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NEW QUESTION: 1
During the workflow design process in IBM SmartCloud Control Desk V7.5, which application stores information about individuals such as users, laborers, asset owners, and supervisors who receive workflow notifications?
A. Supervisors
B. Users
C. Delegates
D. People
Answer: D

NEW QUESTION: 2
An MX Series router has received a BGP prefix from its peer. There are multiple destination routes to the BGP prefix with the default BGP route preference.
In this scenario, what is the first tie-breaker used to select the destination route?
A. shortest cluster list length
B. lowest MED
C. highest local preference
D. lowest origin code
Answer: C
Explanation:
Explanation/Reference:
Understanding BGP Path Selection
For each prefix in the routing table, the routing protocol process selects a single best path. After the best path is selected, the route is installed in the routing table. The best path becomes the active route if the same prefix is not learned by a protocol with a lower (more preferred) global preference value, also known as the administrative distance. The algorithm for determining the active route is as follows:
1. Verify that the next hop can be resolved.
2. Choose the path with the lowest preference value (routing protocol process preference).
Routes that are not eligible to be used for forwarding (for example, because they were rejected by routing policy or because a next hop is inaccessible) have a preference of -1 and are never chosen.
Etc.
3. Prefer the path with higher local preference.
For non-BGP paths, choose the path with the lowest preference2 value.
Etc.
References: https://www.juniper.net/documentation/en_US/junos12.3/topics/reference/general/routing- ptotocols-address-representation.html

NEW QUESTION: 3
Which characteristic can be observable when using a packet sniffer on a network that has a hidden node
issue?
A. excessive retries
B. excessive authentication failures
C. excessive ACK timeouts
D. small fragmentation windows
Answer: C

NEW QUESTION: 4
CORRECT TEXT
Problem Scenario 60 : You have been given below code snippet.
val a = sc.parallelize(List("dog", "salmon", "salmon", "rat", "elephant"}, 3} val b = a.keyBy(_.length) val c = sc.parallelize(List("dog","cat","gnu","salmon","rabbit","turkey","woif","bear","bee"), 3) val d = c.keyBy(_.length) operation1
Write a correct code snippet for operationl which will produce desired output, shown below.
Array[(lnt, (String, String))] = Array((6,(salmon,salmon)), (6,(salmon,rabbit)),
(6,(salmon,turkey)), (6,(salmon,salmon)), (6,(salmon,rabbit)),
(6,(salmon,turkey)), (3,(dog,dog)), (3,(dog,cat)), (3,(dog,gnu)), (3,(dog,bee)), (3,(rat,dog)),
(3,(rat,cat)), (3,(rat,gnu)), (3,(rat,bee)))
Answer:
Explanation:
See the explanation for Step by Step Solution and configuration.
Explanation:
solution:
b.join(d).collect
join [Pair]: Performs an inner join using two key-value RDDs. Please note that the keys must be generally comparable to make this work. keyBy : Constructs two-component tuples
(key-value pairs) by applying a function on each data item. The result of the function becomes the data item becomes the key and the original value of the newly created tuples.