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NEW QUESTION: 1
Which two platforms are supported by Cisco Prime Data Center Network Manager? (Choose two.)
A. Cisco Nexus switches
B. Cisco Aironet
C. Cisco Catalyst switches
D. Cisco MDS switches
E. Cisco ACE
Answer: A,D

NEW QUESTION: 2



A. Option D
B. Option C
C. Option B
D. Option A
Answer: A

NEW QUESTION: 3
대규모 전자 상거래 사이트는 Amazon S3에서 정적 객체를 제공하도록 설계되었습니다. Amazon S3 버킷은 초당 300 회 이상의 GET 요청을 서버에 전달합니다. 성능을 최적화하려면 어떻게 해야 합니까? (2 개 선택)
A. Configure Amazon S3 lifecycle rules.
B. Delete expired Amazon S3 server log files.
C. Enable Amazon S3 cross-region replication.
D. Integrate Amazon CloudFront with Amazon S3.
E. Randomize Amazon S3 key name prefixes.
Answer: D,E
Explanation:
Explanation
CloudWatch definitely. Random key prefixes is still a valid method of improving performance by using parallel reads. It doesn't mention prefix hashing. For instance prefixes 1/,2/,3/,4,5/ could provide 5 x parallel streams for S3 as opposed to all objects being in a single folder/prefix e.g. dev/
https://docs.aws.amazon.com/AmazonS3/latest/dev/optimizing-performance.html
"There are no limits to the number of prefixes in a bucket. You can increase your read or write performance by parallelizing reads. For example, if you create 10 prefixes in an Amazon S3 bucket to parallelize reads, you could scale your read performance to 55,000 read requests per second." The assumption that prefixes don't matter is incorrect, as described by "Amazon S3 performance guidelines recommended randomizing prefix naming with **hashed characters** to optimize performance for frequent data retrievals. You no longer have to randomize prefix naming for performance, and can use sequential date-based naming for your prefixes"

NEW QUESTION: 4
The SAS data set named WORK.TEST is listed below: capacity airplanetype staff
150 Large 10
Which one of the following SAS programs created this data set?
A. data work.test;
capacity = 150;
if 100 le capacity le 200 then
do;
airplanetype = 'Large';
staff = 10;
else
do;
airplanetype = 'Small';
staff = 5;
end;
run;
B. data work.test;
capacity = 150;
if 100 le capacity le 200 then
airplanetype = 'Large' and staff = 10;
else airplanetype = 'Small' and staff = 5;
run;
C. data work.test;
capacity = 150;
if 100 le capacity le 200 then;
airplanetype = 'Small';
staff = 5;
else;
airplanetype = 'Large';
staff = 10;
run;
D. data work.test;
capacity = 150;
if 100 le capacity le 200 then
do;
airplanetype = 'Large';
staff = 10;
end;
else
do;
airplanetype = 'Small';
staff = 5;
end;
run;
Answer: D


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