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NEW QUESTION: 1
효과적인 취약성 관리 프로세스를 만들려면 아래 단계를 주문하십시오.
Answer:
Explanation:
Explanation
NEW QUESTION: 2
Your network contains an Active Directory forest named contoso.com.
The forest contains the domain controllers configured as shown in the following table.
You plan to deploy an Exchange Server 2013 organization to the forest. All Exchange Server 2013 servers will be located in the West site.
A company policy prevents administrators from logging on to DC2 to perform installations.
You need to prepare the Active Directory schema for the planned Exchange Server deployment.
You verify that your user account has all the required permissions to achieve the task. The solution must minimize changes to the Active Directory infrastructure.
What is the best approach to prepare the schema? More than one answer choice may achieve the goal.
Select the BEST answer.
A. Deploy a member server to contoso.com, and then move the server to the West site. Run adprep.exe from the member server.
B. Log on to DC3, and then run Exchange setup.
C. Move the schema master role to DC3, and then run Exchange setup.
D. Log on to DC1 and then run Exchange setup.
Answer: D
NEW QUESTION: 3
Under which circumstance should a network administrator implement one-way NAT?
A. when the network must route UDP traffic
B. when traffic that originates outside the network must be routed to internal hosts
C. when traffic that originates inside the network must be routed to internal hosts
D. when the network has few public IP addresses and many private IP addresses require outside access
Answer: B
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
コマンドuserdel --force --remove bobに関して正しいことは何ですか? (正しい答えを2つ選んでください。)
A. ボブがグループの最後のメンバーだった場合、そのグループは削除されます。
B. bobが所有するすべてのファイルは、マウントされているすべてのファイルシステムから削除されます。
C. ユーザーbobのホームディレクトリが削除されました。
D. ユーザーbobがシステムのユーザーデータベースから削除されます。
E. ロケートデータベースはbobが所有するファイルを削除するように更新されます。
Answer: C,D
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