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NEW QUESTION: 1
The marketing expert can analyze and process data that has been harvested from internal or external channels, in
A. Profile Dashboard
B. Sentiment Engagement
C. Customer Journey Insight
D. Analytics and Reports Gallery
Answer: C
NEW QUESTION: 2
What is a characteristic of the Stakeholder license type in RTC?
A. read and write access to Software Configuration Management
B. read and write access to Change Management
C. read and write access to Planning Capabilities
D. read and write access to Reports as well as the ability to author BIRT reports
Answer: B
NEW QUESTION: 3
セキュリティエンジニアは、SSLを介した悪意のあるWebリクエストから会社のWebサイトを保護するためにWAFをインストールしています。目的を達成するために必要なものは次のうちどれですか？
A. 負荷分散サーバー
Webアプリケーションの前にWAFをデプロイすることにより、Webアプリケーションとインターネットの間にシールドが配置されます。プロキシサーバーは仲介者を使用してクライアントマシンのIDを保護しますが、WAFは一種のリバースプロキシであり、クライアントがサーバーに到達する前にWAFを通過することにより、サーバーを公開から保護します。
WAFは、ポリシーと呼ばれることが多い一連のルールを通じて動作します。これらのポリシーは、悪意のあるトラフィックを除外することにより、アプリケーションの脆弱性から保護することを目的としています。 WAFの価値は、ポリシーの変更を実装できる速度と容易さから部分的に得られ、さまざまな攻撃ベクトルへのより迅速な対応を可能にします。 DDoS攻撃中は、WAFポリシーを変更することで、レート制限をすばやく実装できます。
B. スプリットトンネルVPN
C. 復号化証明書
D. リバースプロキシ
Answer: C
NEW QUESTION: 4
When components in a system are arranged in parallel, system reliability be:
A. The product of the reliability of individual components
B. As good as the best component
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).