In [3]: def isPerfect (x):***Returns whether or not the given number x is perfect.A number is said to be perfect if it is equal to the sum of all itsfactors (for obvious reasons the list of factors being considered doesnot include the number itself).Example: 6 = 3 + 2 + 1, hence 6 is perfect.Example: 28 is another example since 1 + 2 + 4 + 7 + 14 is 28.Note, the number 1 is not a perfect number.your code hereSum - 0for i in range(1, x)if(x1= 0):Sun - Sum + 1if (Sum X)Iprint(" td is a Perfect Number" 1x)print(" td is not a Perfect Number 1x)else:print (isPerfect (6))File "cipython-input-3-b9df3e539e48", line 15for i in range(1, x):IndentationError: unexpected indentIn [411 perfect numbers [6, 28, 496, 8128, 335503361for i in perfect_numbers:assert true (isPerfect (i), str(i) + is perfect')not_perfect_numbers [2, 3, 4, 5, 7, 8, 9, 10,495, 8127, 8129,335503351for i in not perfect_numbers:assert true (not (isPerfect (i)), str(i) is not perfect')#test existence of doestringassert true (len (isPerfect._doc_) > 1, "there is no docstring for isPerfect")print("Success!")

Indentation error occurs when there are extra spaces. for loop should start right below the sum…

In (3) def isPerfect (x) ***Returns whether or not the given number x is perfect. A number is said to be perfect if it is equal to the sum of all its factors (for obvious reasons the list of factors being considered does not include the number itself). Example: 63+2+1, hence 6 is perfect. Example: 20 is another example since 1+2 +4 + 7 + 14 in 28. Note, the number 1 is not a perfect number. *** your code here Sum - 0 for i in range(1, x) LE(0)1 Sun Bum + 1 if (Bum=-X) else! print(" td is a Perfect Number 1x) print(td is not a Perfect Number" 1x) print (isPerfect (6)) File "cipython-input-3-b9df3e539048", line 15 for i in range(1, x) IndentationError: unexpected indent In 1411 perfect numbers 16, 28, 496, 8126, 335503361 for i in perfect numbers: assert true (isPerfect (i), str(i)+ is perfect') not perfect numbers (2, 3, 4, 5, 7, 8, 9, 10, 495, 8127, 8129, 335503351 for i in not perfect numbers assert true (not (isPerfect (i)), str(i) is not perfect') test existence of doestring assert true(len (isPerfect._doc_) 1, "there is no docstring for isPerfect") print("Success!")

In (3) def isPerfect (x) Returns whether or not the given number x is perfect. A number is said to be perfect if it is equal to the sum of all its factors (for obvious reasons the list of factors being considered does not include the number itself). Example: 63+2+1, hence 6 is perfect. Example: 20 is another example since 1+2 +4 + 7 + 14 in 28. Note, the number 1 is not a perfect number. your code here Sum - 0 for i in range(1, x) LE(0)1 Sun Bum + 1 if (Bum=-X) else! print(" td is a Perfect Number 1x) print(td is not a Perfect Number" 1x) print (isPerfect (6)) File "cipython-input-3-b9df3e539048", line 15 for i in range(1, x) IndentationError: unexpected indent In 1411 perfect numbers 16, 28, 496, 8126, 335503361 for i in perfect numbers: assert true (isPerfect (i), str(i)+ is perfect') not perfect numbers (2, 3, 4, 5, 7, 8, 9, 10, 495, 8127, 8129, 335503351 for i in not perfect numbers assert true (not (isPerfect (i)), str(i) is not perfect') test existence of doestring assert true(len (isPerfect._doc_) 1, "there is no docstring for isPerfect") print("Success!")

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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def g11_3(N) :'''Assume that N is a positive integer.Returns the sum from 1 to N inclusive, but omitting the numbers that are of type series1, meaning omit numbers that can be written as 1 + 2 + 3 + 4 + ... + k for some positive integer k. Hint: write a boolean function that checks if an integer is of typeseries1. For example, the first few series1 numbers are 1, 3, 6, 10, ...So, g11_3(10) returns 35since the numbers to add are: 2, 4, 5, 7, 8, 9, where weare not using 1, 3, 6 and 10 because they are of type series1.'''pass# --------------------------------------------------------------# The Testing# --------------------------------------------------------------class myTests(unittest.TestCase):def test1(self):ans = g11_3(1)self.assertEqual(ans, 0)def test2(self):ans = g11_3(2)self.assertEqual(ans, 2)def test3(self):ans = g11_3(10)self.assertEqual(ans, 35)def test4(self):ans = g11_3(100)self.assertEqual(ans, 4595)def test5(self):ans = g11_3(1000)self.assertEqual(ans, 485320)if __name__…
There are five errors in this code. Correct the errors and make the code work. Explain what are the errors found. import java.util.Scanner; //Testing.java //Displays running total of numbers in lines of standard //input correct to two decimal places. //Uses an out of range number (100) to quit. public class TWODPS{ public static void main (String [] args) { Scanner input new Scanner (System.in); double total=0; boolean flag=true; System.out.println ("Use an out of range entry 100 to quit."); while (flag) { System.out.println ("Enter a number on a line:"); double d = input.nextDouble (); if (outofRange (d) ) { flag=false; else{ dispTwoDPs ("The number value is: ",d); total = total + d; dispTwoDPs ("The total is: ",total); System.out.println (); System.out.println ("Next."); } //end of else }//end of while System.out.println ("You quit."); }//end of main static boolean outOfRange (double d) { if (d>-100) return true; if (d>100) return true; return false; }
Mailings Review View Help AaBbC AaBbCcD AaBbCcD AaBbCc Aa Heading 1 T Normal 1No Spac. Heading 2 Hea Paragraph Styles 8. Consider the following equation for the sum of a finite arithmetic series: S, = n(a, ta where n is the number of elements in the series, a is the first element of the series, and a, is the last element of the series. Write a method that returns the sum of the elements stored in the array in 8 above. The array must be passed to the method as an argument. Your solution should implement the equation for the sum S.
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Greatest Common Divisor Two integers A and B have the greatest common divisor which is the largest positive integer that divides A and B, evenly, without a remainder. One way to find the Greatest Common Divisor is to consider the prime factors for A = 372 and B = 84 as shown below. 372 223 * 31 84 22*3*7 GCD (372,84) = 2²*3= 12 In a well-documented Python program, hmwk3Q3.py, query the user for two integers. Use a while-loop to find the remainder R when dividing A by B.Replace the value of A with the value of B and replace the value of B with the value of R.Continue this process within the while-loop until the value of B is zero. At that time the value of A will be the GCD. In a comment, provide the response to your program when A= 98,025,733,547 and B = 2,345,109,894,323.
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1. Divisors: In a class Divisors.java, read in a maximum integer n and use nested loops to print a list of divisors for each of 1 through n as shown in the following transcript. Sample transcript (input underlined): Largest integer? 100 1: 1 2: 1 2 3: 1 3 4: 1 2 4 95: 1 5 19 95 96: 1 2 3 4 6 8 12 16 24 32 48 96 97: 1 97 98: 1 2 7 14 49 98 99: 1 3 9 11 33 99 100: 1 2 4 5 10 20 25 50 100
IN C++. show CODE PLEASE.
Here is a function that is supposed to return True if word (any string) ends with "s" and False otherwise: def ends_with_s(word):if len(word) < 1:return False return word[-1] == "s" Here is the test function I am using to test ends_with_s: def test_ends_with_s(): assert(ends_with_s("cats")) # option 1 assert(ends_with_s("") is False) # option 2 assert(ends_with_s("apple") is False) # option 3 assert(ends_with_s("S")) # option 4 Rewrite ends_with_s so that the test passes
]: Write a piece of code that calculates the uncertainty SP from the error propagation rule for sums, SP = 2√√√(SL)² + (SW)² A few hints: Again, you're translating the above equation into code. • Your result should be stored in a variable uncertainty_P_errorprop • For the square root function, use np. sqrt() • For squares, use ** #YOUR CODE HERE raise Not ImplementedError() ]: ▼ # Print the uncertainty print ("uncertainty of circumference P from error propagation: 11 , uncertainty_P_errorprop)