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EastWind [94]
3 years ago
13

When employees are hired, they agree to only use cell phones during breaks. How would you classify this?

Computers and Technology
2 answers:
lilavasa [31]3 years ago
8 0

Answer:

Workplace Policy

Explanation:

It is usually a policy due to classified things that go on in the work place

AlexFokin [52]3 years ago
5 0

Answer: workplace policy. I had this question

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How can you quickly access the desktop when you have many applications open.
murzikaleks [220]

Answer:

Clear all your cookies and browing data or delete the tabs that u have or do a clean up of your desktop every once in a while it say it will clean my laptop so it throw out all the trash stuff that is in it so baiscally it mean that it'll reset it.

4 0
1 year ago
How do you do this question?
laila [671]

Answer:

(a) 1 to 8

(b) 1 to 6

Explanation:

A "leaf" is a node at the end of a binary tree (in other words, it has no "children").  All other nodes are "non-leaf" nodes.

The smallest number of leaves is 1.  That would be a binary tree that's just a straight line; each node will have only 1 child, until you get to the last node (the leaf).

To find the largest number of leaves, we start drawing a full binary tree.  A complete tree with 15 nodes has 7 non-leaf nodes and 8 leaf nodes.  A full tree with 6 non-leaf nodes can have up to 6 leaf nodes.

6 0
3 years ago
Read 2 more answers
Please create a brute force password cracker using python
serg [7]

Answer:

0

2

I need to make small programs for school to brute force crack different types of passwords; I'm looking to create a brute force python code that will run through every possible combination of alphabetical and alphanumerical passwords and give me the password and the amount of time it took to crack.

I did the same with purely numerical passwords and got this:

import datetime as dt

Password4 = 123456

def crack_password():

   start = dt.datetime.now()

   for n in range(1000000):

       password_guess = '{0:04d}'.format(n)

            if password_guess == str(Password4):

               end = dt.datetime.now()

               print("Password found: {} in {}".format(password_guess, end - start))

              break

   guesses = crack_password()

I then tried to do something somewhat similar for alphabet/alphanumerical passwords but did not work whatever I tried:

   import random

   letters = [str(i) for i in range('a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p')]

   s = [''.join([a,b,c,d,e,f,g,h]) for a in letters for b in letters for c   in letters for d in letters for e in letters for f in letters for g in letters  for h in letters]

   random.shuffle(s)

   real_password = 'aaaaaaaa'

   i = 0

Explanation:

8 0
3 years ago
The process by which information gets into memory storage is
Kay [80]

Answer:

That is the first step and it is called encoding process.

7 0
2 years ago
an IPv6 packet has a 40 byte base header, a 20 byte destination options extension header (which is not used for routing) and 200
NISA [10]

Answer:

Explanation:

Given, total payload is: 2000 bytes

 MTU is: 1500 bytes

 Header is: 20 bytes

 Base header is: 40 bytes

Total number of fragments = Total Payload / MTU

                                            = 2500 / 1500

                                            = 2 fragments

Fragment     Base header    Fragmentation header     Authentication  

                                                                                                     header + Data

1                  40 bytes                  8 bytes                              20 bytes +772bytes

2                 40 bytes                   8 bytes                                 792 bytes

7 0
3 years ago
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