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H12-723-ENU Test Collection - Huawei H12-723-ENU Valid Braindumps Free, Latest Study H12-723-ENU Questions - FreeTorrent

H12-723-ENU

Exam Code: H12-723-ENU

Exam Name: HCIP-Security-CTSS(Huawei Certified ICT Professional -Constructing Terminal Security System)

Version: V22.75

Q & A: 580 Questions and Answers

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NEW QUESTION: 1
Which of the following is NOT a symmetric key algorithm?
A. Triple DES (3DES)
B. RC5
C. Blowfish
D. Digital Signature Standard (DSS)
Answer: D
Explanation:
Explanation/Reference:
Explanation:
Digital Signature Standard (DSS) is not a symmetric key algorithm; it is an asymmetric key algorithm.
Because digital signatures are so important in proving who sent which messages, the U.S. government decided to establish standards pertaining to their functions and acceptable use. In 1991, NIST proposed a federal standard called the Digital Signature Standard (DSS). It was developed for federal departments and agencies, but most vendors also designed their products to meet these specifications. The federal government requires its departments to use DSA, RSA, or the elliptic curve digital signature algorithm (ECDSA) and SHA. SHA creates a 160-bit message digest output, which is then inputted into one of the three mentioned digital signature algorithms. SHA is used to ensure the integrity of the message, and the other algorithms are used to digitally sign the message. This is an example of how two different algorithms are combined to provide the right combination of security services. RSA and DSA are the best known and most widely used digital signature algorithms. DSA was developed by the NSA. Unlike RSA, DSA can be used only for digital signatures, and DSA is slower than RSA in signature verification. RSA can be used for digital signatures, encryption, and secure distribution of symmetric keys.
Incorrect Answers:
A: Blowfish is a block symmetric cipher that uses 64-bit block sizes and variable-length keys.
C: Triple DES is a symmetric cipher that applies DES three times to each block of data during the encryption process.
D: RC5 is a block symmetric cipher that uses variable block sizes (32, 64, 128) and variable-length key sizes (0-2040).
References:
Harris, Shon, All In One CISSP Exam Guide, 6th Edition, McGraw-Hill, 2013, p. 832

NEW QUESTION: 2
Which three settings are required for crypto map configuration? (Choose three.)
A. match address
B. set security-association level per-host
C. set transform-set
D. set peer
E. set security-association lifetime
F. set pfs
Answer: A,C,D

NEW QUESTION: 3
Click the Exhibit button.

An administrator installed UTA2 PCIe cards into an existing FAS8200 HA pair. The new cards will be used for native FC traffic. The administrator wants to configure the UTA2 ports for FC client access.
Referring to the exhibit, which two actions address the requirement? (Choose two.)
A. Use the ucadmin modify command to set the adapter mode to FC.
B. Reboot the node to bring the adapter online after final configuration.
C. Use the network fcp adapter modify command to bring the adapter online.
D. Use the system hardware unified-connect command to verify that the port mode is set to initiator.
Answer: A,B
Explanation:
B: Ports must be converted in pairs, for example, 0c and 0d, after which, a reboot is required, and the ports must be brought back to the up state.
D: CNA ports can be configured into native Fibre Channel (FC) mode or CNA mode.
If the current configuration does not match the desired use (here we want to change from CNA mode to FC mode), enter the following commands to change the configuration as needed:
system node hardware unified-connect modify -node node_name -adapter adapter_name - mode fc|cna -type target|initiator
-mode is the personality type, fc or 10GbE cna.
-type is the FC4 type, target or initiator.
References: https://library.netapp.com/ecmdocs/ECMP1151595/html/GUID-1D06A7C4-F354-4793-A331-68A2CD916084.html

NEW QUESTION: 4

최근 인수를 통해 조직이 성장했습니다. 구매 한 회사 중 두 곳이 동일한 IP CIDR 범위를 사용합니다. AnyCompany A (VPC-A)가 AnyCompany B (VPC-B)에서 IP 주소가 10.0.0.77 인 서버와 통신 할 수 있도록 하는 새로운 단기 요구 사항이 있습니다. AnyCompany A는 AnyCompany C (VPC-C)의 모든 리소스와도 통신해야 합니다. 네트워크 팀이 VPC 피어 링크를 만들었지 만 VPC-A와 VPC-B 간의 통신에 문제가 있습니다. 조사 후 팀은 VPC의 라우팅 테이블이 올바르지 않다고 생각합니다.
AnyCompany A가 AnyCompany B의 데이터베이스 외에 AnyCompany C와 통신 할 수 있는 구성은 무엇입니까?
A. VPC-A에서 VPC 피어 pcx-AB를 통해 VPC-B CIDR 범위 (10.0.0.0/24)에 대한 고정 경로를 생성하고 VPC 피어 pcx-AC에서 10.0.0.0/16의 고정 경로를 생성합니다. VPC-B에서 피어 pcx-AB의 VPC-A CIDR (172.16.0.0/24)에 대한 고정 경로를 생성하고 VPC-C에서 VPC-A CIDR (172.16.0.0/24)에 대한 고정 경로를 생성합니다. 피어 pcx-AC.
B. VPC-A에서 pcx-AB 및 pcx-AC에서 동적 경로 전파를 활성화하고 VPC-B에서 동적 경로 전파를 활성화하고 보안 그룹을 사용하여 VPC 피어 pcx에서 IP 주소 10.0.0.77/32 만 허용합니다. -AB. VPC-C에서 피어 pcx-AC에서 VPC-A를 사용하여 동적 경로 전파를 활성화합니다.
C. VPC-A에서 VPC 피어 pcx-AB를 통해 VPC-B CIDR (10.0.0.77/32) 데이터베이스에 대한 고정 경로를 생성합니다. VPC 피어 pcx-AC에서 VPC-C CIDR에 대한 고정 경로를 생성합니다. VPC-B에서 피어 pcx-AB의 VPC-A CIDR (172.16.0.0/24)에 대한 고정 경로를 생성하고 VPC-C에서 VPC-A CIDR (172.16.0.0/24)에 대한 고정 경로를 생성합니다. 피어 pcx-AC.
D. VPC-A에서 VPC 피어 pcx-AC의 IP 주소 10.0.0.77/32를 차단하는 네트워크 액세스 제어 목록을 생성하고 VPC-A에서 VPC-B CIDR (10.0.0.0/)에 대한 고정 경로를 생성합니다. 24) pcx-AB 및 pcx-AC의 VPC-C CIDR (10.0.0.0/24)에 대한 고정 경로 VPC-B에서 피어를 통해 VPC-A CIDR (172.16.0.0/24)에 대한 고정 경로를 만듭니다. pcx-AB. VPC-C에서 피어 pcx-AC를 통해 VPC-A CIDR (172.16.0.0/24)에 대한 고정 경로를 만듭니다.
Answer: C

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