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ira [324]
2 years ago
14

Write a program that simulates flipping a coin to make decisions. The input is how many decisions are needed, and the output is

either heads or tails. Assume the input is a value greater than 0.
Computers and Technology
1 answer:
Nataly [62]2 years ago
7 0

Answer:

import random

decisions = int(input("How many decisions: "))

for i in range(decisions):

   number = random.randint(0, 1)

   if number == 0:

       print("heads")

   else:

       print("tails")

Explanation:

*The code is in Python.

import the random to be able to generate random numbers

Ask the user to enter the number of decisions

Create a for loop that iterates number of decisions times. For each round; generate a number between 0 and 1 using the randint() method. Check the number. If it is equal to 0, print "heads". Otherwise, print "tails"

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Summary: Given integer values for red, green, and blue, subtract the gray from each value. Computers represent color by combinin
Dmitry_Shevchenko [17]

Answer:

Follows are the code to this question:

#include<iostream>//defining header file

using namespace std;// use package

int main()//main method

{

int red,green,blue,x;//declaring integer variable

cin>> red >>green>>blue;//use input method to input value

if(red<green && red<blue)//defining if block that check red value is greater then green and blue  

{

x = red;//use x variable to store red value

}

else if(green<blue)//defining else if block that check green value greater then blue  

{

x= green; //use x variable to store green value

}

else//defining else block

{

x=blue;//use x variable to store blue value

}

red -= x;//subtract input integer value from x  

green -=x; //subtract input integer value from x

blue -= x;//subtract input integer value from x

cout<<red<<" "<<green<<" "<<blue;//print value

return 0;

}

Output:

130 50 130

80 0 80

Explanation:

In the given code, inside the main method, four integers "red, green, blue, and x" are defined, in which "red, green, and blue" is used for input the value from the user end. In the next step, a conditional statement is used, in the if block, it checks red variable value is greater than then "green and blue" variable. If the condition is true, it will store red variable value in "x", otherwise, it will goto else if block.

  • In this block, it checks the green variable value greater than the blue variable value. if the condition is true it will store the green variable value in x variable.
  • In the next step, else block is defined, that store blue variable value in x variable, at the last step input variable, is used that subtracts the value from x and print its value.      
5 0
2 years ago
A(n) _____ is the highest educational degree available at a community college. master bachelor associate specialist
baherus [9]
<span>An associate's degree requires two years of academic study and is the highest degree available at a community college</span>
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2 years ago
Read 2 more answers
When using the Python shell and code block, what triggers the interpreter to begin evaluating a block of code
finlep [7]

Answer:

The answer is "A blank line".

Explanation:

The blank line initiates the interpreter to start examining the line of statements whenever the Python shell as well as the code block are used.  

  • It is also known as the line that has nothing but spaces or lines without texts or a line.  
  • It prints an empty sheet,  which leaves its performance with such a blank line.

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2 years ago
Use the loop invariant (I) to show that the code below correctly computes the product of all elements in an array A of n integer
NeTakaya

Answer:

Given Loop Variant P = a[0], a[1] ... a[i]

It is product of n terms in array

Explanation:

The Basic Step: i = 0, loop invariant p=a[0], it is true because of 'p' initialized as a[0].

Induction Step: Assume that for i = n - 3, loop invariant p is product of a[0], a[1], a[2] .... a[n - 3].

So, after that multiply a[n - 2] with p, i.e P = a[0], a[1], a[2] .... a[n - 3].a[n - 2].

After execution of while loop, loop variant p becomes: P = a[0], a[1], a[2] .... a[n -3].a[n -2].

for the case i = n-2, invariant p is product of a[0], a[1], a[2] .... a[n-3].a[n-2]. It is the product of (n-1) terms. In while loop, incrementing the value of i, so i=n-1

And multiply a[n-1] with p, i.e P = a[0].a[1].a[2].... a[n-2].

a[n-1]. i.e. P=P.a[n-1]

By the assumption for i=n-3 loop invariant is true, therefore for i=n-2 also it is true.

By induction method proved that for all n > = 1 Code will return product of n array elements.

While loop check the condition i < n - 1. therefore the conditional statement is n - i > 1

If i = n , n - i = 0 , it will violate condition of while loop, so, the while loop will terminate at i = n at this time loop invariant P = a[0].a[1].a[2]....a[n-2].a[n-1]

6 0
2 years ago
Without using any additional variables, and without changing the values of ndays or the elements of the parkingTickets array, wr
dusya [7]

Answer:

Explanation:

mostTickets=0;

for (k=0; k< ndays; k++){

if (parkingTickets[k]>mostTickets) mostTickets=parkingTickets[k];

}

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