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Exam Code: JN0-649
Exam Name: Enterprise Routing and Switching, Professional (JNCIP-ENT)
Vendor: Juniper

60 Questions & Answers
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NEW QUESTION: 1
A company has the following Microsoft SQL Server instances Instance1 and Instance2. You plan to enable Always Encrypted for both instances.
You need to configure the instances to meet the following requirements:
* Instance1 must use an initialization vector that is different each time the instance is initiated.
* Instance2 must use an initialization vector that is derived from an algorithm.
In the table below, identify the encryption type that must be used for each instance.
NOTE: Make only one selection in each column. Each correct selection is worth one point.

Answer:
Explanation:

Explanation

Always Encrypted supports two types of encryption: randomized encryption and deterministic encryption.
* Randomized encryption uses a method that encrypts data in a less predictable manner. Randomized encryption is more secure, but prevents searching, grouping, indexing, and joining on encrypted columns.
* Deterministic encryption always generates the same encrypted value for any given plain text value.
Using deterministic encryption allows point lookups, equality joins, grouping and indexing on encrypted columns. However, but may also allow unauthorized users to guess information about encrypted values by examining patterns in the encrypted column, especially if there is a small set of possible encrypted values, such as True/False, or North/South/East/West region.
References:
https://docs.microsoft.com/en-us/sql/relational-databases/security/encryption/always-encrypted-database-engine

NEW QUESTION: 2
Evaluate the following SQL commands:

The command to create a table fails. Identify the two reasons for the SQL statement failure?
A. You cannot use the NEXTVAL sequence value as a default value for a column.
B. You cannot use ORD_NO and ITEM_NO columns as a composite primary key because ORD_NO is also the foreign key.
C. You cannot use SYSDATE in the condition of a check constraint.
D. You cannot use the BETWEEN clause in the condition of a check constraint.
Answer: A,C
Explanation:
CHECK Constraint The CHECK constraint defines a condition that each row must satisfy. The condition can use the same constructs as the query conditions, with the following exceptions: References to the CURRVAL, NEXTVAL, LEVEL, and ROWNUM pseudocolumns Calls to SYSDATE, UID, USER, and USERENV functions Queries that refer to other values in other rows A single column can have multiple CHECK constraints that refer to the column in its definition. There is no limit to the number of CHECK constraints that you can define on a column. CHECK constraints can be defined at the column level or table level. CREATE TABLE employees (...
Salary NUMBER(8, 2) CONSTRAINT emp_salary_min CHECK (salary > 0),

NEW QUESTION: 3
An administrator has to determine host operating systems on the network and has deployed a transparent proxy. Which of the following fingerprint types would this solution use?
A. Packet
B. Active
C. Passive
D. Port
Answer: C


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