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Huawei H19-491_V1.0 - HCSE-Presales-Finance V1.0 Exam Braindumps

Huawei H19-491_V1.0 - HCSE-Presales-Finance V1.0 Exam Braindumps

  • Certification Provider:Huawei
  • Exam Code:H19-491_V1.0
  • Exam Name:HCSE-Presales-Finance V1.0 Exam Exam
  • Total Questions:276 Questions and Answers
  • Product Format: PDF & Test Engine Software Version
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NEW QUESTION: 1
フォレンジックアナリストは、サーバーで進行中のネットワーク攻撃に対応するよう求められます。フォレンジックアナリストが保存する正しい順序で、以下のリストのアイテムを配置します。

Answer:
Explanation:

Explanation:
When dealing with multiple issues, address them in order of volatility (OOV); always deal with the most volatile first. Volatility can be thought of as the amount of time that you have to collect certain data before a window of opportunity is gone. Naturally, in an investigation you want to collect everything, but some data will exist longer than others, and you cannot possibly collect all of it once. As an example, the OOV in an investigation may be RAM, hard drive data, CDs/DVDs, and printouts.
Order of volatility: Capture system images as a snapshot of what exists, look at network traffic and logs, capture any relevant video/screenshots/hashes, record time offset on the systems, talk to witnesses, and track total man-hours and expenses associated with the investigation.

NEW QUESTION: 2

A. Option A
B. Option B
C. Option D
D. Option C
Answer: D

NEW QUESTION: 3
Under which circumstance should a network administrator implement one-way NAT?
A. when traffic that originates outside the network must be routed to internal hosts
B. when traffic that originates inside the network must be routed to internal hosts
C. when the network must route UDP traffic
D. when the network has few public IP addresses and many private IP addresses require outside access
Answer: A
Explanation:
NAT operation is typically transparent to both the internal and external hosts. Typically the internal host is aware of the true IP address and TCP or UDP port of the external host. Typically the NAT device may function as the default gateway for the internal host. However the external host is only aware of the public IP address for the NAT device and the particular port being used to communicate on behalf of a specific internal host.
NAT and TCP/UDP
"Pure NAT", operating on IP alone, may or may not correctly parse protocols that are totally concerned with IP information, such as ICMP, depending on whether the payload is interpreted by a host on the "inside" or "outside" of translation. As soon as the protocol stack is traversed, even with such basic protocols as TCP and UDP, the protocols will break unless NAT takes action beyond the network layer. IP packets have a checksum in each packet header, which provides error detection only for the header. IP datagrams may become fragmented and it is necessary for a NAT to reassemble these fragments to allow correct recalculation of higher-level checksums and correct tracking of which packets belong to which connection. The major transport layer protocols, TCP and UDP, have a checksum that covers all the data they carry, as well as the TCP/UDP header, plus a "pseudo-header" that contains the source and destination IP addresses of the packet carrying the TCP/UDP header. For an originating NAT to pass TCP or UDP successfully, it must recompute the TCP/UDP header checksum based on the translated IP addresses, not the original ones, and put that checksum into the TCP/UDP header of the first packet of the fragmented set of packets. The receiving NAT must recompute the IP checksum on every packet it passes to the destination host, and also recognize and recompute the TCP/UDP header using the retranslated addresses and pseudo-header. This is not a completely solved problem. One solution is for the receiving NAT to reassemble the entire segment and then recompute a checksum calculated across all packets.
The originating host may perform Maximum transmission unit (MTU) path discovery to determine the packet size that can be transmitted without fragmentation, and then set the don't fragment (DF) bit in the appropriate packet header field. Of course, this is only a one- way solution, because the responding host can send packets of any size, which may be fragmented before reaching the NAT.

NEW QUESTION: 4
Judgement rating is used in case of
A. Health Insurance
B. Life Insurance
C. Fire Insurance
D. Ocean Marine Insurance
Answer: D

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