Exam Code: H31-341_V2.5
Exam Name: HCIP-Transmission V2.5
Total Questions: 132 Q&A
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NEW QUESTION: 1
A recently hired security administrator is advising developers about the secure integration of a legacy in-house application with a new cloud based processing system. The systems must exchange large amounts of fixed format data such as names, addresses, and phone numbers, as well as occasional chunks of data in unpredictable formats. The developers want to construct a new data format and create custom tools to parse and process the data. The security administrator instead suggests that the developers:
A. Use well formed standard compliant XML and strict schemas.
B. Create a custom standard to define the data.
C. Implement a de facto corporate standard for all analyzed data.
D. Only document the data format in the parsing application code.
To ensure the successful parsing of the data, the XML code containing the data should be well-formed. We can use strict schemas to ensure the correct formatting of the data.
XML has two main advantages: first, it offers a standard way of structuring data, and, second, we can specify the vocabulary the data uses. We can define the vocabulary (what elements and attributes an XML document can use) using either a document type definition (DTD) or the XML Schema language.
Schemas provide the ability to define an element's type (string, integer, etc.) and much finer constraints (a positive integer, a string starting with an uppercase letter, etc.). DTDs enforce a strict ordering of elements; schemas have a more flexible range of options. Finally schemas are written in XML, whereas DTDs have their own syntax.
For an application to accept an XML document, it must be both well formed and valid.
A document that is not well formed is not really XML and doesn't conform to the W3C's stipulations for an XML document. A parser will fail when given that document, even if validation is turned off.
NEW QUESTION: 2
In the vCenter HA configuration below, drag the two correct components to the blank boxes in this diagram. The same component may be used more twice. (Choose two.)
NEW QUESTION: 3
Refer to the exhibit.
At the end of an RSTP election process, which access layer switch port will assume the discarding role?
A. Switch3, port Gi0/2
B. Switch3, port fa0/1
C. Switch3, port fa0/12
D. Switch4, port fa0/11
E. Switch4, port fa0/2
F. Switch3, port Gi0/1
In this question, we only care about the Access Layer switches (Switch3 & 4). Switch 3 has a lower bridge ID than
Switch 4 (because the MAC of Switch3 is smaller than that of Switch4) so both ports of Switch3 will be in forwarding
state. The alternative port will surely belong to Switch4.
Switch4 will need to block one of its ports to avoid a bridging loop between the two switches. But how does Switch4
select its blocked port? Well, the answer is based on the BPDUs it receives from Switch3. A BPDU is superior to
another if it has:
1. A lower Root Bridge ID
2. A lower path cost to the Root
3. A lower Sending Bridge ID
4. A lower Sending Port ID
These four parameters are examined in order. In this specific case, all the BPDUs sent by Switch3 have the same Root
Bridge ID, the same path cost to the Root and the same Sending Bridge ID. The only parameter left to select the best
one is the Sending Port ID (Port ID = port priority + port index). In this case the port priorities are equal because they
use the default value, so Switch4 will compare port index values, which are unique to each port on the switch, and
because Fa0/12 is inferior to Fa0/1, Switch4 will select the port connected with Fa0/1 (of Switch3) as its root port and
block the other port -> Port fa0/11 of Switch4 will be blocked (discarding role).
NEW QUESTION: 4
s3 기반 AMI에 의해 지원되는 EC2 인스턴스가 종료 될 때. 루트 볼륨의 데이터는 어떻게 됩니까?
A. 데이터가 자동으로 EBS 볼륨으로 저장됩니다.
B. 데이터가 자동으로 EBS 스냅 샷으로 저장됩니다.
C. 데이터가 자동으로 삭제됩니다.
D. 인스턴스를 다시 시작할 때까지 데이터를 사용할 수 없습니다.
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