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NEW QUESTION: 1
Which statement is true about link aggregation?
A. It facilitates control plane redundancy through link aggregation
B. It combines multiple Ethernet interfaces into a single link-layer interface
C. It combines multiple Ethernet interfaces into a multilink-layer interface.
D. It provides an active-passive failover mechanism for redundant Layer 2 links
Answer: B
Explanation:
Explanation/Reference:
Explanation:
The IEEE 802.3ad link aggregation specification enables multiple Ethernet interfaces to be grouped together and form a single link layer interface, also known as a link aggregation group (LAG) or bundle.
The physical links participating in a LAG are known as member links. LAGs are commonly used to aggregate trunk links between access and aggregation switches.

NEW QUESTION: 2
Content is migrated to Tivoli Storage Manager in aggregation mode.
Which of the following changes to settings will enable partial retrieval for Tivoli Storage Manager aggregate object members?
A. TSM_PARTIAL_RETRIEVE_DISABLEDin RMCONFIGURATION tables set to NO, and neither Tivoli Storage Manager client compression nor Tivoli Storage Manager client encryption is used
B. TSM_PARTIAL_RETRIEVE_DISABLED in RMCONFIGURATION table is set to NO. with Tivoli Storage Manager client encryption is used.
C. TSM_PARTIAI__RETRIEVE_D<SABLED in RMCONFIGURATION table is set to YES.
D. TSM_PARTIAL_RETR1EVE_DISABLED in RMCONFIGURATION table is set to NO. with Tivoli Storage Manager client compression is used.
Answer: C

NEW QUESTION: 3
A Unity system LUN has a daily snapshot schedule with a Retention Policy of seven days. A hardware failure results in a double fault of the LUN. Which snapshot behavior will be experienced?
A. All of the snapshots will be unavailable to restore data from
B. Data restoration is available from the oldest snapshot, all others are unavailable
C. All of the snapshots will be available to restore data from
D. Data restoration is available from the newest snapshot, all others are unavailable
Answer: A

NEW QUESTION: 4

data work.test;
First = 'Ipswich, England';
City_Country = substr(First,1,7)!!', '!!'England';
run;

A. 0
B. 1
C. 2
D. 3
Answer: A