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NEW QUESTION: 1
As a station moves away from the access point to which it is associated, it changes its data rate from 600 Mbps to 540 Mbps and then to 450 Mbps.
What IEEE 802.11 term or phase is used to describe this functionality?
A. Multirate Control
B. Modulation and Coding Selection
C. Dynamic Rate Switching
D. Rate Set Selectivity
Answer: C

NEW QUESTION: 2
Which statement indicates something that can cause an inbound PSTN call to an H.323 gateway that is configured in Cisco Unified Communications Manager to fail to ring an IP phone?
A. The gateway IP address that is configured in Cisco Unified Communications Manager does not match the IP address that is configured at the gateway in the h323-gateway voip bind srcaddr command.
B. The Cisco Unified Communications Manager does not have a matching route pattern to match the called number.
C. The gateway is not registered in Cisco Unified Communications Manager.
D. The gateway is missing the command allow-connections h323 to h323 under the voice service voip configuration.
Answer: A

NEW QUESTION: 3
Der Datenbankadministrator (DBA) schlägt vor, die Datenbankeffizienz durch Denormalisierung einiger Tabellen zu verbessern. Dies hätte zur Folge:
A. Vertraulichkeitsverlust.
B. Anwendungsfehler.
C. nicht autorisierte Zugriffe.
D. erhöhte Redundanz.
Answer: D
Explanation:
Erläuterung:
Die Normalisierung ist ein Entwurfs- oder Optimierungsprozess für eine relationale Datenbank (DB), der die Redundanz minimiert. Eine Denormalisierung würde daher die Redundanz erhöhen. Redundanz, die normalerweise als positiv angesehen wird, wenn es um die Verfügbarkeit von Ressourcen geht, ist in einer Datenbankumgebung negativ, da sie zusätzlichen und ansonsten unnötigen Aufwand für die Datenverarbeitung erfordert. Eine Denormalisierung ist manchmal aus funktionellen Gründen ratsam. Es sollte nicht zu Vertraulichkeitsverlusten, unbefugten Zugriffen oder Anwendungsfehlern führen.

NEW QUESTION: 4
What output will the following command sequence produce?
echo '1 2 3 4 5 6' | while read a b c; do echo result: $c $b $a; done
A. result: 6 5 4
B. result: 3 2 1
C. result: 1 2 3 4 5 6
D. result: 6 5 4 3 2 1
E. result: 3 4 5 6 2 1
Answer: E