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NEW QUESTION: 1
What is the datastore size?
A. The amount of CPU available for optimization use
B. The amount of disk space available for optimization use
C. The amount of disk space available for Scalable Data Referencing (SDR) use
D. The amount of memory available for optimization use
E. The amount of memory available for SDR use
Answer: C

NEW QUESTION: 2



A. Option D
B. Option B
C. Option C
D. Option E
E. Option A
Answer: A
Explanation:
Explanation
The IP address 192.168.1.17 255.255.255.0 specifies that the address is part of the 192.168.1.0/24 subnet
24 mask bits = 255.255.255.0
28 mask bits = 255.255.255.240
192.168.1.0/24 subnet has a host range of 192.168.1.1 to 192.168.1.254 (0 being network and 255 being broadcoast)
192.168.1.17/28 subnet has a host range of 192.168.1.17 to 192.168.1.30 (16 being network and 31 being broadcast)
192.168.1.65/28 subnet has a host range of 192.168.1.65 - 192.168.1.78 (64 being network and 79 being broadcast) if fa0/0 was left as /24, you can see that the host range includes the host range of 192.168.1.64/28 which conflicts. Simply speaking, you can't overlap the subnets. By changing the subnet mask of fa0/0 to
255.255.255.240, these networks would no longer overlap.

NEW QUESTION: 3
CORRECT TEXT
Problem Scenario 49 : You have been given below code snippet (do a sum of values by key}, with intermediate output.
val keysWithValuesList = Array("foo=A", "foo=A", "foo=A", "foo=A", "foo=B", "bar=C",
"bar=D", "bar=D")
val data = sc.parallelize(keysWithValuesl_ist}
//Create key value pairs
val kv = data.map(_.split("=")).map(v => (v(0), v(l))).cache()
val initialCount = 0;
val countByKey = kv.aggregateByKey(initialCount)(addToCounts, sumPartitionCounts)
Now define two functions (addToCounts, sumPartitionCounts) such, which will produce following results.
Output 1
countByKey.collect
res3: Array[(String, Int)] = Array((foo,5), (bar,3))
import scala.collection._
val initialSet = scala.collection.mutable.HashSet.empty[String]
val uniqueByKey = kv.aggregateByKey(initialSet)(addToSet, mergePartitionSets)
Now define two functions (addToSet, mergePartitionSets) such, which will produce following results.
Output 2:
uniqueByKey.collect
res4: Array[(String, scala.collection.mutable.HashSet[String])] = Array((foo,Set(B, A}},
(bar,Set(C, D}}}
Answer:
Explanation:
See the explanation for Step by Step Solution and configuration.
Explanation:
Solution :
val addToCounts = (n: Int, v: String) => n + 1
val sumPartitionCounts = (p1: Int, p2: Int} => p1 + p2
val addToSet = (s: mutable.HashSet[String], v: String) => s += v
val mergePartitionSets = (p1: mutable.HashSet[String], p2: mutable.HashSet[String]) => p1
+ += p2