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Databricks Databricks-Generative-AI-Engineer-Associate exam
  • Exam Code: Databricks-Generative-AI-Engineer-Associate
  • Exam Name: Databricks Certified Generative AI Engineer Associate
  • Version: V12.35
  • Q & A: 70 Questions and Answers
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NEW QUESTION: 1
どの3つの ESPのフィールドは、送信時に暗号化されることができますか? (3を選択してください。)
A. シーケンス番号
B. セキュリティパラメータインデックス
C. パディング長
D. 次のヘッダー
E. Macアドレス
F. パディング
Answer: C,D,F
Explanation:
Explanation
The packet begins with two 4-byte fields (Security Parameters Index (SPI) and Sequence Number). Following these fields is the Payload Data, which has substructure that depends on the choice of encryption algorithm and mode, and on the use of TFC padding, which is examined in more detail later. Following the Payload Data are Padding and Pad Length fields, and the Next Header field. The optional Integrity Check Value (ICV) field completes the packet.
Source: https://tools.ietf.org/html/rfc4303#page-14

NEW QUESTION: 2
You have three devices enrolled .n Microsoft Intune as shown .n the following table.

For each of the following statements, select Yes if the statement is true. Otherwise, select No.
NOTE: Each correct selection is worth one point.

Answer:
Explanation:

Explanation


NEW QUESTION: 3
What has the highest precedence for evaluation of the access control list?
A. User principals
B. Group principals
C. Last in multi-group permission
D. First in multi-group permission
Answer: A

NEW QUESTION: 4
Which one of the following queries selects the customer whose order has the highest total price?
A. CriteriaBuilder cb = ...
CriteriaQuery<Customer> cq = cb.cteateQuery(Customer.class);
Root<Customer> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer__.orders);
cq.select(c).distinct(true);
Subquery<Double> sq = cq.subquery(Double.class);
Root<Order> subo = cq.correlate(o);
sq.select(cb.max(subo.get(Order_.totalPrice)));
cq.where(cb.equal(o.get(Order_.totalPrice), cb.all(sq)));
B. CriteriaBuilder cb = ...
CriteriaQuery<Customer> cq = cb.createquery(customer.class)
Root<Customer> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer__.orders);
cq.select(c).distinct(true);
Subquery<Double> sq = cq.subquery(Double.class);
Root<Order> subo = cq.correlate(o);
sq.select(cb.max(subo.get(Order_.totalPrice)));
cq.where(cb.equal(o.get(Order_.totalPrice), cb.all(sq)));
C. CriteriaBuilder cb = ...
CriteriaQuery<Customer> cq = cb.createQuery(Customer.class);
Root<Customer> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer_.orders);
cq.select(c).distinct(true);
Subquery<Double> sq = cq.subquery(Double.class);
Root<Order> subo = sq.from(Order.class);
sq. select (ci: . max ( subo . get (Order_ . Total Price) ) ) ;
cq.where(sq.all(o.gei(Order_.totalPrice)));
D. CriteriaBuilder cb = ...
Criteria Query <Customer> cq = cb.create Query (Customer.class);
Root<Customer> c = cq.from(Customer.class);
Join<Customer, Order> o = c.join(Customer__.orders);
cq.select(c).distinct(true);
Subquery<Double> sq = cq.subquery(Double.class);
Root<Order> subo = cq.correlate(o);
sq.select(cb.max(subo.get(Order_.totalPrice)));
cq.where(cb.equal(o.get(Order_.totalPrice), cb.all(sq)));
Answer: B

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