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NEW QUESTION: 1
The Backup ports in RSTP may alternatively failed root port.
A. False
B. True
Answer: A

NEW QUESTION: 2
은행은 메인 프레임 기반 신용 카드 승인 처리 애플리케이션을 AWS 클라우드의 클라우드 네이티브 애플리케이션으로 재구성합니다.
새 애플리케이션은 최대로드시 초당 최대 1,000 개의 요청을 수신합니다. 각 트랜잭션에는 여러 단계가 있으며 각 단계는 이전 단계의 결과를 수신해야 합니다. 전체 요청은 데이터 손실없이 2 초 이내에 인증 응답을 반환해야 합니다.
모든 요청은 응답을 받아야 합니다. 솔루션은 PCI DSS (Payment Card Industry Data Security Standard)를 준수해야합니다.
규정 준수 요구 사항을 충족하면서도 가장 작은 복잡성과 최저 비용으로 은행의 모든 목표를 달성 할 수있는 옵션은 무엇입니까?
A. 대상 그룹의 Amazon EC2 전용 인스턴스에서 Amazon ECS 클러스터를 사용하여 Application Load Balancer를 생성하여 수신 요청을 처리합니다. Auto Scaling을 사용하여 평균 CPU 사용률에 따라 클러스터를 확장 / 확장하십시오. 모든 승인 단계를 처리하고 승인 상태의 JSON 오브젝트를 리턴하는 웹 서비스를 배치하십시오.
B. 전용 인스턴스의 Amazon EC2에 애플리케이션을 배포하십시오. Auto Scaling 그룹의 애플리케이션 서버 팜 앞에서 Elastic Load Balancer를 사용하여 수신 요청을 처리하십시오. 단일 인스턴스의 용량을 측정 한 후 초당 인바운드 요청 수에 대한 사용자 지정 Amazon CloudWatch 지표를 기반으로 스케일 아웃 / 인합니다.
C. 여러 단계를 수행하고 승인 상태의 JSON 객체를 반환하는 단일 AWS Lambda 작업을 사용하여 인바운드 요청을 처리 할 Amazon API Gateway를 생성합니다. 새 애플리케이션으로 인해 활동이 급증 할 수 있도록 동시 Lambdas 수의 한계를 늘리려면 지원 사례를 여십시오.
D. 각각 Amazon SQS 입력 대기열이있는 일련의 AWS Lambda 프로세스를 사용하여 인바운드 요청을 처리하도록 Amazon API Gateway를 생성합니다. 각 단계가 완료되면 다음 단계 큐에 결과가 기록됩니다. 마지막 단계는 승인 상태의 JSON 객체를 반환합니다. 새 애플리케이션으로 인해 활동이 급증 할 수 있도록 동시 Lambdas 수의 한계를 늘리려면 지원 사례를여십시오.
Answer: B

NEW QUESTION: 3
public class Two {
public static void main(String[] args) {
try {
doStuff();
system.out.println("1");
}
catch {
system.out.println("2");
}}
public static void do Stuff() {
if (Math.random() > 0.5) throw new RunTimeException(); doMoreStuff();
System.out.println("3 ");
}
public static void doMoreStuff() {
System.out.println("4");
}
}
Which two are possible outputs?
A. 4 3 1
B. 0
C. 1
D. 1 2
Answer: A,C
Explanation:
A: Output is 2 if Math.random() is greater than 0.5.
B: If Math.random() returns a value lessequalto0.5, the code won't throw an exception, it will continue with the doMore() method which will println "4" after which the program will continue with the doStuff() method and will println "3", after that we will be back in main() and the program will print "1".