Question 25 Given the code below: def frequency (theList): counters = {} for item in theList: if item in counters + else: counters [ return counters counters: 1 = 1 ] += 1 grades [95, 96, 100, 85, 95, 90, 95, 100, 100] print (frequency (grades)) (95: 3, 96: 1, 100: 3, 85: 1, 901) Fill in the blanks in the code above so that the resulting output looks as follows:

Question 25 Given the code below: def frequency (theList): counters = {} for item in theList: if item in counters + else: counters [ return counters counters: 1 = 1 ] += 1 grades [95, 96, 100, 85, 95, 90, 95, 100, 100] print (frequency (grades)) (95: 3, 96: 1, 100: 3, 85: 1, 901) Fill in the blanks in the code above so that the resulting output looks as follows:

New Perspectives on HTML5, CSS3, and JavaScript

6th Edition

ISBN:9781305503922

Author:Patrick M. Carey

Publisher:Patrick M. Carey

Chapter12: Working With Document Nodes And Style Sheets: Creating A Dynamic Document Outline

Section12.2: Visual Overview: Exploring Attribute Nodes

Problem 2QC

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def city_dict(adict): """ Question 6 -Given a dictionary that maps a person to a list of countries they want to travel to, return a dictionary with the value being the list sorted by the last letter in each countries name. If two countries have the same last letter, sort by the first letter. -For an extra challenge, try doing this in one line (Optional). Args: adict (dict) Returns: dict >>> city_dict({"Pablo": ["Belgium", "Canada", "Germany"], "Athena": ["Italy", "France", "Egypt"], "Liv": ["Japan", "Bolivia", "Greece"]}) {'Pablo': ['Canada', 'Belgium', 'Germany'], 'Athena': ['France', 'Egypt', 'Italy'], 'Liv': ['Bolivia', 'Greece', 'Japan']} >>> city_dict({"Jacob": ["Bahamas", "Brazil", "Chile"], "Lexi": ["Colombia", "Finland", "Panama"], "Emily": ["Ireland", "Russia", "Kenya", "Jordan"]}) {'Jacob': ['Chile', 'Brazil', 'Bahamas'], 'Lexi': ['Colombia', 'Panama', 'Finland'], 'Emily': ['Kenya',…
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Question 20 A list is a collection with additional index- and iteration- related operations. True False Question 21 O(N) is the order of growth execution time of the size operation when using the SortedArrayCollection class, assuming a collection size of N. True False Question 22 If N represents the number of elements in the list, then the index-based set method of the ABList class is O(1). True False Question 23 O(N) is the order of growth execution time of the remove operation when using the LinkedCollection class, assuming a collection size of N. True False Question 24 It is not possible to use an array to implement a linked list. True False Question 25 O(N) is the order of growth execution time of the remove operation when using the ArrayCollection class, assuming a collection size of N. True False Question 26 Our linked implementation of lists implements a bounded list. True False Question 27 O(N) is the order of growth execution time of the contains operation…
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Question 30 If N represents the number of elements in the list, then the index-based add method of the LBList class is O(N). True False Question 31 A header node does not contain actual list information. True False Question 32 Any class that implements the Comparable interface must provide a compareTo method. True False Question 33 A SortedABList list can only be kept sorted based on the "natural order" of its elements. True False Question 34 O(N) is the order of growth execution time of the add operation when using the SortedArrayCollection class, assuming a collection size of N. True False Question 35 The iterator operation is required by the Iterable interface. Group of answer choices True False Question 36 O(N) is the order of growth execution time of the remove operation when using the SortedArrayCollection class, assuming a collection size of N. True False Question 37 O(N) is the order of growth execution time of the index-based add operation when…