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kvasek [131]
2 years ago
14

Address the FIXME comments. Move the respective code from the while-loop to the created function. The add_grade function has alr

eady been created.# FIXME: Create add_grade functiondef add_grade(student_grades):print('Entering grade. \n')name, grade = input(grade_prompt).split()student_grades[name] = grade# FIXME: Create delete_name function# FIXME: Create print_grades functionstudent_grades = {} # Create an empty dictgrade_prompt = "Enter name and grade (Ex. 'Bob A+'):\n"delete_prompt = "Enter name to delete:\n"menu_prompt = ("1. Add/modify student grade\n""2. Delete student grade\n""3. Print student grades\n""4. Quit\n\n")command = input(menu_prompt).lower().strip()while command != '4': # Exit when user enters '4'if command == '1':add_grade(student_grades)elif command == '2':# FIXME: Only call delete_name() hereprint('Deleting grade.\n')name = input(delete_prompt)del student_grades[name]elif command == '3':# FIXME: Only call print_grades() hereprint('Printing grades.\n') for name, grade in student_grades.items(): print(name, 'has a', grade) else: print('Unrecognized command.\n') command = input().lower().strip()
Computers and Technology
1 answer:
Sever21 [200]2 years ago
4 0

Answer:

The Python code is given below with appropriate comments

Explanation:

# FIXME: Create add_grade function

def add_grade(student_grades):

print('Entering grade. \n')

name, grade = input(grade_prompt).split()

student_grades[name] = grade

# FIXME: Create delete_name function

def delete_name(student_grades):

print('Deleting grade.\n')

name = input(delete_prompt)

del student_grades[name]

 

# FIXME: Create print_grades function

def print_grades(student_grades):

print('Printing grades.\n')    

for name, grade in student_grades.items():

print(name, 'has a', grade)

 

student_grades = {} # Create an empty dict

grade_prompt = "Enter name and grade (Ex. 'Bob A+'):\n"

delete_prompt = "Enter name to delete:\n"

menu_prompt = ("1. Add/modify student grade\n"

"2. Delete student grade\n"

"3. Print student grades\n"

"4. Quit\n\n")

command = input(menu_prompt).lower().strip()

while command != '4': # Exit when user enters '4'

if command == '1':

add_grade(student_grades)

elif command == '2':

# FIXME: Only call delete_name() here

delete_name(student_grades)

elif command == '3':

# FIXME: Only call print_grades() here

print_grades(student_grades)

else:

print('Unrecognized command.\n')

command = input().lower().strip()

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Answer:

Explanation:

import java.util.Scanner;

public class KboatTriangleAngle

{

  public static void main(String args[]) {

      Scanner in = new Scanner(System.in);

      System.out.print("Enter first angle: ");

      int a1 = in.nextInt();

      System.out.print("Enter second angle: ");

      int a2 = in.nextInt();

      System.out.print("Enter third angle: ");

      int a3 = in.nextInt();

      int angleSum = a1 + a2 + a3;

       

      if (angleSum == 180 && a1 > 0 && a2 > 0 && a3 > 0) {

          if (a1 < 90 && a2 < 90 && a3 < 90) {

              System.out.println("Acute-angled Triangle");

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2 years ago
Locker doors There are n lockers in a hallway, numbered sequentially from 1 to n. Initially, all the locker doors are closed. Yo
kow [346]

Answer:

// here is code in C++

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

   // variables

   int n,no_open=0;

   cout<<"enter the number of lockers:";

   // read the number of lockers

   cin>>n;

   // initialize all lockers with 0, 0 for locked and 1 for open

   int lock[n]={};

   // toggle the locks

   // in each pass toggle every ith lock

   // if open close it and vice versa

   for(int i=1;i<=n;i++)

   {

       for(int a=0;a<n;a++)

       {

           if((a+1)%i==0)

           {

               if(lock[a]==0)

               lock[a]=1;

               else if(lock[a]==1)

               lock[a]=0;

           }

       }

   }

   cout<<"After last pass status of all locks:"<<endl;

   // print the status of all locks

   for(int x=0;x<n;x++)

   {

       if(lock[x]==0)

       {

           cout<<"lock "<<x+1<<" is close."<<endl;

       }

       else if(lock[x]==1)

       {

           cout<<"lock "<<x+1<<" is open."<<endl;

           // count the open locks

           no_open++;

       }

   }

   // print the open locks

   cout<<"total open locks are :"<<no_open<<endl;

return 0;

}

Explanation:

First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.

Output:

enter the number of lockers:9

After last pass status of all locks:

lock 1 is open.

lock 2 is close.

lock 3 is close.

lock 4 is open.

lock 5 is close.

lock 6 is close.

lock 7 is close.

lock 8 is close.

lock 9 is open.

total open locks are :3

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