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Veeam Certified Engineer v12

  • Exam Number/Code : VMCE_v12
  • Exam Name : Veeam Certified Engineer v12
  • Questions and Answers : 213 Q&As
  • Update Time: 2019-01-10
  • Price: $ 99.00 $ 39.00

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NEW QUESTION: 1

A. Option C
B. Option A
C. Option D
D. Option B
Answer: B,C
Explanation:
You have to run esxcli network ip interface list command to list all the IPs. Then you need to make sure that MTU size is correctly set in the VMkernel interface properties in the vsphere client.

NEW QUESTION: 2
완벽한 정보의 예상 가치는 다음과 같습니다.
A. 확실성 하에서 예상되는 이익과 불확실성 하에서 최선의 행동으로 예상되는 금전적 가치의 차이.
B. 확실하게 예상되는 이익과 동일합니다.
C. 각 행위의 최고 사건에 대한 조건부 이익 손실의 합)에 각 사건 발생 확률의 곱하기.
D. 확실성에 따른 예상 이익과 예상 기회 손실의 차이.
Answer: A
Explanation:
Perfect information permits certainty that a future state of nature will occur. The expected value of perfect information determines the maximum amount a decision maker is willing to pay for information. It is the difference between the expected value without perfect information, that is, the expected value of the best action under uncertainty and the expected value under certainty. Under certainty, a decision maker knows in each case which state of nature will occur and can act accordingly.

NEW QUESTION: 3
A customer wants to implement a BigInsights solution using the open source Hadoop Distributed File System (HDFS) on a Power Systems server. Which operating system is appropriate?
A. zOS
B. IBM i
C. AIX
D. Linux
Answer: D

NEW QUESTION: 4
Which two statements describe the IP address 10.16.3.65/23? (Choose two.)
A. The broadcast address of the subnet is 10.16.3.255 255.255.254.0.
B. The last valid host address in the subnet is 10.16.2.254 255.255.254.0
C. The subnet address is 10.16.3.0 255.255.254.0.
D. The lowest host address in the subnet is 10.16.2.1 255.255.254.0.
E. The network is not subnetted.
Answer: A,D
Explanation:
Explanation/Reference:
The mask 255.255.254.0 (/23) used with a Class A address means that there are 15 subnet bits and 9 host bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in 0, 2, 4, 6, etc., all the way to 254. The host 10.16.3.65 is in the 2.0 subnet. The next subnet is 4.0, so the broadcast address for the 2.0 subnet is 3.255. The valid host addresses are 2.1 through 3.254