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C_C4H63_2411 Online Test & C_C4H63_2411 Examsfragen - SAP Certified Associate - Implementation Consultant - SAP Customer Data Platform Pruefungssimulationen - Photoexperienceacademy

  • Name: SAP Certified Associate - Implementation Consultant - SAP Customer Data Platform
  • Exam Code: C_C4H63_2411
  • Certification: C_C4H63_2411
  • Vendor: SAP
  • Total Question: 630
  • Price: 49$

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NEW QUESTION: 1
Which of the following answers is described as a random value used in cryptographic algorithms to ensure that patterns are not created during the encryption process?
A. IV - Initialization Vector
B. OTP - One Time Pad
C. Ciphertext
D. Stream Cipher
Answer: A
Explanation:
The basic power in cryptography is randomness. This uncertainty is why encrypted data is unusable to someone without the key to decrypt.
Initialization Vectors are a used with encryption keys to add an extra layer of randomness to encrypted data. If no IV is used the attacker can possibly break the keyspace because of patterns resulting in the encryption process. Implementation such as DES in Code Book Mode (CBC) would allow frequency analysis attack to take place.
In cryptography, an initialization vector (IV) or starting variable (SV)is a fixed-size input to a cryptographic primitive that is typically required to be random or pseudorandom.
Randomization is crucial for encryption schemes to achieve semantic security, a property
whereby repeated usage of the scheme under the same key does not allow an attacker to
infer relationships between segments of the encrypted message. For block ciphers, the use
of an IV is described by so-called modes of operation. Randomization is also required for
other primitives, such as universal hash functions and message authentication codes
based thereon.
It is define by TechTarget as:
An initialization vector (IV) is an arbitrary number that can be used along with a secret key
for data encryption. This number, also called a nonce, is employed only one time in any
session.
The use of an IV prevents repetition in data encryption, making it more difficult for a hacker
using a dictionary attack to find patterns and break a cipher. For example, a sequence
might appear twice or more within the body of a message. If there are repeated sequences
in encrypted data, an attacker could assume that the corresponding sequences in the
message were also identical. The IV prevents the appearance of corresponding duplicate
character sequences in the ciphertext.
The following answers are incorrect:
-Stream Cipher: This isn't correct. A stream cipher is a symmetric key cipher where plaintext digits are combined with pseudorandom key stream to product cipher text.
-OTP - One Time Pad: This isn't correct but OTP is made up of random values used as key material. (Encryption key) It is considered by most to be unbreakable but must be changed with a new key after it is used which makes it impractical for common use.
-Ciphertext: Sorry, incorrect answer. Ciphertext is basically text that has been encrypted with key material (Encryption key)
The following reference(s) was used to create this question:
For more details on this TOPIC and other questions of the Security+ CBK, subscribe to our Holistic Computer Based Tutorial (CBT) at http://www.cccure.tv and whatis.techtarget.com/definition/initialization-vector-IV and en.wikipedia.org/wiki/Initialization_vector

NEW QUESTION: 2
SIMULATION
A network associate is configuring a router for the weaver company to provide internet access. The ISP has provided
the company six public IP addresses of 198.18.184.105 198.18.184.110. The company has 14 hosts that need to access
the internet simultaneously. The hosts in the company LAN have been assigned private space addresses in the range
of 192.168.100.17 - 192.168.100.30.
C_C4H63_2411 Online Test
C_C4H63_2411 Online Test
C_C4H63_2411 Online Test
Answer:
Explanation:
Here are the Steps for this Lab Solution:
Explanation:
The above named organization has 14 hosts that need to access the internet simultaneously but were provided with
just 6 public IP addresses from198.18.184.105 to 198.18.184.110/29.
In this case, you have to consider using NAT Overload (or PAT)
Double Click on the Weaver router to access the CLI
Router> enable
Router# configure terminal
First you should change the router's name to Weaver:
Router(config)#hostname Weaver
Create a NAT pool of global addresses to be allocated with their netmask:
Weaver(config)# ip nat pool mypool 198.18.184.105 198.18.184.110 netmask 255.255.255.248
Create a standard access control list that permits the addresses that are to be translated:
Weaver(config)#access-list 1 permit 192.168.100.16 0.0.0.15
Establish dynamic source translation, specifying the access list that was defined in the prior step:
Weaver(config)#ip nat inside source list 1 pool mypool overload
Finally, we should save all your work with the following command:
Weaver#copy running-config startup-config (Don't forget this)
Check your configuration by going to "Host for testing" and type:
C : \ >ping 192.0.2.114
The ping should work well and you will be replied from 192.0.2.114
This command translates all source addresses that pass access list 1, which means a source address from
192.168.100.17 to 192.168.100.30, into an address from the pool named mypool (the pool contains addresses from
198.18.184.105 to 198.18.184.110) Overload keyword allows to map multiple IP addresses to a single registered IP
address (many-to- one) by using different ports.
The question said that appropriate interfaces have been configured for NAT inside and NAT outside statements.
This is how to configure the NAT inside and NAT outside, just for your understanding:
Weaver(config)#interface fa0/0
Weaver(config-if)#ip nat inside
Weaver(config-if)#exit
Weaver(config)#interface s0/0
Weaver(config-if)#ip nat outside
Weaver(config-if)#end

NEW QUESTION: 3
C_C4H63_2411 Online Test
C_C4H63_2411 Online Test
C_C4H63_2411 Online Test
A. Option C
B. Option A
C. Option D
D. Option B
Answer: C


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