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Tpy6a [65]
3 years ago
6

Zeke used primarily Web sources for his informative speech about gun control. However, his over-reliance on the website sponsore

d by the National Rifle Association caused his speech to be slanted in one direction, rather than presenting multiple sides of the issue. Which of the six criteria for evaluating Web sources Zeke ignored? a. objectivity b. accuracy c. usability d. accountability
Computers and Technology
1 answer:
Andreas93 [3]3 years ago
7 0

Answer:

a. Objectivity

Explanation:

Zeke ignored objectivity as a criteria for evaluating Web sources.

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I recommend that he use this thesis statement :

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Is a list of instructions an arguments?<br> Yes or No
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Explanation:

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Define an iterative function named append_ordered; it is passed two arguments: a linked list (ll) whose values are ordered from
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Answer:

A python programming language (code) was used for this exercise.

The code is shown below in the explanation section

Explanation:

Solution

Python code:

class LN:

def __init__(self,value,next=None):

self.value = value

self.next = next

  def list_to_ll(self,l):

if l == []:

return None

front = rear = LN(l[0])

for v in l[1:]:

rear.next = LN(v)

rear = rear.next

return front

  def str_ll(self,ll):

answer = ''

while ll != None:

answer += str(ll.value)+'->'

ll = ll.next

return answer + 'None'

def append_ordered(self,ll,v):

ln1=LN(v); #create a node with the new value by creating an instance of the LN class

if ll is None: #if the linked list is empty

ll=ln1 #return the new created node

elif(ll.value >= ln1.value): #if the value to be add is smallest, append the linked list to it

ln1.next=ll

ll=ln1

else:

current=ll; #crate a temporary pointer to iterate over linked list

#iterate till the next node value is smaller than the new value and list is not over

while (current.next is not None and current.next.value < ln1.value):

current=current.next

ln1.next=current.next #save the current node in the next of the new node

current.next=ln1 #save new node in the next of current node

return ll

ln=LN(0); #create an instance of class LN

ll1=ln.list_to_ll([1,2,4,5]); #create a linked list of the list

answer=ln.str_ll(ll1);    

print(answer); #print the linked list

ll2=ln.append_ordered(ll1,3); #append the value 3 in the linked list

answer=ln.str_ll(ll2)

print(answer); #print the new linked list

Now,

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Note: Kindly find the output python code attached to the question below.

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Answer:

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