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nika2105 [10]
3 years ago
9

Write a class Example() such that it has a method that gives the difference between the size of strings when the '-' (subtractio

n) symbol is used between the two objects of the class. Additionally, implement a method that returns True if object 1 is greater than object 2 and False otherwise when the (>) (greater than) symbol is used. For example: obj1
Computers and Technology
1 answer:
Ratling [72]3 years ago
7 0

Answer:

Here the code is given as follows,

Explanation:

class Example:

   def _init_(self, val):

       self.val = val

   def _gt_(self, other):

       return self.val > other.val

   def _sub_(self,other):

       return abs(len(self.val) - len(other.val))

def main():

   obj1 = Example('this is a string')

   obj2 = Example('this is another one')

   print(obj1 > obj2)

   print(obj1 - obj2)

main()

Output:-

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2) [13 points] You’ve been hired by Banker Bisons to write a C++ console application that determines the number of months to rep
Paha777 [63]

Answer:

Check the explanation

Explanation:

CODE:

#include <iostream>

#include <iomanip>

using namespace std;

double getLoanAmount() { // functions as per asked in question

double loan;

while(true){ // while loop is used to re prompt

cout << "Enter the car loan amount ($2,500-7,500): ";

cin >> loan;

if (loan>=2500 && loan <= 7500){ // if the condition is fulfilled then

break; // break statement is use to come out of while loop

}

else{ // else error message is printed

cout << "Error: $"<< loan << " is an invalid loan amount." << endl;

}

}

return loan;

}

double getMonthlyPayment(){ // functions as per asked in question

double monthpay;

while(true){ // while loop is used to re prompt

cout << "Enter the monthly payment ($50-750): ";

cin >> monthpay;

if (monthpay>=50 && monthpay <= 750){ // if the condition is fulfilled then

break; // break statement is use to come out of while loop

}

else{ // else error message is printed

cout << "Error: $"<< monthpay << " is an invalid monthly payment." << endl;

}

}

return monthpay;

}

double getInterestRate(){ // functions as per asked in question

double rate;

while(true){ // while loop is used to re prompt

cout << "Enter the annual interest rate (1-6%): ";

cin >> rate;

if (rate>=1 && rate <= 6){ // if the condition is fulfilled then

break; // break statement is use to come out of while loop

}

else{ // else error message is printed

cout << "Error: "<< rate << "% is an invalid annual interest rate." << endl;

}

}

return rate;

}

int main() {

cout << setprecision(2) << fixed; // to print with 2 decimal places

int month = 0; // initializing month

// calling functions and storing the returned value

double balance= getLoanAmount();

double payment= getMonthlyPayment();

double rate= getInterestRate();

rate = rate /12 /100; // as per question

// printing as per required in question

cout << "Month Balance($) Payment($) Interest($) Principal($)"<< endl;

while(balance>0){ // while the balance is more than zero

month = month + 1; // counting Months

// calculations as per questions

double interest = balance * rate;

double principal = payment - interest;

balance = balance - principal;

// printing required info with proper spacing

cout <<" "<< month;

cout <<" "<< balance;

cout <<" "<< payment;

cout <<" "<< interest;

cout <<" "<< principal << endl;

}

cout << "Months to repay loan: " << month << endl; // displaying month

cout << "End of Banker Bisons";

Kindly check the output in the attached image below.

5 0
3 years ago
Naynar kis dhrm se sambandhit hai​
Inessa05 [86]

Answer:

Islam dharm se sabandhit

5 0
3 years ago
Read 2 more answers
Write a script that calculates the common factors between 8 and 24. To find a common factor, you can use the modulo operator (%)
AnnyKZ [126]

Answer:

  1. common = []
  2. num1 = 8
  3. num2 = 24
  4. for i in range(1, num1 + 1):
  5.    if(num1 % i == 0 and num2 % i == 0):
  6.        common.append(i)
  7. print(common)

Explanation:

The solution is written in Python 3.

Firstly create a common list to hold a list of the common factor between 8 and 24 (Line 1).

Create two variables num1, and num2 and set 8 and 24 as their values, respectively (Line 3 - 4).

Create a for loop to traverse through the number from 1 to 8 and use modulus operator to check if num1 and num2 are divisible by current i value. If so the remainder of both num1%i and num2%i  will be zero and the if block will run to append the current i value to common list (Line 6-8).

After the loop, print the common list and we shall get [1, 2, 4, 8]

8 0
4 years ago
A datagram network allows routers to drop packets whenever they need to. The probability of a router discarding a packetis p. Co
tresset_1 [31]

Answer:

a.) k² - 3k + 3

b.) 1/(1 - k)²

c.) k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

Explanation:

a.) A packet can make 1,2 or 3 hops

probability of 1 hop = k  ...(1)

probability of 2 hops = k(1-k)  ...(2)

probability of 3 hops = (1-k)²...(3)

Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)

                                                       = (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)

                                                       = k + 2k - 2k² + 3(1 + k² - 2k)

∴mean number of hops                = k² - 3k + 3

b.) from (a) above, the mean number of hops when transmitting a packet is k² - 3k + 3

if k = 0 then number of hops is 3

if k = 1 then number of hops is (1 - 3 + 3) = 1

multiple transmissions can be needed if K is between 0 and 1

The probability of successful transmissions through the entire path is (1 - k)²

for one transmission, the probility of success is (1 - k)²

for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)

for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on

∴ for transmitting a single packet, it makes:

     ∞                             n-1

T = ∑ n(1 - k)²(1 - (1 - k)²)

    n-1

   = 1/(1 - k)²

c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)

from (a) above, mean number of hops when transmitting a packet =  k² - 3k + 3

from (b) above, mean number of transmissions by a packet = 1/(1 - k)²

substituting: mean number of required packet =  k^{2}  - 3k + 3 * \frac{1}{(1 - k)^{2} }\\\\= \frac{k^{2} - 3k + 3 }{(1-k)^{2} }

6 0
3 years ago
List 5 ways by which Artificial intelligence (AI) can be used to drive our business.​
andreyandreev [35.5K]

Answer:

start a website

Explanation:

true

5 0
3 years ago
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