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Leokris [45]
3 years ago
5

The work day has just started and you receive reports that the inventory management server is not accessible on your company's n

etwork. You recall that the new network administration assistant was working on that server last night. Which tool can you use to determine if the network administration assistant left that server's NIC disabled?
Computers and Technology
1 answer:
Reptile [31]3 years ago
5 0

Answer:

Server Manager is the correct answer.

Explanation:

The user uses the Server Manager to determine if the admin of the network helper left the following server because that user received the reports related to the inventory management that is unable to access on their organizations' network. That's why he recall the new admin of the network. So, the following answer is correct.

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True

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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
enot [183]

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,

  • If Linked list is empty then make the node as head and return it.
  • If value of the node to be inserted is smaller than value of head node, then insert the node at start and make it head.
  • In a loop, find the appropriate node after which the input node ( 10 ) is to be inserted.To find the appropriate node start from head,keep moving until you reach a node GN (12 ) who's value is greater than the input node. The node just before GN is the appropriate node (8).
  • Insert the node (10) after the appropriate node (8) found in step 3.

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

3 0
4 years ago
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