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H13-624_V5.5 pdf
  • Exam Code: H13-624_V5.5
  • Exam Name: HCIP-Storage V5.5
  • Version: V17.95
  • Q & A: 400 Questions and Answers
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  • Exam Code: H13-624_V5.5
  • Exam Name: HCIP-Storage V5.5
  • Version: V17.95
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  • Exam Code: H13-624_V5.5
  • Exam Name: HCIP-Storage V5.5
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NEW QUESTION: 1
The business analysis approach generally defines all of the following attributes except for which one?
A. Team roles
B. Deliverables
C. Requirements for solutions acceptance
D. Analysis technique
Answer: C
Explanation:
Explanation/Reference:
Explanation:

NEW QUESTION: 2
Your network contains two subnets. The subnets are configured as shown in the following table.

You have a server named Server1 that runs Windows Server 2012. Server1 is connected to LAN1.
You run the route print command as shown in the exhibit. (Click the Exhibit button.)

You need to ensure that Server1 can communicate with the client computers on LAN2.
What should you do?
A. Set the state of the Microsoft ISATAP Adapter #2 interface to disable.
B. Change the default gateway address.
C. Change the metric of the 10.10.1.0 route.
D. Set the state of the Teredo interface to disable.
Answer: B
Explanation:
In general, the first and last addresses in a subnet are used as the network identifier and broadcast address,respectively. All other addresses in the subnet can be assigned to hosts on that subnet. For example, IP addresses of networks with subnet masks of at least 24 bits ending in .0 or .255 can never beassigned to hosts. Such "last" addresses of a subnet are considered "broadcast" addresses and all hosts onthe corresponding subnet will respond to it.
Theoretically, there could be situations where you can assign an address ending in .0: for example, if you havea subnet like 192.168.0.0/255.255.0.0, you are allowed to assign a host the address 192.168.1.0. It couldcreate confusion though, so it's not a very common practice. Example10.6.43.0 with subnet 255.255.252.0 (22 bit subnet mask) means subnet ID 10.6.40.0, a host address range from 10.6.40.1 to 10.6.43.254 and a broadcast address10.6.43.255. So in theory, your example 10.6.43.0 would be allowed as a valid host address. The default gateway address should not end in .0 with the /24 address
http://tools.ietf.org/html/rfc4632 http://en.wikipedia.org/wiki/IPv4#Addresses_ending_in_0_or_255

NEW QUESTION: 3
ネットワーク技術者は、ベンダーにネットワークの脆弱性を確認した後、ネットワークで使用されている特定のハードウェアに対して新しいエクスプロイトがリリースされたという更新をRSSフィードで受け取ります。技術者は、脆弱性がほぼ2年前にパッチされたことを発見しました。将来、古いファームウェアを発見するために実装する必要がある手法は次のうちどれですか?
A. アプリケーションのファジング
B. マルウェアの署名管理
C. tcpdumpを使用したネットワークトラフィック分析
D. 脆弱性スキャン
Answer: D

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