Write a function that accepts a positive random number as a parameter and returns the sum of the random number's digits. Write a program that generates random numbers until the sum of the random number's digits divides the random number without remainder. assembly language

Answers

Answer 1

Answer:

Explanation:

The following Python program has a function called addDigits which takes a number as a parameter and returns the sum of the digits. Then the program creates a loop that keeps creating random numbers between 222 and 1000 and divides it by the value returned from the function addDigits. If the remainder is 0 is prints out a statement and breaks the loop, ending the program. The picture below shows the output of the program.

import random

def addDigits(num):

   sum = 0

   for x in str(num):

       sum += int(x)

   return sum

sum = addDigits(random.randint(222, 1000))

while True:

   myRandomNum = random.randint(2, 99)

   if (sum % myRandomNum) == 0:

       print("No remainder between: " + str(sum) + " and " + str(myRandomNum))

       break

Write A Function That Accepts A Positive Random Number As A Parameter And Returns The Sum Of The Random

Related Questions

17. Which of the following keyboard shortcut is used to copy the selected text?
a. Ctrl+c
b. Ctrl+V
c. Ctrl+X
d. Ctrl+Z​

Answers

Answer:

ctrl+c is used to copy the selected text

Answer:

A. Ctrl+c

Explanation:

I took the test, also you can use Command c and Command v to copy and paste.

Let L be the set of exactly those strings over the alphabet \Sigma = {a, b, c, g}, that satisfy all of the following properties: the length of the string is equal to 5n+3, for some natural number n greater or equal than 0; all of the first (leftmost) 2n symbols are elements of the set {b, c, g}; all of the last (rightmost) 3n symbols are elements of the set {a, b}; the symbols at position 2n+1, 2n+2, and 2n+3 (from the left, i.e., after the first 2n symbols but before the last 3n symbols) are elements of the set {c, g}; Write a complete formal definition of a context free grammar that generates L. If such a context free grammar does not exist, state that the context free grammar does not exist, and prove it.

Answers

Answer:

S -> LSR

S -> M

M -> YYY

Y -> c | g

L -> XX

X -> b | c | g

R -> ZZZ

Z -> a | b

Explanation:

This is a long time ago, but I think this does what you want.

Start symbol S expands to LSR and allows you to grow L and R on either side as much as you want. Ultimately S must be replaced by M. Then you have a pattern like LLLLLMRRRRR.

We can then further break down L into 2 times b, c or g, and similar for M and R.

In this lab, you complete a C++ program that uses an array to store data for the village of Marengo.
The program is described in Chapter 8, Exercise 5, in Programming Logic and Design. The program should allow the user to enter each household size and determine the mean and median household size in Marengo. The program should output the mean and median household size in Marengo. The file provided for this lab contains the necessary variable declarations and input statements. You need to write the code that sorts the household sizes in ascending order using a bubble sort and then prints the mean and median household size in Marengo. Comments in the code tell you where to write your statements.
Instructions
Make sure that the file HouseholdSize.cpp is selected and open.
Write the bubble sort.
Output the mean and median household size in Marengo.
Execute the program by clicking the Run button and the bottom of the screen.
Enter the following input, and ensure the output is correct. Household sizes: 4, 1, 2, 4, 3, 3, 2, 2, 2, 4, 5, 6 followed by 999 to exit the program.
Here is my code so far. I need help finishing it (printing the mean & median, etc). Thanks so much:
// HouseholdSize.cpp - This program uses a bubble sort to arrange up to 300 household sizes in
// descending order and then prints the mean and median household size.
// Input: Interactive.
// Output: Mean and median household size.
#include
#include
using namespace std;
int main()
{
// Declare variables.
const int SIZE = 300; // Number of household sizes
int householdSizes[SIZE]; // Array used to store 300 household sizes
int x;
int limit = SIZE;
int householdSize = 0;
int pairsToCompare;
bool switchOccurred;
int temp;
double sum = 0;
double mean = 0;
double median = 0;
// Input household size
cout << "Enter household size or 999 to quit: ";
cin >> householdSize;
// This is the work done in the fillArray() function
x = 0;
while(x < limit && householdSize != 999)
{
// Place value in array.
householdSizes[x] = householdSize;
// Calculate total of household sizes
sum+= householdSizes[x];
x++; // Get ready for next input item.
cout << "Enter household size or 999 to quit: ";
cin >> householdSize;
} // End of input loop.
limit = x;
// Find the mean
// This is the work done in the sortArray() function
pairsToCompare = limit - 1;
switchOccured = true;
while(switchOccured == true)
{
x = 0;
switchOccured == false;
while (x < pairsToCompare)
{
if(householdSizes[x]) > householdSizes[x+1])
{
//perform switch
}
x++;
}
pairsToCompare--;
}
// This is the work done in the displayArray() function
//Print the mean
// Find the median
median = (limit-1) / 2;
if (limit % 2 ==0)
{
cout << "Median is: " << (householdSizes[(int)median] + householdSizes[(int)median + 1]) / 2.0 << endl;
}
else {
// Print the median household size
}
// Print the median
return 0;
} // End of main function

Answers

Answer:

For the mean, do the following:

mean = sum/limit;

cout<<"Mean: "<<mean;

For the median do the following:

for(int i = 0; i<limit; i++) {

  for(int j = i+1; j<limit; j++){

     if(householdSizes[j] < householdSizes[i]){

        temp = householdSizes[i];

        householdSizes[i] = householdSizes[j];

        householdSizes[j] = temp;       }    } }

median= (householdSizes[(limit-1)/2]+householdSizes[1+(limit-1)/2])/2.0;

if((limit - 1)%2==0){

   median = householdSizes[limit/2];

}

cout<<endl<<"Median: "<<median;

Explanation:

The bubble sort algorithm in your program is not well implemented;

So, I replaced the one in your program with another.

Also, some variable declarations were removed (as they were no longer needed) --- See attachment for complete program

Calculate mean

mean = sum/limit;

Print mean

cout<<"Mean: "<<mean;

Iterate through each element

for(int i = 0; i<limit; i++) {

Iterate through every other elements forward

  for(int j = i+1; j<limit; j++){

Compare both elements

     if(householdSizes[j] < householdSizes[i]){

Reposition the elements if not properly sorted

        temp = householdSizes[i];

        householdSizes[i] = householdSizes[j];

        householdSizes[j] = temp;       }    } }

Calculate the median for even elements

median= (householdSizes[(limit-1)/2]+householdSizes[1+(limit-1)/2])/2.0;

Calculate the median for odd elements

if((limit - 1)%2==0){

   median = householdSizes[limit/2];

}

Print median

cout<<endl<<"Median: "<<median;

In the header element, insert a navigation list containing an unordered list with the items: Home, Race Info, and FAQ. Link the items to the dr _index.html, dr_info.html, and dr_ faq.html files respectively.
Home
Race info
FAQ

Answers

Answer:

Explanation:

The following is a barebones HTML document that contains the Unordered List inside the header tag as requested. The list items are also wrapped around links so that they link to the requested pages and take you there when they are clicked. These html files need to be in the same folder as this code in order for it to take you to those pages. The code can be seen below.

<!DOCTYPE html>

<html>

<head>

<ul>

   <li><a href="dr _index.html">Home</a></li>

   <li><a href="dr_info.html">Race Info</a></li>

   <li><a href="dr_ faq.html">FAQ</a></li>

</ul>

</head>

<body>

</body>

</html>

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