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NEW QUESTION: 1
Which statement about H.323 Gatekeeper Call Admission Control is true?
A. Gatekeeper Call Admission Control applies to centralized Cisco Unified Communications deployments that use locations based Call Admission Control.
B. Gatekeeper Call Admission Control applies to distributed Cisco Unified Communications deployments.
C. Gatekeeper Call Admission Control setting is configured in Cisco Unified Communications Manager.
D. Gatekeeper Call Admission Control applies only to distributed Cisco Unified Communications Express deployments.
Answer: B
Explanation:
in distributed call processing deployments on a simple hub-and-spoke topology, you can implement call admission control with a Cisco IOS gatekeeper. In this design, the call processing agent (which could be a Unified CM cluster, Cisco Unified Communications Manager Express (Unified CME), or an H.323 gateway) registers with the Cisco IOS gatekeeper and queries it each time the agent wants to place an IP WAN call.
Link: http://www.cisco.com/en/US/docs/voice_ip_comm/cucm/srnd/7x/cac.html#wp1044743

NEW QUESTION: 2
Drag and Drop Question
Drag and drop the IPv4 network subnets from the left onto the correct usable host ranges on the right Select and Place:

Answer:
Explanation:

Explanation:
This subnet question requires us to grasp how to subnet very well. To quickly find out the subnet range, we have to find out the increment and the network address of each subnet. Let's take an example with the subnet
172.28.228.144/18:
From the /18 (= 1100 0000 in the 3rd octet), we find out the increment is 64. Therefore the network address of this subnet must be the greatest multiple of the increment but not greater than the value in the 3rd octet (228). We can find out the 3rd octet of the network address is 192 (because 192 = 64 * 3 and 192 < 228) -> The network address is 172.28.192.0. So the first usable host should be 172.28.192.1 and it matches with the 5th answer on the right. In this case we don't need to calculate the broadcast address because we found the correct answer.
Let's take another example with subnet 172.28.228.144/23 -> The increment is 2 (as /23 = 1111 1110 in 3rd octet) -> The 3rd octet of the network address is 228 (because 228 is the multiply of 2 and equal to the 3rd octet) -> The network address is 172.28.228.0 -> The first usable host is 172.28.228.1. It is not necessary but if we want to find out the broadcast address of this subnet, we can find out the next network address, which is 172.28.(228 + the increment number).0 or 172.28.230.0 then reduce 1 bit -> 172.28.229.255 is the broadcast address of our subnet. Therefore the last usable host is 172.28.229.254.

NEW QUESTION: 3
Refer to the exhibit.

Which collections and legal hold capabilities will users assigned to the customized user role shown in the exhibit have?
A. Case Admin access to collections and case manager access to legal holds
B. Case User access to collections and case manager access to legal holds
C. read-only access to collections and full access to legal holds
D. no access to collections but read access and write access to legal holds
Answer: C

NEW QUESTION: 4
注：この質問は、同じシナリオを提示する一連の質問の一部です。 シリーズの各質問には、記載された目標を達成できる独自の解決策が含まれています。 質問セットの中には複数の正しい解決策があるものもあれば、正しい解決策がないものもあります。
このセクションの質問に答えた後、あなたはそれに戻ることができません。 その結果、これらの質問はレビュー画面に表示されません。
ネットワークにcontoso.comという名前のActive Directoryフォレストが含まれています。
どのサーバーがスキーママスターであるかを識別する必要があります。
解決方法：Active Directoryドメインと信頼を開き、Active Directoryを右クリックします
[ドメインと信頼]をクリックし、[操作マスタ]をクリックします。
これは目標を満たしていますか？
A. いいえ
B. はい
Answer: A
Explanation:
This solution only shows the Domain Naming Master.

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