C++.

Line by line explanation is as follows;

This declares name as character of maximum size of 100 characters

char name[100];

This declares the grading items as float

float classp, test, assgn, exam, prctscore,ave;

This prompts the user for student name

cout<<"Student Name: ";

This gets the student name using getline

cin.getline(name,100);

This prompts the user for class participation. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

cout<<"Class Participation: "; cin>>classp;

while(classp <0 || classp > 100){ cout<<"Class Participation: "; cin>>classp; }

This prompts the user for test. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

cout<<"Test: "; cin>>test;

while(test <0 || test > 100){ cout<<"Test: "; cin>>test; }

This prompts the user for assignment. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

cout<<"Assignment: "; cin>>assgn;

while(assgn <0 || assgn > 100){ cout<<"Assignment: "; cin>>assgn; }

This prompts the user for examination. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

cout<<"Examination: "; cin>>exam;

while(exam <0 || exam > 100){ cout<<"Examination: "; cin>>exam; }

This prompts the user for practice score. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

cout<<"Practice Score: "; cin>>prctscore;

while(prctscore <0 || prctscore > 100){ cout<<"Practice Score: "; cin>>prctscore; }

This calculates the average of the grading items

ave = (int)(classp + test + assgn + exam + prctscore)/5;

This prints the calculated average

cout <<setprecision(1)<<fixed<<"The average score is "<<ave;

8 0

3 years ago

4. In this problem, we consider sending real-time voice from Host A to Host B over a packet-switchednetwork (VoIP). Host A conve

mina [271]

Answer:

The time elapsed is 0.017224 s

Solution:

As per the question:

Analog signal to digital bit stream conversion by Host A =64 kbps

Byte packets obtained by Host A = 56 bytes

Rate of transmission = 2 Mbps

Propagation delay = 10 ms = 0.01 s

Now,

Considering the packets' first bit, as its transmission is only after the generation of all the bits in the packet.

Time taken to generate and convert all the bits into digital signal is given by;

t = $\frac{Total\ No.\ of\ packets}{A/D\ bit\ stream\ conversion}$

t = $\frac{56\times 8}{64\times 10^{3}}$ (Since, 1 byte = 8 bits)

t = 7 ms = 0.007 s

Time Required for transmission of the packet, t':

$t' = \frac{Total\ No.\ of\ packets}{Transmission\ rate}$

$t' = \frac{56\times 8}{2\times 10^{6}} = 2.24\times 10^{- 4} s$

Now, the time elapse between the bit creation and its decoding is given by:

t + t' + propagation delay= 0.007 + $2.24\times 10^{- 4} s$ + 0.01= 0.017224 s

8 0

3 years ago

Triangle O N M is cut by line segment L K. Line segment L K goes from side N O to side N M. The length of N L is x, the length o

Fantom [35]

Answer:

D) x=8

Explanation:

4 0

4 years ago

Read 2 more answers

What do you call a firewall that is connected to the internet, the internal network, and the dmz?

Firlakuza [10]

Three-pronged firewall hope this helps!

3 0

3 years ago

In JAVA, answer the following:

erica [24]

Answer:

The JAVA program is as follows.

import java.util.Scanner;

public class Program

{

static int n;

public static void main(String[] args) {

//scanner object

Scanner stdin = new Scanner(System.in);

//loop executes till number entered is in the range of 1 to 10

do

{

System.out.print("Enter any number: ");

n = stdin.nextInt();

}while(n<1 || n>10);

System.out.println("Number is valid. Exiting...");

}

OUTPUT

Enter any number: 0

Enter any number: 23

Enter any number: 10

Number is valid. Exiting...

Explanation:

The program is explained.

1. The integer variable, n, is declared static since the variable should be accessible inside main() which is static.

2. An object of Scanner, stdin, is created inside main().

3. Inside do-while loop, the user is asked input any value for n.

4. The loop continues till user inputs a number which is not in the given range, beginning from 1 to 10.

5. The loop will not be terminated till the user enters a valid input.

6. Once a valid input is entered, the message is displayed to the console and the program ends.

7. The variable, n, is declared at the class level and is a class variable. Class variables are declared inside the class but outside all the methods in that class.

8. Since JAVA is a purely object-oriented language, all the code is written inside class.

9. The object of class is not created since only one class is included in the program. If more than one class is included, then the object of the class which does not has the main() method will be created inside main().

10. User input for integer is taken via scanner object and using nextInt() method.

11. The methods, print() and println() display the messages to the console. The only difference is that println() inserts a new line after displaying the message.

12. The name of the class having main() and the name of the program should be the same.

7 0

3 years ago