Which best explains the relationship among these three facts question 8

p25: File Write and Read1) User enters a file name (such as "myMovies.txt").2) User enters the titles of 4 of their favorite mov

SCORPION-xisa [38]

Answer:

<em>The program goes as follows </em>

<em>Comments are used to explain difficult lines</em>

#include<iostream>

#include<fstream>

#include<sstream>

#include<string>

using namespace std;

int main()

{

//Declare String to accept file name

string nm;

//Prompt user for file name

cout<<"Enter File Name: ";

getline(cin,nm);

//Create File

ofstream Myfile(nm.c_str());

//Prompt user for 4 names of movies

string movie;

for(int i = 1;i<5;i++)

{

cout<<"Please enter a movie title #"<<i<<": ";

cin>>movie;

//Write to file, the names of each movies

Myfile<<movie<<endl;

}

Myfile.close(); // Close file

//Create an Array for four elements

string myArr[4];

//Prepare to read from file to array

ifstream file(nm.c_str());

//Open File

if(file.is_open())

{

for(int i =0;i<4;i++)

{

//Read each line of the file to array (line by line)

file>>myArr[i];

}

file.close(); // Close file

//Create a reverseOrder.txt file

nm = "reverseOrder.txt";

//Prepare to read into file

ofstream Myfile2(nm.c_str());

for(int j = 3;j>=0;j--)

{

//Read each line of the file to reverseOrder.txt (line by line)

Myfile2<<myArr[j]<<endl;

}

Myfile2.close(); //Close File

return 0;

}

See attached for .cpp program file

Download cpp

6 0

3 years ago

Write a program that asks the user to enter a number of seconds. There are 60 seconds in a minute. If the number of seconds ente

frez [133]

Answer:

// here is code in c.

#include <stdio.h>

// main function

int main()

{

// variable to store seconds

long long int second;

printf("enter seconds:");

// read the seconds

scanf("%lld",&second);

// if seconds is in between 60 and 3600

if(second>=60&& second<3600)

{

// find the minutes

int min=second/60;

printf("there are %d minutes in %lld seconds.",min,second);

}

// if seconds is in between 3600 and 86400

else if(second>=3600&&second<86400)

{

// find the hours

int hours=second/3600;

printf("there are %d minutes in %lld seconds.",hours,second);

}

// if seconds is greater than 86400

else if(second>86400)

{

// find the days

int days=second/86400;

printf("there are %d minutes in %lld seconds.",days,second);

}

return 0;

}

Explanation:

Read the seconds from user.If the seconds is in between 60 and 3600 then find the minutes by dividing seconds with 60 and print it.If seconds if in between 3600 and 86400 then find the hours by dividing second with 3600 and print it. If the seconds is greater than 86400 then find the days by dividing it with 86400 and print it.

Output:

enter seconds:89

there are 1 minutes in 89 seconds.

enter seconds:890000

there are 10 days in 890000 seconds.

8 0

2 years ago

A large population of ALOHA users manages to generate 50 requests/sec, including both originals and retransmissions. Time is slo

krok68 [10]

Answer:

The answer is below

Explanation:

Given that:

Frame transmission time (X) = 40 ms

Requests = 50 requests/sec, Therefore the arrival rate for frame (G) = 50 request * 40 ms = 2 request

a) Probability that there is success on the first attempt = $e^{-G}G^k$ but k = 0, therefore Probability that there is success on the first attempt = $e^{-G}=e^{-2}=0.135$