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VMariaS [17]
3 years ago
9

Can some one fix this input ("Enter a number: ") print (num * 8)

Computers and Technology
1 answer:
Kaylis [27]3 years ago
6 0
If you save the input as num,
this will print the input 8 times.

num = input("Enter a number: ")
print(num * 8)

If you want to do actual math calculations,
then the input needs to be a number.

num = float(input("Enter a number: "))
print(num * 8)

This doesn't account for any errors in which the user doesn't input a number, but I don't think that's what you were looking for anyway :)
You might be interested in
The measure of how quickly things may be converted to something of value is called.
Airida [17]

Answer:

liquidity. is a measure of how quickly things can be converted to something of value like cash. commodity money. based on some item of value.

Hope this helps! If so please mark brainliest and rate/heart to help my account if it did!!

3 0
3 years ago
Read 2 more answers
Write a program that reads numbers from a file (or allow user to input data) and creates an ordered binary tree. The program sho
IgorLugansk [536]

Answer:

See explaination

Explanation:

include<bits/stdc++.h>

using namespace std;

typedef struct Node

{

int data;

struct Node *left,*right;

}Node;

bool search(Node *root,int data)

{

if(root==NULL)

return false;

if(root->data==data)

return true;

queue<Node*> q;

q.push(root);

while(!q.empty())

{

Node *temp=q.front();

q.pop();

if(temp->data==data)

return true;

if(temp->left)

q.push(temp->left);

if(temp->right)

q.push(temp->right);

}

return false;

}

Node *insert(Node *root,int data)

{

if(root==NULL)

{

Node *temp=new Node();

temp->data=data;

temp->left=NULL;

temp->right=NULL;

return temp;

}

if(data < root->data)

root->left=insert(root->left,data);

if(data>root->data)

root->right=insert(root->right,data);

return root;

}

Node *get_smallest_element_right_subtree(Node *root)

{

while(root && root->left!=NULL)

root=root->left;

return root;

}

Node *delete_node(Node *root,int data)

{

if(root==NULL)

return root;

if(data < root->data)

root->left=delete_node(root->left,data);

else if(data > root->data)

root->right=delete_node(root->right,data);

else

{

if(root->left==NULL) //If right only presents means - delete the curr node and return right node

{

Node *temp=root->right;

free(root);

return temp;

}

else if(root->right==NULL) //If left only presents means - delete the curr node and return let node

{

Node *temp=root->left;

free(root);

return temp;

}

else

{

Node *temp=get_smallest_element_right_subtree(root->right);

root->data=temp->data;

root->right=delete_node(root->right,temp->data);

}

return root;

}

}

void inorder(Node *root)

{

if(root!=NULL)

{

inorder(root->left);

cout<<root->data<<" ";

inorder(root->right);

}

}

void postorder(Node *root)

{

if(root!=NULL)

{

inorder(root->left);

inorder(root->right);

cout<<root->data<<" ";

}

}

void preorder(Node *root)

{

if(root!=NULL)

{

cout<<root->data<<" ";

inorder(root->left);

inorder(root->right);

}

}

int main()

{

fstream f;

string filename;

cout<<"\n\n1 - Input through File ";

cout<<"\n\n2 - Input through your Hand";

int h;

cout<<"\n\n\nEnter Your Choice : ";

cin>>h;

Node *root=NULL; // Tree Declaration

if(h==1)

{

cout<<"\n\nEnter the Input File Name : ";

cin>>filename;

f.open(filename.c_str());

if(!f)

cout<<"\n\nError in Opening a file !";

else

{

cout<<"\n\nFile is Being Read ........";

string num;

int value;

int node=0;

while(f>> num)

{

value=stoi(num);

root=insert(root,value);

node++;

}

cout<<"\n\nTree has been successfully created with : "<<node<<" Nodes"<<endl;

}

}

if(h==2)

{

int y;

cout<<"\n\nEnter the Total No of Input :";

cin>>y;

int i=1,g;

while(i!=y+1)

{

cout<<"\n\nEnter Input "<<i<<" : ";

cin>>g;

root=insert(root,g);

i++;

}

cout<<"\n\nTree has been successfully created with : "<<y<<" Nodes"<<endl;

}

if(h>=3)

{

cout<<"\n\nInvalid Choice !!! ";

return 0;

}

int n=0;

while(n!=6)

{

cout<<"\n\n\n1 - Insert Element";

cout<<"\n\n2 - Remove Element";

cout<<"\n\n3 - Inorder (LNR) Display ";

cout<<"\n\n4 - Pre (NLR) Order Display";

cout<<"\n\n5 - Post (LRN) Order Display";

cout<<"\n\n6 - Quit";

cout<<"\n\nEnter Your Choice : ";

cin>>n;

switch(n)

{

case 1:

{

int k;

cout<<"\n\nEnter Element to insert : ";cin>>k;

root=insert(root,k);

cout<<"\n\nElement Sucessfully Inserted !!!!!";

break;

}

case 2:

{

int k;

cout<<"\n\nEnter Element to Remove : ";

cin>>k;

if(search(root,k))

{

root=delete_node(root,k);

cout<<"\n\nValue Successfully Deleted !!!";

}

else

cout<<"\n\n!!!!!!!!!!!!!!!!!!!! No Such Element !!!!!!!!!!!!!!!!!!!!!!";

break;

}

case 3:

{

cout<<"\n\nThe Elements (LNR) are : ";

inorder(root);

break;

}

case 4:

{

cout<<"\n\nThe Elements (NLR) are : ";

preorder(root);

break;

}

case 5:

{

cout<<"\n\nThe Elements (LRN) are : ";

postorder(root);

break;

}

case 6:

{

break;

}

}

}

cout<<"\n\nBye!!!! See You !!!"<<endl;

7 0
3 years ago
Consider sending a 10000-byte datagram into a link that has an MTU of 4468 bytes. Suppose the original datagram is stamped with
AfilCa [17]

Answer:

Number of fragments is 3

Explanation:

The maximum size of data field in each fragment = 4468 - 20(IP Header)

= 4448 bytes

Hence, the number of required fragment = (10000 - 20)/4448

= 3

Fragment 1

Id = 218

offset = 0

total length = 4468 bytes

flag = 1

Fragment 2

Id = 218

offset = 556

total length = 4468 bytes

flag = 1

Fragment 3

Id = 218

offset = 1112

total length = 1144 bytes

flag = 0

5 0
2 years ago
you want to implement a protocol on your network that allows computers to find the ip address of a host from a logical name. whi
aniked [119]

The protocol that should be implemented is that one needs to enable hosts on the network to find the IP address.

<h3>What is IP address?</h3>

It should be noted that IP address simply means a unique address that defines a device on the internet.

In this case, the protocol that should be implemented is that one needs to enable hosts on the network to find the IP address.

Learn more about IP address on:

brainly.com/question/24930846

#SPJ12

7 0
1 year ago
Please debug the below code in Java please.
lisabon 2012 [21]

Answer:

Check the explanation

Explanation:

//Bugs are highlighted in bold text

class Invoice

Declarations

private num invoiceNumber

private string customer

private num balanceDue

private num tax

public void setInvoiceNUMBER(num number)

Declarations

num LOW_NUM = 1000

num HIGH_NUM = 9999

if number > HIGH_NUM then

invoiceNumber = 0

else

if number < LOW_NUM then

invoiceNumber = 0

else

invoiceNumber = num

endif

return

public void setCustomer(string cust)

customer = cust

return

public void setBalanceDue(num balance)

//Bug balanceDue is Invoice class varible

//but it is assigned to balance .it gives error

balance = balanceDue

setTax()

return

private void setTax()

Declarations

//Bug TAX_RATE is declared as string

//but assigned a double value

string TAX_RATE = 0.07

tax = tax * TAX_RATE

return

public void displayInvoice()

output "Invoice #", invoiceNumber

output "Customer: ", customer

output "Due: ", balanceDue

output "Tax: ", taxDue

//Bug

//Invoice class has no variable called balance .it should be balanceDue

output "Total ", balance + taxDue

return

endClass

start

Declarations

Invoice inv1

Invoice inv2

Invoice inv3

//Warning

//it gives warning object taken but not initilaized

Invoice inv4

inv1.setInvoiceNumber(1244)

inv1.setCustomer("Brown")

inv1.setBalanceDue(1000.00)

inv1.displayInvoice()

inv2.setInvoiceNumber(77777)

inv2.setCustomer("Jenkins")

inv2.setBalanceDue(2000.00)

inv2.displayInvoice()

inv3.setInvoiceNumber(888)

inv3.setCustomer("Russell")

inv3.setBalanceDue(3000.00)

//Bug

//setTax method of Invioce doesnot take any arguments

inv3.setTax(210.00)

inv3.displayInvoice()

stop

//Here is the complete program in c++

//Run the program using Microsoft visual studio 2010 vc++

#include<iostream>

#include<iomanip>

#include<string>

using namespace std;

class Invoice

{

//class varibales

private:

           int invoiceNumber;

           string customer;

           double balanceDue;

           double tax;

//class methods

public:

           void setCustomer(string cus);

           void displayInvoice();

           void setBalanceDue(double balance);

           void setInvoiceNUMBER(int number);

           void setTax();

};

void Invoice::displayInvoice()

{

cout<< setw(10)<<"Invoice #"<<setw(5)<<invoiceNumber<<endl;

cout<<setw(10)<<"Customer: "<<setw(5)<<customer<<endl;

cout<<setw(10)<<"Due: "<<setw(5)<<balanceDue<<endl;

cout<<setw(10)<<"Tax: "<<setw(5)<<tax<<endl;

//Bug

//Invoice class has no variable called balance .it should be balanceDue

cout<< "Total "<< balanceDue + tax<<endl;

}

void Invoice::setCustomer(string cust)

{

customer = cust;

}

void Invoice::setInvoiceNUMBER(int number)

{

const int LOW_NUM = 1000;

const int HIGH_NUM = 9999;

if( number > HIGH_NUM )

invoiceNumber = 0;

else

if (number < LOW_NUM )

       invoiceNumber = 0;

else

   invoiceNumber = number;

}

void Invoice::setBalanceDue(double balance)

{

balanceDue = balance;

}

void Invoice::setTax()

{

double TAX_RATE = 0.07;

tax = balanceDue * TAX_RATE;

}

int main()

{

Invoice inv1;

Invoice inv2;

Invoice inv3;

inv1.setInvoiceNUMBER(1244);

inv1.setCustomer("Brown");

inv1.setBalanceDue(1000.00);

inv1.setTax();

inv1.displayInvoice();

inv2.setInvoiceNUMBER(77777);

inv2.setCustomer("Jenkins");

inv2.setBalanceDue(2000.00);

inv2.setTax();

inv2.displayInvoice();

inv3.setInvoiceNUMBER(888);

inv3.setCustomer("Russell");

inv3.setBalanceDue(3000.00);

inv3.setTax();

inv3.displayInvoice();

system("pause");

return 0;

}

Kindly check the output image below.

5 0
3 years ago
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