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svlad2 [7]
2 years ago
13

40 POINTS IF YOu ANSWER FOR EACH STEP.

Computers and Technology
1 answer:
NISA [10]2 years ago
8 0

Answer:

4 beeps indicate a Memory Read / Write failure. Try re-seating the Memory module by removing and reinserting it in the slot. This could mean it could've just jiggled loose after a while or dust, there could be a hundred other problems but those are the two most common.n

Explanation:

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A patient asks her doctor to send a copy of her records to another medical office. However, the records are not transmitted beca
bezimeni [28]
 This type of security failure is called WIN32K_SECURITY_FAILURE. <span> This indicates a </span>security failure was detected in win32k. <span>According to a new report issued by Dark Reading, there are a number of key </span>security failures<span> that cybercriminals take advantage of.</span>
5 0
2 years ago
Read 2 more answers
An employee sets up an automation that transfers files in a specific folder on their PC to a remote drive for archiving, provide
Yuri [45]

Answer:

Rule based automation

Explanation:

8 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
Modify the guessing-game program so that the user thinks of a number that the computer must guess.
Triss [41]

Answer:

import random

import math

smaller = int(input("Enter the smaller number: "))

larger = int(input("Enter the larger number: "))

count = 0

print()

while True:

   count += 1

   myNumber = (smaller + larger) // 2

   print('%d %d' % (smaller, larger))

   print('Your number is %d' % myNumber)

   choice = input('Enter =, <, or >: ')

   if choice == '=':

       print("Hooray, I've got it in %d tries" % count)

       break

   elif smaller == larger:

       print("I'm out of guesses, and you cheated")

       break

   elif choice == '<':

       larger = myNumber - 1

   else:

       smaller = myNumber + 1

Explanation:

The code to be Modified is as indicated below:

# Modify the code below:

import random

smaller = int(input("Enter the smaller number: "))

larger = int(input("Enter the larger number: "))

myNumber = random.randint(smaller, larger)

count = 0

while True:

   count += 1

   userNumber = int(input("Enter your guess: "))

   if userNumber < myNumber:

       print("Too small")

   elif userNumber > myNumber:

       print("Too large")

   else:

       print("You've got it in", count, "tries!")

       break

Detailed Explanation:

Line 3 and 4 of the codes prompts the user to enter a smaller and larger number. Line 5 we initiate the formula for minimum number of guesses. Line 14 and 17 prints out the output  I'm out of guesses, and you cheated and Hooray, I've got it in X after validating the conditions.

7 0
2 years ago
Why is project scope management so challenging in IT projects? What suggestions do you have for preventing scope creep in projec
seropon [69]

Answer:

Explanation:

<u>Ways to Avoid Scope Creep</u>

Scope creep is what happens when changes are made to the scope of a project without any control. Changes happen to projects all the time without been notify ontime as a project manager. It is that very rare project that ends up delivering exactly what was asked for on the first day. However, without there being some control over the changes, a project manager has little chance of keeping on top of the work and managing the project effectively.

Generally, scope creep is when new requirements are added after the project has commence. These changes are not properly reviewed. The project team is expected to deliver them with the same resources and in the same time as the original scope.

On the other hand, as a project manager you could end up with a project with lots of approved, considered changes, that never ends because every time you think you have finished a new requirement arrives in your inbox and you have to make more changes.

The following are five ways to keep control of your project.

<em>1-Document the Requirements</em>

<em>2-Set up Change Control Processes</em>

<em>3-Create a Clear Project Schedule</em>

<em>4-Verify the Scope with the Stakeholders</em>

<em>5-Engage the Project Team</em>

6 0
2 years ago
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