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NEW QUESTION: 1
あなたの会社は、Azure API Managementを使用して、そのサービス用のAPIを公開する予定です。
サービス応答にAspNet-Versionヘッダーが含まれていることがわかります。
公開されているAPIの応答からAspNet-Versionを削除するための解決策を推奨する必要があります。
あなたは推薦に何を含めるべきですか？
A. 新しい方針
B. 新商品
C. URLスキームの修正
D. 新しいリビジョン
Answer: A
Explanation:
References:
https://docs.microsoft.com/en-us/azure/api-management/transform-api

NEW QUESTION: 2
Which two statements about HSRP are true? (Choose two.)
A. Its default configuration allows for pre-emption.
B. It supports unique virtual MAC addresses.
C. Its virtual MAC is 0000.0C07.Acxx.
D. It supports tracking.
E. Its multicast virtual MAC is 0000.5E00.01xx.
Answer: C,D
Explanation:
Default HSRP Configuration Feature Default Setting
HSRP version Version 1 HSRP groups None configured Standby group number 0
Standby MAC address System assigned as: 0000.0c07.acXX, where XX is the HSRP group number
Standby priority 100 Standby delay 0 (no delay) Standby track interface priority 10 Standby hello time 3 seconds Standby holdtime 10 seconds
- The standby track interface configuration command ties the router hot standby priority to the availability of its interfaces and is useful for tracking interfaces that are not configured for HSRP. When a tracked interface fails, the hot standby priority on the device on which tracking has been configured decreases by 10. If an interface is not tracked, its state changes do not affect the hot standby priority of the configured device. For each interface configured for hot standby, you can configure a separate list of interfaces to be tracked.
- The standby track interface-priority interface configuration command specifies how much to decrement the hot standby priority when a tracked interface goes down. When the interface comes back up, the priority is incremented by the same amount.
Reference: http://www.cisco.com/c/en/us/td/docs/switches/lan/catalyst3750x_3560x/software/release/1 2-2_55_se/configuration/guide/3750xscg/swhsrp.html

NEW QUESTION: 3
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. 8 kHz x 8-bit code words
C. 2 x 4 kHz x 8-bit code words
D. 2 x 4-bit code words x 8 kHz
Answer: C
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."

NEW QUESTION: 4
For the wavelength division system, configure the 1st-20th wave, use the NS4 single board, and test the M40 output port with the spectrum analyzer during the commissioning. It is found that the optical power of the 20th wave is significantly lower than the other 19 waves. The possible reasons are as follows: (Multiple Choice)
A. The 20th channel corresponding to the M40V has a large insertion loss.
B. The 20th wave OTU's RX port is incorrectly added with 5dB fixed light decay.
C. The 20th wave OTU center is long drifted
D. The 20th wave OTU wave side light is too small
Answer: A,C,D