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2 years ago

Task 2

Computers and Technology

1 answer:

adelina 88 [10]2 years ago

3 0

The program is an illustration of file manipulations

<h3>What are file manipulations?</h3>

File manipulations are program statements that are used to write & append to file, and also read from the file

<h3>The actual program in Python</h3>

Assume the file name is top10.txt, the program in Python where comments are used to explain each line is as follows:

#This opens the file

a_file = open("top10.txt")

#This reads the file contents

file_contents = a_file.read()

#This prints the contents

print(file_contents)

Read more about file manipulations at:

brainly.com/question/15683939

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What does editing refer to

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Editing is the process of working over and revising a piece of work, whether that piece of work is in writing, audio, or visual mediums. Editing in computer terminology can mean to search for errors and revise.

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Tim is a project manager. He is brainstorming with his team about problems that could arise during the next phase of the project

Grace [21]

Answer:

Explanation:

just took the quiz

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2 years ago

A user is attempting to access a shared file locally on their workstation. The file has Full control NTFS permissions assigned,

BaLLatris [955]

Answer:

The answer is read only permission

Explanation:

The effective permission for the user is the read only permission.

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3 years ago

The birthday problem is as follows: given a group of n people in a room, what is the probability that two or more of them have t

Julli [10]

Answer:

import random

def calc_birthday_probability(num_people):

num_tries = 1e6

num_duplicates = 0

tries = 0

while tries < num_tries:

total_birthday = {}

for i in range(1, 366):

total_birthday[i] = 0

for count in range(num_people):

birthday = random.randint(1, 365)

if total_birthday[birthday] != 0:

num_duplicates += 1

break

total_birthday[birthday] += 1

tries += 1

return num_duplicates/num_tries

def main():

num_people = 10

p = calc_birthday_probability (num_people)

print(p)

num_people = 20

p = calc_birthday_probability (num_people)

print(p)

return

if __name__ == "__main__":

random.seed(2020)

main()

Explanation:

Run the loop until the value of tries is less than num_tries to do number of trials.
Run the loop to create the desired number of birthdays by generating random numbers.
Increase the num_duplicates variable by 1, if the birthday has already occurred.
Lastly define and then call the main function.

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4 years ago

Here is the problem specification: An Internet service provider has three different subscription packages for its customers: Pac

Andrej [43]

Answer:

In C++:

#include <iostream>

using namespace std;

int main(){

int hour; char pkg; float bill = 0;

cout<<"Package: "; cin>>pkg;

cout<<"Hour: "; cin>>hour;

if(hour<=744 && hour >=0){

switch (pkg) {

case 'A':

bill = hour * 9.95;

if(hour >10){bill = 10 * 9.95 + (hour - 10) * 2;}

break;

case 'B':

bill = hour * 14.5;

if(hour >10){bill = 20 * 14.5 + (hour - 20) * 1;}

break;

case 'C':

bill = 19.95;

break;

default:

cout << "Package must be A, B or C";}

cout<<"Total Bills: $"<<bill; }

else{ cout<<"Hour must be 0 - 744"; }

return 0;

}

Explanation:

This declares all variables: int hour; char pkg; float bill=0;

This prompts the user for package type: cout<<"Package: "; cin>>pkg;

This prompts the user for number of hours: cout<<"Hour: "; cin>>hour;

This checks if hour is between 0 and 744 (inclusive)

if(hour<=744 && hour >=0){

If true, the following is executed

A switch statement to check valid input for package

switch (pkg) {

For 'A' package

case 'A':

Calculate the bill

End of A package: break;

For 'B' package

case 'B':

Calculate the bill

End of B package:<em> </em>break;

For C package

case 'C':

Calculate bill: bill = 19.95;

End of C package: break;

If package is not A, or B or C

default:

Prompt the user for valid package cout << "Package must be A, B or C";}

Print total bills: cout<<"Total Bills: $"<<bill; }

If hour is not 0 to 744: <em>else{ cout<<"Hour must be 0 - 744"; }</em>

6 0

3 years ago