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NEW QUESTION: 1
You have created a user in SAP NetWeaver AS Java. In which log file can you find the user information?
A. applications.log
B. security.log
C. default Trace.trc
D. server.log
Answer: B

NEW QUESTION: 2
회사는 AWS Identity and Access Management (IAM)와 SAML 연동을 사용하여 내부 사용자에게 AWS 계정에 대한 SSO를 제공합니다. 회사의 ID 공급자 인증서는 정상적인 수명주기의 일부로 교체되었습니다. 얼마 후 사용자가 로그인을 시도 할 때 다음 오류가 표시되기 시작했습니다.
"오류 : 응답 서명이 잘못되었습니다 (서비스 : AWSSecuntyTokenService, 상태 코드 : 400, 오류 코드 : InvalidldentltyToken)"보안 엔지니어는 즉각적인 문제를 해결하고 문제가 다시 발생하지 않도록 해야 합니다.
이를 수행하기 위해 보안 엔지니어는 어떤 단계 조합을 취해야 합니까? (2 개 선택)
A. 자격 증명 공급자로부터 SAML 메타 데이터 파일의 새 복사본을 다운로드합니다. 새
B. 다음 인증서 교체 기간 동안 현재 인증서가 만료되기 전에 새 인증서를 ID 공급자의 보조 인증서로 추가합니다. 새 메타 데이터 파일을 생성하고 IAM 자격 증명 공급자 엔터티에 업로드합니다. 필요한 경우 인증서의 자동 또는 수동 순환을 수행합니다.
C. 자격 증명 공급자로부터 SAML 메타 데이터 파일의 새 복사본을 다운로드 합니다. 새 IAM 자격 증명 공급자 엔터티를 생성합니다. 새 IAM 자격 증명 공급자 엔터티에 새 메타 데이터 파일을 업로드 합니다.
D. 자격 증명 공급자로부터 SAML 메타 데이터 파일의 새 사본을 다운로드합니다. 문제의 SAML 통합을 위해 구성된 IAM 자격 증명 공급자 엔터티에 새 메타 데이터를 업로드합니다.
E. 다음 인증서 교체 기간 동안 현재 인증서가 만료되기 전에 새 인증서를 ID 공급자의 보조 인증서로 추가합니다. 메타 데이터 파일의 새 복사본을 생성하고 새 IAM 자격 증명 공급자 엔터티를 생성합니다. 새 IAM 자격 증명 공급자 엔터티에 메타 데이터 파일을 업로드합니다. 필요한 경우 인증서의 자동 또는 수동 순환을 수행합니다.
Answer: C,E

NEW QUESTION: 3
When components in a system are arranged in parallel, system reliability be:
A. As good as the best component
B. As good as the weakest component
C. The product of the reliability of individual components
D. Better than the best component
Answer: D
Explanation:
Regarding parallel and serial reliability: for components arranged in parallel, the reliability is better than the best component and for components arranged in serial, the reliability of the system is worse than the worst component. When components are arranged in parallel, only one of the units need to be successful for the whole system to be a success (i.e. the system fails only when every unit fails). There is redundancy built into the system. In a series arrangement, the failure of any one of the components leads to the failure of the whole system. Here is an example: Suppose two components with reliability 0.3 and 0.6 are arranged in parallel; then the reliability of the system is given by: R = 1 - [(1-R1)(1-R2)] = 1 - (0.7)(0.4) = 0.72, which is better than both 0.7 and 0.4.If two components with reliability 0.3 and 0.6 are arranged in series, then the reliability of the system is given by: R = R1 * R2 = 0.3 * 0.6 = 0.18, which is lower than both 0.3 and 0.6.Also, a parallel configuration (because of its redundancy) gives you a higher reliability (.72)than a similar serial configuration (.18). [Another way to look into this is: If, 1stS -> 1st being successful, 1stF -> 1st being unsuccessful, and 2ndS -> 2nd being successful, 2ndF -> 2nd being unsuccessful, then Probability of 1stS + 2ndS = 0.3 * 0.6 = 0.18 Probability of 1stS + 2ndF = 0.3 * 0.4 = 0.12 Probability of 1stF + 2ndS = 0.7 * 0.6 = 0.42 Probability of 1stF + 2ndF =
0.7 * 0.4 = 0.28 So, when they are in parallel, Probability of success = Probability of at least one being successful = 0.18 + 0.12 + 0.42 = 0.72 And, when they are in series, Probability of success = probability of both being successful = 0.18] So, the answer to this question is (a).