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NEW QUESTION: 1
The embedded MCU of TE60 supports ()
A. 9V + 3A
B. 4V + 3A
C. 6V + 3A
Answer: A

NEW QUESTION: 2
You have configured location-based Call Admission Control on a Cisco Unified Communications Manager and Cisco IOS Gateway. After the policy below is configured, the calls now fail. All calls are using g729r8 as the codec. Which configuration resolves the issue?

A:

B:

C:

D:

A. Option A
B. Option C
C. Option B
D. Option D
Answer: D

NEW QUESTION: 3
When components in a system are arranged in parallel, system reliability be:
A. As good as the best component
B. As good as the weakest component
C. The product of the reliability of individual components
D. Better than the best component
Answer: D
Explanation:
Regarding parallel and serial reliability: for components arranged in parallel, the reliability is better than the best component and for components arranged in serial, the reliability of the system is worse than the worst component. When components are arranged in parallel, only one of the units need to be successful for the whole system to be a success (i.e. the system fails only when every unit fails). There is redundancy built into the system. In a series arrangement, the failure of any one of the components leads to the failure of the whole system. Here is an example: Suppose two components with reliability 0.3 and 0.6 are arranged in parallel; then the reliability of the system is given by: R = 1 - [(1-R1)(1-R2)] = 1 - (0.7)(0.4) = 0.72, which is better than both 0.7 and 0.4.If two components with reliability 0.3 and 0.6 are arranged in series, then the reliability of the system is given by: R = R1 * R2 = 0.3 * 0.6 = 0.18, which is lower than both 0.3 and 0.6.Also, a parallel configuration (because of its redundancy) gives you a higher reliability (.72)than a similar serial configuration (.18). [Another way to look into this is: If, 1stS -> 1st being successful, 1stF -> 1st being unsuccessful, and 2ndS -> 2nd being successful, 2ndF -> 2nd being unsuccessful, then Probability of 1stS + 2ndS = 0.3 * 0.6 = 0.18 Probability of 1stS + 2ndF = 0.3 * 0.4 = 0.12 Probability of 1stF + 2ndS = 0.7 * 0.6 = 0.42 Probability of 1stF + 2ndF =
0.7 * 0.4 = 0.28 So, when they are in parallel, Probability of success = Probability of at least one being successful = 0.18 + 0.12 + 0.42 = 0.72 And, when they are in series, Probability of success = probability of both being successful = 0.18] So, the answer to this question is (a).

NEW QUESTION: 4
What levels are not included in the TCP/IPv4 model? (Multiple choice)
A. Transport layer
B. Session layer
C. Presentation layer
D. Network layer
E. Application layer
Answer: B,C

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