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NEW QUESTION: 1
Your network contains a server named Server1 that runs Windows Server 2012. Server1 has the Hyper-V server role installed.
Server1 hosts four virtual machines named VM1, VM2, VM3, and VM4. Server1 is configured as shown in the following table.

VM3 is used to test Applications.
You need to prevent VM3 from synchronizing its clock to Server1.
What should you configure?
A. Automatic Start Action
B. the VLAN ID
C. virtual Machine Chimney
D. the startup order
E. resource metering
F. Resource control
G. Single-root I/O virtualization
H. NUMA topology
I. Integration Services
J. Processor Compatibility
K. Port mirroring
Answer: I
Explanation:
By default when you install the Integration Services/Components you get time synchronization with the host OS, here is how to disable ongoing time synchronization.
When you install the integration services/components in Hyper-V virtual machine you get a set of services installed and enabled by default.
Operating system shutdown Time synchronization Data exchange heartbeat
Backup via VSS If you do not want the virtual machine to continuously synch its time to the Hyper-V host using the integration service, you can disable the integration service from the Hyper-V manager.
Open up the settings for the VM Under Management, highlight the Integration Services option and you will get a list of the Integration Services installed and enabled Uncheck the Time Synchronization service and press App1y.
The virtual machine will now not sync its time with the Hyper-V host on a continuous basis....BUT it will always sync once at power on. This is required to boot strap the timer inside the virtual machine http://www.virtualizationadmin.com/kbase/VirtualizationTips/ServerVirtualization/MicrosoftH yper-VTips/PerformanceandScalability/DisablingTimeSyncinaVM.html http://blogs.technet.com/b/virtualization/archive/2008/08/29/backing-up-hyper-v-virtualmachines.aspx

NEW QUESTION: 2
What statement is true regarding database and index reorganization in relation to directory-container storage pools?
A. Offline Index reorganization is not necessary because the server completes inline index reorganizations with a CLEANUP ONLY technique
B. Database reorganization will automatically be started when the reorg_db_threshold is reached
C. Database reorganization is not necessary because the inventory component of the server will no longer become fragmented
D. Index reorganization is still necessary but not so often as with legacy file based deduplication pools
Answer: A
Explanation:
Explanation/Reference:
Explanation:
Do I need to run offline index reorganization if I use directory-container storage pools?
No. The server completes inline index reorganizations with a CLEANUP ONLY technique. No other type of index reorganization is required.
References: https://www.ibm.com/developerworks/community/wikis/home?lang=en#!/wiki/Tivoli%
20Storage%20Manager/page/Directory-container%20storage%20pools%20FAQs?section=q13

NEW QUESTION: 3
정부 고객은 기밀 데이터 중 일부에 대해 안전한 암호화 키 저장소를 설정해야 합니다. AWS CloudHSM이 이를 위한 최상의 서비스라고 결정합니다.
그러나이 작업을 수행하기 전에 몇 가지 전제 조건이 있는 것으로 보입니다 (특정 포트가 열린 보안 그룹 중 하나). 해당 보안 그룹과 관련하여 다음 중 올바른 것은 무엇입니까?
A. 네트워크에 포트 22 (SSH) 또는 포트 3389 (RDP)가 열려있는 보안 그룹.
B. 네트워크에 열려있는 포트가없는 보안 그룹.
C. 포트 3389 (RDP 용) 만 네트워크에 열려있는 보안 그룹.
D. 네트워크에 포트 22 (SSH 용) 만 열려있는 보안 그룹.
Answer: A
Explanation:
설명:
AWS CloudHSM은 AWS 클라우드에서 하드웨어 보안 모듈 (HSM)을 사용 가능하게 하여 고객에게 안전한 암호화 키 스토리지를 제공합니다.
HSM 어플라이언스를 프로비저닝하려면 AWS CloudHSM에 다음 환경이 필요합니다. AWS CloudHSM 서비스를 원하는 리전의 가상 프라이빗 클라우드 (VPC). VPC에 하나의 프라이빗 서브넷 (인터넷 게이트웨이가없는 서브넷). HSM 어플라이언스는이 서브넷에 프로비저닝됩니다.
하나의 퍼블릭 서브넷 (인터넷 게이트웨이가 연결된 서브넷) 제어 인스턴스가이 서브넷에 연결되어 있습니다.
AWS 리소스에 대한 액세스 권한을 AWS CloudHSM에 위임하는 AWS Identity and Access Management (IAM) 역할.
SafeNet 클라이언트 소프트웨어가 설치된 HSM 어플라이언스와 동일한 VPC에있는 EC2 인스턴스. 이 인스턴스를 제어 인스턴스라고하며 HSM 어플라이언스에 연결하고 관리하는데 사용됩니다.
포트 22 (SSH의 경우) 또는 포트 3389 (RDP의 경우)가 네트워크에 열려있는 보안 그룹. 이 보안 그룹은 제어 인스턴스에 연결되어 있으므로 원격으로 액세스 할 수 있습니다.

NEW QUESTION: 4
Assuming that the following code has been executed successfully, selected the expression which evaluate to True (Select two answers) Def f (x, y); Nom, denom = x, y def g ( ) ; Return nom / denom A = f (1, 2) B = f (3, 4)
A. a ! = b
B. a ( ) == 4
C. b ( ) == 4
D. a is not None
Answer: A,D