EC-COUNCIL 212-81 Exam Pass Guide Please email if you need to use more than 5 (five) computers, EC-COUNCIL 212-81 Exam Pass Guide It will be very easy for you to pass the exam and get the certification, In order to meet all demands of all customers, our company has employed a lot of excellent experts and professors in the field to design and compile the 212-81 test dump with a high quality, Our 212-81 exam simulator can help you and alleviate you from those problems with all content based on the real exam and the most effective high-quality and accuracy knowledge.
Database Design for Mere Mortals has earned worldwide respect as Exam 212-81 Pass Guide the simplest way to learn relational database design, You know it can be an important part of your online marketing strategy;
You will see how we use this later to create hyperlinks, You will well know the ability of our 212-81 test training vce clearly, We decided that sites needed to be more involved in security evaluations, enabling 212-81 Latest Real Exam them to learn about their security processes and participate in developing improvement recommendations.
Please email if you need to use more than 5 (five) computers, (https://www.exam4labs.com/212-81-practice-torrent.html) It will be very easy for you to pass the exam and get the certification, In order to meet all demandsof all customers, our company has employed a lot of excellent experts and professors in the field to design and compile the 212-81 test dump with a high quality.
Pass Guaranteed Quiz EC-COUNCIL - High-quality 212-81 Exam Pass Guide
Our 212-81 exam simulator can help you and alleviate you from those problems with all content based on the real exam and the most effective high-quality and accuracy knowledge.
Most of my friends were not even able to pass the 212-81 Latest Learning Material EC-COUNCIL exam on their first attempt because they only studied with books, You will find these amazing 212-81 test dumps highly compatible with your needs as well as quite in line with the real 212-81 exam questions.
Because of its high efficiency, you can achieve remarkable results, People always do things that will benefit them, so as get a certificate of the 212-81 test dumps.
We do not have hot lines, We offer you the best 212-81 Dumps PDF high quality and cost-effective Certified Encryption Specialist real exam dumps for you, you won't find any better one available, Our Certified Encryption Specialist free dumps can not only save time and money, but also help you pass 212-81 prep4sure exam with high pass rate.
So you can rely on us without any doubt.
Download Certified Encryption Specialist Exam Dumps
NEW QUESTION 48
Which one of the following attempts to hide data in plain view?
- A. Asymmetric cryptography
- B. Steganography
- C. Cryptography
- D. Substitution
Answer: B
Explanation:
Steganography
https://en.wikipedia.org/wiki/Steganography
Steganography is the practice of concealing a file, message, image, or video within another file, message, image, or video. The word steganography comes from Greek steganographia, which combines the words steganos , meaning "covered or concealed", and -graphia meaning "writing".
NEW QUESTION 49
During the process of encryption and decryption, what keys are shared?
- A. User passwords
- B. Public keys
- C. Private keys
- D. Public and private keys
Answer: B
Explanation:
Public keys
https://en.wikipedia.org/wiki/Public-key_cryptography
Public-key cryptography, or asymmetric cryptography, is a cryptographic system that uses pairs of keys: public keys, which may be disseminated widely, and private keys, which are known only to the owner. The generation of such keys depends on cryptographic algorithms based on mathematical problems to produce one-way functions. Effective security only requires keeping the private key private; the public key can be openly distributed without compromising security.
In such a system, any person can encrypt a message using the receiver's public key, but that encrypted message can only be decrypted with the receiver's private key.
Alice and Bob have two keys of their own - just to be clear, that's four keys total. Each party has their own public key, which they share with the world, and their own private key which they well, which they keep private, of course but, more than that, which they keep as a closely guarded secret. The magic of public key cryptography is that a message encrypted with the public key can only be decrypted with the private key. Alice will encrypt her message with Bob's public key, and even though Eve knows she used Bob's public key, and even though Eve knows Bob's public key herself, she is unable to decrypt the message. Only Bob, using his secret key, can decrypt the message assuming he's kept it secret, of course.
Alice and Bob do not need to plan anything ahead of time to communicate securely: they generate their public-private key pairs independently, and happily broadcast their public keys to the world at large. Alice can rest assured that only Bob can decrypt the message she sends because she has encrypted it with his public key.
NEW QUESTION 50
John is responsible for VPNs at his company. He is using IPSec because it has two different modes. He can choose the mode appropriate for a given situation. What are the two modes of IPSec? (Choose two)
- A. Transport mode
- B. Decrypt mode
- C. Tunnel mode
- D. Encrypt mode
Answer: A,C
Explanation:
Correct answers: Transport mode and Tunnel mode
https://en.wikipedia.org/wiki/IPsec#Modes_of_operation
The IPsec protocols AH and ESP can be implemented in a host-to-host transport mode, as well as in a network tunneling mode.
NEW QUESTION 51
What is the formula m^e %n related to?
- A. Decrypting with RSA
- B. Encrypting with EC
- C. Encrypting with RSA
- D. Generating Mersenne primes
Answer: C
Explanation:
Encrypting with RSA
https://en.wikipedia.org/wiki/RSA_(cryptosystem)
RSA Encrypting a message m (number) with the public key (n, e) is calculated:
M' := m^e %n
Incorrect answers:
Decrypting with RSA:
M'' := m^d %n
Generation Mersenne primes:
Mn = 2^n - 1
Encrypting with Elliptic Curve (EC):
y^2 = x^3 + ax + b
NEW QUESTION 52
Cylinder tool. Wrap leather around to decode. The diameter is the key. Used in 7th century BC by greek poet Archilochus.
- A. Enigma machine
- B. Caesar cipher
- C. Cipher disk
- D. Scytale
Answer: D
Explanation:
Scytale
https://en.wikipedia.org/wiki/Scytale
A scytale is a tool used to perform a transposition cipher, consisting of a cylinder with a strip of parchment wound around it on which is written a message. The ancient Greeks, and the Spartans in particular, are said to have used this cipher in 7th century BC to communicate during military campaigns.
The recipient uses a rod of the same diameter on which the parchment is wrapped to read the message. It has the advantage of being fast and not prone to mistakes-a necessary property when on the battlefield. It can, however, be easily broken. Since the strip of parchment hints strongly at the method, the ciphertext would have to be transferred to something less suggestive, somewhat reducing the advantage noted.
Incorrect answers:
Cipher disk - is an enciphering and deciphering tool developed in 1470 by the Italian architect and author Leon Battista Alberti. He constructed a device, (eponymously called the Alberti cipher disk) consisting of two concentric circular plates mounted one on top of the other. The larger plate is called the "stationary" and the smaller one the "moveable" since the smaller one could move on top of the "stationary".
Enigma machine - is an encryption device developed and used in the early- to mid-20th century to protect commercial, diplomatic and military communication. It was employed extensively by Nazi Germany during World War II, in all branches of the German military.
Caesar cipher - (also known as Caesar's cipher, the shift cipher, Caesar's code or Caesar shift) is one of the simplest and most widely known encryption techniques. It is a type of substitution cipher in which each letter in the plaintext is replaced by a letter some fixed number of positions down the alphabet. For example, with a left shift of 3, D would be replaced by A, E would become B, and so on. The method is named after Julius Caesar, who used it in his private correspondence.
NEW QUESTION 53
......
