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NEW QUESTION: 1
A Machine Learning Specialist is working with a large cybersecurity company that manages security events in real time for companies around the world. The cybersecurity company wants to design a solution that will allow it to use machine learning to score malicious events as anomalies on the data as it is being ingested. The company also wants be able to save the results in its data lake for later processing and analysis.
What is the MOST efficient way to accomplish these tasks?
A. Ingest the data into Apache Spark Streaming using Amazon EMR, and use Spark MLlib with k- means to perform anomaly detection. Then store the results in an Apache Hadoop Distributed File System (HDFS) using Amazon EMR with a replication factor of three as the data lake.
B. Ingest the data and store it in Amazon S3. Have an AWS Glue job that is triggered on demand transform the new data. Then use the built-in Random Cut Forest (RCF) model within Amazon SageMaker to detect anomalies in the data.
C. Ingest the data and store it in Amazon S3. Use AWS Batch along with the AWS Deep Learning AMIs to train a k-means model using TensorFlow on the data in Amazon S3.
D. Ingest the data using Amazon Kinesis Data Firehose, and use Amazon Kinesis Data Analytics Random Cut Forest (RCF) for anomaly detection. Then use Kinesis Data Firehose to stream the results to Amazon S3.
Answer: D
Explanation:
https://aws.amazon.com/tw/blogs/machine-learning/use-the-built-in-amazon-sagemaker-random- cut-forest-algorithm-for-anomaly-detection/

NEW QUESTION: 2
Refer to the exhibit.

Which statement about the configuration between the Default and BR regions is true?
A. Calls between the two regions can use either 64 kbps or 8 kbps.
B. Only 64 kbps will be used between the two regions because the link is "lossy".
C. Calls between the two regions can use only the G.729 codec.
D. Both codecs can be used depending on the loss statistics of the link. When lossy conditions are high, the G.711 codec will be used.
Answer: C

NEW QUESTION: 3

A. ff02:a:b:c::l/64
B. ff02:33ab:l:32::2/128
C. 2001:000a:lb2c::/64
D. 2001:aaaa: 1234:456c: 1/64
E. 2fff:f:f:f::f/64
F. 2001:bad:2345:a:b::cef/128
Answer: D,F
Explanation:
Explanation: Option B is valid, assuming theres a faulty colon : in the IPv6 Address, just before the last 1, that is: 2001:aaaa:1234:456c::1/64Option F is valid, dispite its odd network mask (128 bits), sometimes used in tunnel links.
Incorrect answer:
Option A is invalid, since it is a Multicast addressOption C seems to be invalid because the
3rd group of characters includes an l (lb2c), but if it is a 1 instead of an l (faulty scan) and the required options are 3 instead of 2, then this address is still valid (2001:000a:1b2c::/64), because the 4th group of characters would be 0000 (remember that we can represent a continuous sequence of zeros by ::).Option D is definitely invalid since it is a reserved address. As states the IANA online :document about the IPv6 Unicast Global Addresses, the range below is reserved, not allocated to any RIR (Regional Internet Registry):
2E00:0000::/7 IANA 1999-07-01 RESERVED
Reference: http://www.iana.org/assignments/ipv6-unicast-address-assignments/ipv6- unicast-address-assignments.txt