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NEW QUESTION: 1
ソリューションアーキテクトは、Amazon RDSにデータベースを含めるソリューションを設計しています。
企業のセキュリティポリシーでは、データベース、そのログ、およびバックアップをすべて暗号化することが義務付けられています。
Amazon RDSを使用してセキュリティポリシーを満たす最も効率的なオプションはどれですか？
A. Amazon RDSインスタンスを起動します。バックアップの暗号化を有効にします。データベースエンジン機能を使用してログを暗号化します。
B. 暗号化を有効にしてAmazon RDSインスタンスを起動します。ログとバックアップは自動的に暗号化されます。
C. Amazon RDSインスタンスを起動します。データベース、ログ、バックアップの暗号化を有効にします。
D. 暗号化を有効にしてAmazon RDSインスタンスを起動します。ログとバックアップの暗号化を有効にします。
Answer: B
Explanation:
Explanation
https://docs.aws.amazon.com/AmazonRDS/latest/UserGuide/Overview.Encryption.html

NEW QUESTION: 2
When components in a system are arranged in parallel, system reliability be:
A. As good as the best component
B. The product of the reliability of individual components
C. As good as the weakest component
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).

NEW QUESTION: 3
Which KPI types are supported out of the box by contextualized SAP HANA-based analytics? There are 3 correct answers to this question.
A. Change document-related KPIs
B. Duplicate-related KPIs
C. Request status-related KPIs
D. Processing time-related KPIs
E. Completeness-related KPIs
Answer: A,B,E

NEW QUESTION: 4
Which sample, non-intrusive process should a technical specialist recommend to obtain an overview of the storage environment?
A. Visual Storage Intelligence study
B. IBM Spectrum Connect
C. IBM Spectrum Control Advanced
D. Disk Magic study
Answer: A