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NEW QUESTION: 1
In the .Fusioninsight HD system, which is the LDAP data synchronization method?
A. Isolate differences
B. Two-way peers
C. One-way synchronization
D. Data cross-peer
Answer: C

NEW QUESTION: 2
다음 중 고위 경영진이 IT 성과를 검토 한 가장 좋은 증거는 무엇입니까?
A. 집행위원회 회의록
B. 밸런스드 스코어 카드
C. IT 젖산 계획
D. 주요 IT 성과 지표
Answer: A

NEW QUESTION: 3
A Primary Fault Tolerant VM is performing poorly despite its ESXi host having adequate resources and CPUs
are idle. Other hosts in the cluster are much more heavily utilized.
What is the cause?
A. The Secondary VM is assigned to a different portgroup than the Primary.
B. The Secondary VM is running on an ESXi host that is overcommitted.
C. The vMotion network has not been configured for Jumbo Frames.
D. The Secondary VM is running more applications than the Primary.
Answer: B

NEW QUESTION: 4
Which of these is the equation used to derive a 64 Kbps bit rate?
A. 2 x 8 kHz x 4-bit code words
B. 2 x 4 kHz x 8-bit code words
C. 2 x 4-bit code words x 8 kHz
D. 8 kHz x 8-bit code words
Answer: B
Explanation:
Explanation/Reference:
Explanation:
While the human ear can sense sounds from 20 to 20, 000 Hz, and speech encompasses sounds from about 200 to 9000 Hz, the telephone channel was designed to operate at about 300 to 3400 Hz. This economical range carries enough fidelity to allow callers to identify the party at the far end and sense their mood. Nyquist decided to extend the digitization to 4000 Hz, to capture higher-frequency sounds that the telephone channel may deliver. Therefore, the highest frequency for voice is 4000 Hz. According to Nyquist theory, we must double the highest frequency, so 2x4kHz = 8kHz.
Each sample will be encoded into a 8-bit code. Therefore 8kHz x 8-bit code = 64 Kbps (notice about the unit Kbps: 8kHz = 8000 samples per second so 8000 x 8-bit = 64000 bit per second = 64 Kilobit per second = 64 Kbps)
Link: http://encyclopedia2.thefreedictionary.com/Nyquist+theorem
Note:
Nyquist theory:
"When sampling a signal (e.g., converting from an analog signal to digital), the sampling frequency must be greater than twice the bandwidth of the input signal in order to be able to reconstruct the original perfectly from the sampled version."