CWNA-108 Practice Questions

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NEW QUESTION: 1
HOTSPOT
Your network contains an Active Directory named contoso.com.
The domain contains the computers configured as shown in the following table.

Server1 has a share named Share1 with the following configurations:-

Server1, Computer1, and Computer2 have the connection security rules configured as shown in follow:-

Please Select the correct statement as below:

Answer:
Explanation:

Explanation:
When Computer1 accesses Share1, SMB encryption will be used: YES
When Computer2 accesses Share1, SMB encryption will be used: YES
When Server1 accesses a shared folder on Computer1, IPsec encryption will be used: NO
The shared folder "Share1" is configured with "EncryptData : True", no matter which network the client resides,SMB 3 communication will be encrypted.When Server1 access
Computer1 over network, the original packet L3 IP Header is as follow:-172.16.1.30 ->
172.16.10.60These traffic does not match the enabled IPSec rule "Rule2" nor "Rule3", and the only matching rule "Rule1" isdisabled. So, no IPsec encryption will be achieved.

NEW QUESTION: 2
Sie haben einen Windows Server 2016-Failovercluster mit dem Namen Cluster1, der drei Knoten mit den Namen Server1, Server2 und Server3 enthält. Jeder Knoten hostet mehrere virtuelle Maschinen. Die virtuellen Maschinen sind so konfiguriert, dass ein Failover auf einen anderen Knoten in Cluster1 erfolgt, wenn der Hostknoten ausfällt.
Sie müssen sicherstellen, dass die virtuelle Maschine dieses Knotens sofort ein Failover ausführt, wenn der Clusterdienst auf einem der Knoten ausfällt.
Welche Einstellung solltest du konfigurieren?
A. ResiliencyLevel
B. ResiliencyPeriod
C. FailureConditionaLevel
D. QuarantäneDauer
Answer: A

NEW QUESTION: 3
Which CREATE TABLE statement is valid?
A. CREATE TABLE ord_details(ord_no NUMBER(2) PRIMARY KEY,item_no NUMBER(3) PRIMARY KEY,ord_date DATE NOT NULL);
B. CREATE TABLE ord_details(ord_no NUMBER(2) ,item_no NUMBER(3),ord_date DATE DEFAULT NOT NULL,CONSTRAINT ord_uq UNIQUE (ord_no),CONSTRAINT ord_pk PRIMARY KEY (ord_no));
C. CREATE TABLE ord_details(ord_no NUMBER(2),item_no NUMBER(3),ord_date DATE DEFAULT SYSDATE NOT NULL,CONSTRAINT ord_pk PRIMARY KEY (ord_no, item_no));
D. CREATE TABLE ord_details(ord_no NUMBER(2) UNIQUE, NOT NULL,item_no NUMBER(3),ord_date DATE DEFAULT SYSDATE NOT NULL);
Answer: C
Explanation:
Explanation/Reference:
Explanation:
PRIMARY KEY Constraint
A PRIMARY KEY constraint creates a primary key for the table. Only one primary key can be created for each table. The PRIMARY KEY constraint is a column or a set of columns that uniquely identifies each row in a table. This constraint enforces the uniqueness of the column or column combination and ensures that no column that is part of the primary key can contain a null value.
Note: Because uniqueness is part of the primary key constraint definition, the Oracle server enforces the uniqueness by implicitly creating a unique index on the primary key column or columns.

NEW QUESTION: 4
Which of the following services is not provided by PaaS?
A. Security
B. Database
C. Development tool
D. Middleware
Answer: A