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Scorpion4ik [409]
2 years ago
14

A small company is moving towards sharing printers to reduce the number of printers used within the company. The technician has

security concerns. What security risk is associated with sharing a printer?
Computers and Technology
1 answer:
Misha Larkins [42]2 years ago
8 0

Answer:

Security risk associated with sharing a printer are

  1. Printer Attacks
  2. Theft
  3. Breach of data
  4. Vulnerable Network

Explanation:

Printer Attacks

A network printer can be used for a DDoS attack.As printer are not very secured and are a weak link in network these can be easily exploited by the hackers for any kind of malicious activities and even lanching a DDoS attack.

DDoS attack is denial of service attack in which network is flooded with malicious traffic which cause it to choke and make it inaccessible for users.

Theft

Physical theft of document can be an issue.Anyone can just took printed pages from printer tray by any one.

Breach of Data

The documents which are printed are usually stored in printer cache for some time which can be accessed by any one connected to the network. Any document containing confidential information which are printed on network printer can fall in wrong hands.

Vulnerable Network

As mentioned a single unsecured network printer can pose great threat to entire network as it can provide a way into the network.

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#Write a function called string_finder. string_finder should #take two parameters: a target string and a search string. #The fun
adoni [48]

Answer:

I am writing a Python program:

def string_finder(target,search): #function that takes two parameters i.e. target string and a search string

   position=(target.find(search))# returns lowest index of search if it is found in target string

   if position==0: # if value of position is 0 means lowers index

       return "Beginning" #the search string in the beginning of target string

   elif position== len(target) - len(search): #if position is equal to the difference between lengths of the target and search strings

       return "End" # returns end

   elif position > 0 and position < len(target) -1: #if value of position is greater than 0 and it is less than length of target -1

       return "Middle" #returns middle        

   else: #if none of above conditions is true return not found

       return "not found"

#you can add an elif condition instead of else for not found condition as:

#elif position==-1    

#returns "not found"

#tests the data for the following cases      

print(string_finder("Georgia Tech", "Georgia"))

print(string_finder("Georgia Tech", "gia"))

print(string_finder("Georgia Tech", "Tech"))

print(string_finder("Georgia Tech", "Idaho"))

Explanation:

The program can also be written in by using string methods.

def string_finder(target,search):  #method definition that takes target string and string to be searched

       if target.startswith(search):  #startswith() method scans the target string and checks if the (substring) search is present at the start of target string

           return "Beginning"  #if above condition it true return Beginning

       elif target.endswith(search):  #endswith() method scans the target string and checks if the (substring) search is present at the end of target string

           return "End" #if above elif condition it true return End

       elif target.find(search) != -1:  #find method returns -1 if the search string is not in target string so if find method does not return -1 then it means that the search string is within the target string and two above conditions evaluated to false so search string must be in the middle of target string

           return "Middle"  #if above elif condition it true return End

       else:  #if none of the above conditions is true then returns Not Found

           return "Not Found"

6 0
1 year ago
WILL MARK BRAINLIEST HELP
hichkok12 [17]

My guess would be A.

3 0
2 years ago
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A regional bank implemented an automated solution to streamline their operations for receiving and processing checks/cheques. Th
Sphinxa [80]

Answer: Machine learning

Explanation:

The technology that could be combined with the current solution to do this is the machine learning.

Machine learning refers to the use and development of the computer systems which can learn and adapt without them following explicit instructions. This is done through the use of statistical models and algorithms in order to analyse inferences from the patterns in data.

Since the bank wants to streamline their operations for the receiving and processing checks while also enhancing the solution to recognize signs of potential check fraud, then the machine learning can be used.

3 0
1 year ago
assume that name is a variable of type stirng that has been assigned a value write an expression whose value is the last charact
Snezhnost [94]

Answer:

The most straight forward way to do it: in general string are zero index based array of characters, so you need to get the length of the string, subtract one and that will be the last character, some expressions in concrete languages would be:

In Python:

name = "blair"

name[len(name) - 1]

In JavaScript:

name = "blair"

name[name.length - 1]

In C++:

#include <string>

string name = "blair";

name[name.length() - 1];

7 0
1 year ago
Generating a signature with RSA alone on a long message would be too slow (presumably using cipher block chaining). Suppose we c
boyakko [2]

Answer:

Following are the algorithm to this question:

Explanation:

In the RSA algorithm can be defined as follows:  

In this algorithm, we select two separate prime numbers that are the "P and Q", To protection purposes, both p and q combines are supposed to become dynamically chosen but must be similar in scale but 'unique in length' so render it easier to influence. Its value can be found by the main analysis effectively.  

Computing N = PQ.  

In this, N can be used for key pair, that is public and private together as the unit and the Length was its key length, normally is spoken bits. Measure,

\lambda (N) = \ lcm( \lambda  (P), \lambda (Q)) = \ lcm(P- 1, Q - 1)  where \lambda is the total function of Carmichaels. It is a privately held value. Selecting the integer E to be relatively prime from 1and gcd(E,  \lambda (N) ) = 1; that is E \ \ and  \ \ \lambda (N).  D was its complex number equivalent to E (modulo \lambda (N) ); that is d was its design multiplicative equivalent of E-1.  

It's more evident as a fix for d provided of DE ≡ 1 (modulo \lambda (N) ).E with an automatic warning latitude or little mass of bigging contribute most frequently to 216 + 1 = 65,537 more qualified encrypted data.

In some situations it's was shown that far lower E values (such as 3) are less stable.  

E is eligible as a supporter of the public key.  

D is retained as the personal supporter of its key.  

Its digital signature was its module N and the assistance for the community (or authentication). Its secret key includes that modulus N and coded (or decoding) sponsor D, that must be kept private. P, Q, and \lambda (N) will also be confined as they can be used in measuring D. The Euler totient operates \varphi (N) = (P-1)(Q - 1) however, could even, as mentioned throughout the initial RSA paper, have been used to compute the private exponent D rather than λ(N).

It applies because \varphi (N), which can always be split into  \lambda (N), and thus any D satisfying DE ≡ 1, it can also satisfy (mod  \lambda (N)). It works because \varphi (N), will always be divided by \varphi (N),. That d issue, in this case, measurement provides a result which is larger than necessary (i.e. D >   \lambda (N) ) for time - to - time). Many RSA frameworks assume notation are generated either by methodology, however, some concepts like fips, 186-4, may demand that D<   \lambda (N). if they use a private follower D, rather than by streamlined decoding method mostly based on a china rest theorem. Every sensitive "over-sized" exponential which does not cooperate may always be reduced to a shorter corresponding exponential by modulo  \lambda (N).

As there are common threads (P− 1) and (Q – 1) which are present throughout the N-1 = PQ-1 = (P -1)(Q - 1)+ (P-1) + (Q- 1)), it's also possible, if there are any, for all the common factors (P -1) \ \ \ and \ \ (Q - 1)to become very small, if necessary.  

Indication: Its original writers of RSA articles conduct their main age range by choosing E as a modular D-reverse (module \varphi (N)) multiplying. Because a low value (e.g. 65,537) is beneficial for E to improve the testing purpose, existing RSA implementation, such as PKCS#1, rather use E and compute D.

8 0
2 years ago
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