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NEW QUESTION: 1
Which two options must be verified on the PCS device so that only trusted SSL Web sites can be viewed?
(Choose two.)
A. Verify that Allow browsing untrusted SSL websites is not selected under Web options.
B. Verify that the certificate is associated with a virtual port.
C. Verify that the certificate is installed under Trusted Server CAs.
D. Verify that the certificate is installed under Trusted Client CAs.
Answer: B,D

NEW QUESTION: 2
What is the maximum data retention period for a Kinesis Stream? Choose the correct answer:
A. 7 days
B. 14 days
C. 1 day
D. 1 hour
Answer: A
Explanation:
Data records are accessible for a default of 24 hours from the time they are added to a stream.
This time frame is called the retention period and is configurable in hourly increments from 24 to
168 hours (1 to 7 days).
Reference:
http://docs.aws.amazon.com/streams/latest/dev/service-sizes-and-limits.html

NEW QUESTION: 3
사용량이 많지 않은 시간에 개발자는 일반 작업 부하에 영향을 주지 않으면 서 전체 Amazon DynamoDB 테이블 검색의 실행 시간을 최소화하려고 합니다. 작업량은 비 피크 시간 동안 강하게 일관성 있는 읽기 용량 단위의 평균을 나타냅니다.
개발자가이 스캔을 어떻게 최적화합니까?
A. 스캔 작업 중에 일관성을 유지하도록 일관성을 변경합니다.
B. 스캔 작업 중 읽기 용량 단위 늘리기
C. 비율을 제한하면서 병렬 스캔 사용
D. 순차적 스캔 사용
Answer: C
Explanation:
설명
병렬 스캔은 스캔 성능을 향상시키는 데 도움이되며 제한 매개 변수를 사용하면 제공된 처리량이 제한 될 수 있습니다.
AWS 설명서 참조 - DynamoDB 병렬 스캔
기본적으로 스캔 작업은 데이터를 순차적으로 처리합니다. DynamoDB는 데이터를 1MB 단위로 응용 프로그램에 반환하고 응용 프로그램은 다음 1MB의 데이터를 검색하기 위해 추가 작업을 수행합니다.
스캔 할 테이블이나 인덱스가 클수록 테이크 시간이 길어집니다. 또한 순차적 검색은 준비된 읽기 처리량을 항상 충분히 활용하지 못할 수 있습니다. DynamoDB가 여러 물리적 파티션에 큰 테이블의 데이터를 배포하더라도 스캔 작업은 한 번에 하나의 파티션 만 읽을 수 있습니다. 이러한 이유 때문에 a의 처리량은 단일 파티션의 최대 처리량으로 인해 제한됩니다.
이러한 문제를 해결하기 위해 스캔 작업은 표 또는 보조 색인을 논리적으로 여러 세그먼트로 나눌 수 있으며 여러 응용 프로그램 작업자가 세그먼트를 병렬로 검색합니다. 각 작업자는 스레드 (멀티 스레딩을 지원하는 프로그래밍 언어에서) 또는 운영 체제 프로세스가 될 수 있습니다.
다수의 작업자가 있는 병렬 스캔은 스캔중인 테이블 또는 인덱스에 대해 프로비저닝 된 모든 처리량을 쉽게 소비 할 수 있습니다. 테이블 또는 색인이 다른 응용 프로그램에서 많은 읽기 또는 쓰기 활동을 초래할 경우 이러한 스캔을 피하는 것이 가장 좋습니다.
요청 당 반환되는 데이터의 양을 제어하려면 Limit 매개 변수를 사용하십시오. 이렇게하면 한 명의 작업자가 다른 모든 작업자를 희생시키면서 모든 처리량을 소비하는 상황을 방지 할 수 있습니다.

NEW QUESTION: 4
Electronic Company has a reliability of 0.90. 2 of these components arranged in series in a sub-assembly. The reliability for the sub-assembly is:
A. 0.99
B. 1.80
C. 0.81
D. 0.9
Answer: C
Explanation:
If the connection is in series then multiply the reliability. If the connection is in parallel then
add the reliabilities.
Hence, the answer would be C (= 0.9 x 0.9 = 0.81).Theory For Serial reliability (events
mutually dependents):
Total = Rt = R1 * R2 = 0.9 * 0.9 = 0.81 For Parallel Reliability (events independents and
redundant): Total = Rt = 1 (1-R1)* (1-R2) = 1 - 0.1 * 0.1 = 0.99.