VMware 2V0-17.25 Q&A - in .pdf

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  • Exam Code: 2V0-17.25
  • Exam Name: VMware Cloud Foundation 9.0 Administrator
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  • Q & A: 365 Questions and Answers
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NEW QUESTION: 1
Which two logical router components span across all transport nodes? (Choose two.)
A. TIER0_DISTRIBUTED_ROUTER
B. DISTRIBUTED_ROUTER_TIER0
C. DISTRIBUTED_ROUTER_TIER1
D. SERVICE_ROUTER_TIER1
E. SERVICE_ROUTER_TIER0
Answer: C,E

NEW QUESTION: 2
ある会社は、フォーラムでメタデータを維持するためにAmazonDynamoDBを使用したいと考えています。下の画像のサンプルデータセットを参照してください。

DevOpsエンジニアは、パーティションキー、ソートキー、ローカルセカンダリインデックス、投影された属性、およびフェッチ操作を使用してテーブルスキーマを定義する必要があります。
スキーマは、コストを最小限に抑えるために、プロビジョニングが最も少ない読み取り容量単位を使用した次の検索例をサポートする必要があります。
-ForumName内で、件名が「a」で始まるアイテムを検索します。
-指定されたLastPostDateTime時間枠内のフォーラムを検索します。
-LastPostDateTimeが過去3か月以内のスレッド値を返します。
どのスキーマが要件を満たしていますか?
A. 主キーとしてSubjectを使用し、ソートキーとしてForumNameを使用します。 LastPostDateTimeをソートキーとして使用するLSIと、スレッドのフェッチ操作を用意します。
B. 主キーとしてForumNameを使用し、ソートキーとしてSubjectを使用します。ソートキーとしてスレッドを持ち、投影された属性LastPostDateTimeを持つLSIを用意します。
C. 主キーとしてSubjectを使用し、ソートキーとしてForumNameを使用します。 LastPostDateTimeのソートキーとフェッチ操作としてスレッドを使用するLSIを用意します。
D. 主キーとしてForumNameを使用し、ソートキーとしてSubjectを使用します。 LastPostDateTimeをソートキーおよび投影された属性スレッドとして持つLSIを使用します。
Answer: D
Explanation:
Explanation
https://docs.aws.amazon.com/amazondynamodb/latest/developerguide/LSI.html

NEW QUESTION: 3
Given the following requirements:
Create a table to contain employee data, with a unique numeric identifier automatically assigned when a row is added, has an EDLEVEL column that permits only the values 'C', 'H' and 'N', and permits inserts only when a corresponding value for the employee's department exists in the DEPARTMENT table.
Which of the following CREATE statements will successfully create this table?
A. CREATE TABLEemp (
empno SMALLINT NOT NULL GENERATED BY DEFAULT AS IDENTITY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
workdept CHAR(3),
edlevel CHAR(1),
CONSTRAINT emp_pk PRIMARY KEY (empno),
CONSTRAINT emp_workdept_fk FOREIGN KEY (workdept) REFERENCES department
(deptno),
CONSTRAINT edlevel_ck CHECK edlevel VALUES ('C','H','N')
);
B. CREATE TABLEemp (
empno SMALLINT NEXTVAL GENERATED BY DEFAULT AS IDENTITY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
workdept CHAR(3) NOT NULL,
edlevel CHAR(1) CHECK IN ('C','H','N')),
CONSTRAINT emp_pk PRIMARY KEY (empno),
CONSTRAINT emp_workdept_fk FOREIGN KEY department (deptno) REFERENCES
(workdept)
);
C. CREATE TABLEemp (
empno SMALLINT NOT NULL GENERATED BY DEFAULT AS IDENTITY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
workdept CHAR(3),
edlevel CHAR(1),
CONSTRAINT emp_pk PRIMARY KEY (empno),
CONSTRAINT emp_workdept_fk FOREIGN KEY (workdept) REFERENCES department
(deptno),
CONSTRAINT edlevel_ck CHECK (edlevel IN ('C','H','N'))
);
D. CREATE TABLEemp (
empno SMALLINT NEXTVAL GENERATED ALWAYS AS IDENTITY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
workdept CHAR(3) NOT NULL,
edlevel CHAR(1),
PRIMARY KEY emp_pk (empno),
FOREIGN KEY emp_workdept_fk ON (workdept) REFERENCES department (deptno),
CHECK edlevel_ck VALUES (edlevel IN ('C','H','N')),
);
Answer: C

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