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Anit [1.1K]
3 years ago
12

____ hack systems to conduct terrorist activities through network or Internet pathways

Computers and Technology
1 answer:
Advocard [28]3 years ago
6 0

Answer:

Cyberterrorists

Explanation:

Cyberterrorists -

These are the people who illegally , without any authentication , are able access some other person's personal system or internet server , are referred to as cyber terrorists.  

Internet service is used in this process , which can lead to loss of life , harm , threat or some illegal practice.  

Hence, from the statement of the question,  

The correct term is cyberterrorists.

You might be interested in
Write a function wordcount() that takes the name of a text file as input and prints the number of occurrences of every word in t
o-na [289]

Answer:

I am writing a Python program. Let me know if you want the program in some other programming language.        

import string  #to use string related functions

def wordcount(filename):  # function that takes a text file name as parameter and returns the number of occurrences of every word in file

   file = open(filename, "r")  # open the file in read mode

   wc = dict()  # creates a dictionary

   for sentence in file:  # loop through each line of the file

       sentence = sentence.strip()  #returns the text, removing empty spaces

       sentence=sentence.lower() #converts each line to lowercase to avoid case sensitivity

       sentence = sentence.translate(sentence.maketrans("", "", string.punctuation))  #removes punctuation from every line of the text file

       words = sentence.split(" ")  # split the lines into a list of words

       for word in words:  #loops through each word of the file

           if len(word)>2:  #checks if the length of the word is greater than 2

               if word in wc:  # if the word is already in dictionary

                   wc[word] = wc[word] + 1  #if the word is already present in dict wc then add 1 to the count of that word

               else:  #if the word is not already present

                   wc[word] = 1  # word is added to the wc dict and assign 1 to the count of that word                

   for w in list(wc.keys()):  #prints the list of words and their number of occurrences

       print(w, wc[w])  #prints word: occurrences in key:value format of dict        

wordcount("file.txt") #calls wordcount method and passes name of the file to that method

Explanation:

The program has a function wordcount that takes the name of a text file (filename) as parameter.

open() method is used to open the file in read mode. "r" represents the mode and it means read mode. Then a dictionary is created and named as wc. The first for loop, iterates through each line (sentence) of the text file. strip() method is used to remove extra empty spaces or new line character from each sentence of the file, then each sentence is converted to lower case using lower() method to avoid case sensitivity. Now the words "hello" and "Hello" are treated as the same word.

sentence = sentence.translate(sentence.maketrans("", "", string.punctuation))  statement uses two methods i.e. maketrans() and translate(). maketrans() specifies the punctuation characters that are to be deleted from the sentences and returns a translation table. translate() method uses the table that maketrans() returns in order to replace a character to its mapped character and returns the lines of text file after performing these translations.

Next the split() method is used to break these sentences into a list of words. Second for loop iterates through each word of the text file. As its given to ignore words of length 2 or less, so an IF statement is used to check if the length of word is greater than 2. If this statement evaluates to true then next statement: if word in wc:   is executed which checks if the word is already present in dictionary. If this statement evaluates to true then 1 is added to the count of that word. If the word is not already present  then the word is added to the wc dictionary and 1 s assigned to the count of that word.

Next the words along with their occurrences is printed. The program and its output are attached as screenshot. Since the frankenstein.txt' is not provided so I am using my own text file.

4 0
3 years ago
In which scenario would instant messaging be more useful than other forms of communication?
ivann1987 [24]
When you're busy doing things and/or be too sick for over the phone communication
7 0
3 years ago
How much time does a gold chest take to open
MaRussiya [10]

Gold chest takes 8 hours to open or unlock.

<u>Explanation:</u>

Clash royale is a game which has been developed and published with the help of a super cell. It is a video game. Many players can play this game at the same time. In the sense, clash royale is a multi player game.

In this game, the players who are playing the game are awarded with the gold chest. They are also awarded with the gems. In order to unlock or open a gold chest eight hours are needed. Apart from this there is an other way also with which the gold chest can be unlocked. It can be unlocked with the help of forty eight gems.

7 0
3 years ago
Good business ethics is a good marketing strategy. Discuss
Arada [10]
Good business ethics are the key to good marketing and success. Ethics portray whether an individual has good judgment, intelligence, and many more things that are essential to business. Without good ethics business as a whole would be ruthless. Some business ethics are even required by the law in some places and there are rules and regulation to trade. 

Hope that answered your question!
6 0
3 years ago
You are a visitor at a political convention with delegates; each delegate is a member of exactly one political party. It is impo
Fed [463]

Answer:

The algorithm is as follows:

Step 1: Start

Step 2: Parties = [All delegates in the party]

Step 3: Lent = Count(Parties)

Step 4: Individual = 0

Step 5: Index = 1

Step 6: For I in Lent:

Step 6.1: If Parties[Individual] == Parties[I]:

Step 6.1.1: Index = Index + 1

Step 6.2: Else:

Step 6.2.1 If Index == 0:

Step 6.2.2: Individual = I

Step 6.2.3: Index = 1

Step 7: Else

Step 7.1: Index = Index - 1

Step 8: Print(Party[Individual])

Step 9: Stop

Explanation:

The algorithm begins here

Step 1: Start

This gets the political parties as a list

Step 2: Parties = [All delegates in the party]

This counts the number of delegates i.e. the length of the list

Step 3: Lent = Count(Parties)

This initializes the first individual you come in contact with, to delegate 0 [list index begins from 0]

Step 4: Individual = 0

The next person on the list is set to index 1

Step 5: Index = 1

This begins an iteration

Step 6: For I in Lent:

If Parties[Individual] greets, shakes or smile to Party[i]

Step 6.1: If Parties[Individual] == Parties[I]:

Then they belong to the same party. Increment count by 1

Step 6.1.1: Index = Index + 1

If otherwise

Step 6.2: Else:

This checks if the first person is still in check

Step 6.2.1 If Index == 0:

If yes, the iteration is shifted up

Step 6.2.2: Individual = I

Step 6.2.3: Index = 1

If the first person is not being checked

Step 7: Else

The index is reduced by 1

Step 7.1: Index = Index - 1

This prints the highest occurrence party

Step 8: Print(Party[Individual])

This ends the algorithm

Step 9: Stop

The algorithm, implemented in Python is added as an attachment

<em>Because there is an iteration which performs repetitive operation, the algorithm running time is: O(n) </em>

Download txt
5 0
3 years ago
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