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  • PEGACPRSA25V1 pdf
  • Exam Code: PEGACPRSA25V1
  • Exam Name: Certified Pega Robotics System Architect 25
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NEW QUESTION: 1
AWS Key Management Service(KMS)を使用するためのどのオプションが、将来のキーの危殆化によって露出される可能性のあるデータの範囲を最小限に抑えることに重点を置いたキー管理のベストプラクティスをサポートしていますか?
A. CMKアクセス許可を変更して、キーをプロビジョニングできる個人が、キーを使用できる同じ個人ではないようにします。
B. 新しい顧客マスターキー(CMK)を生成し、新しいCMKで既存のデータをすべて再暗号化し、将来のすべての暗号化操作に使用します。
C. CMKエイリアスを90日ごとに変更し、キー呼び出しアプリケーションを新しいキーエイリアスで更新します。
D. KMS自動キーローテーションを使用してマスターキーを置き換え、以前に暗号化されたデータを再暗号化することなく、今後の暗号化操作にこの新しいマスターキーを使用します。
Answer: B
Explanation:
Explanation
"automatic key rotation has no effect on the data that the CMK protects. It does not rotate the data keys that the CMK generated or re-encrypt any data protected by the CMK, and it will not mitigate the effect of a compromised data key. You might decide to create a new CMK and use it in place of the original CMK. This has the same effect as rotating the key material in an existing CMK, so it's often thought of as manually rotating the key." https://docs.aws.amazon.com/kms/latest/developerguide/rotate-keys.html

NEW QUESTION: 2
You are concerned about how recovery from a failed Exadata Storage Server would work. Which statement is true about the Exadata CELLBOOT USB?
A. Regular Exadata Storage Server patches make sure that the internal USB is updated so that it can be used for recovery.
B. Exadata automatically copies OS binaries and configuration files from another cell during patching to sync the internal USB.
C. Changes to network configuration files are possible without using ipconf.plas long as all relevant files are updated.
D. Each patch that is performed on Exadata storage servers requires manual resync of the CELLBOOT USB.
E. Some space from the CELLBOOT USB is used for cache metadata when write back flash cache is enabled.
Answer: A
Explanation:
Explanation/Reference:
Explanation:

NEW QUESTION: 3
What will happen when you attempt to compile and run the code below, assuming you enter the following sequence: 1 2 3<enter>?
#include <iostream>
using namespace std;
int main ()
{
int a,b,c;
cin>>a>>b>>c;
cout<<a<<b<<c<<endl;
return 0;
}
Program will output:
A. 0
B. 1
C. compilation error
D. the result is unspecified
E. 1 2 3
Answer: B

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