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NEW QUESTION: 1
Your network contains one Active Directory domain named contoso.com. The domain contains three users named User1, User2, and User3. You need to ensure that the users can log on to the domain by using the user principal names (UPNs) shown in the following table.
What should you use?
A. the Set-ADDomain cmdlet
B. the Setspn command
C. the Set-ADUser cmdlet
D. the Add-DNSServerSecondaryZone cmdlet
Answer: C
Explanation:
Explanation/Reference:
References:
https://technet.microsoft.com/en-us/library/ee617215.aspx
NEW QUESTION: 2
Click the Exhibit button. Referring to the exhibit, you are asked to ensure that host-1 can communicate with host-3 while also allowing host-2 to communicate with host-4.
What should you do to enable this behavior?
A. Configure the native-vlan-id default statement under the ge-0/0/1 port settings on Switch-1.
B. Configure the native-vlan-id default statement under the ge-0/0/1 port settings on both Switch-1 and Switch-2
C. Use the all keyword when defining the member VLANs for the ge-0/0/1 interface on both Switch-1 and Switch-2.
D. Use the all keyword when defining the member VLANs for the ge-0/0/1 interface on Switch-1.
Answer: B
Explanation:
Explanation/Reference:
Explanation:
Configuring Mixed Tagging Support for Untagged Packets For 1-, 4-, and 8-port Gigabit Ethernet IQ2 and IQ2-E PICs, for 1-port 10-Gigabit Ethernet IQ2 and IQ2-E PICs, for all MX Series router Gigabit Ethernet, Tri-Rate Ethernet copper, and 10-Gigabit Ethernet interfaces configured for 802.1Q flexible VLAN tagging, and for aggregated Ethernet interfaces on IQ2 and IQ2-E PICs or MX Series DPCs, you can configure mixed tagging support for untagged packets on a port.
Untagged packets are accepted on the same mixed VLAN-tagged port. To accept untagged packets, include the native-vlan-id statement and the flexible-vlan-tagging statement at the [edit interfaces interface- name] hierarchy level: [edit interfaces ge-fpc/pic/port] flexible-vlan-tagging; native-vlan-id number; Reference:
https://www.juniper.net/documentation/en_US/junos13.3/topics/usage-guidelines/interfaces- enabling-vla n-tagging.html
NEW QUESTION: 3
Examine the exhibit to view the query and its execution plan.
Identify the two correct interpretations that can be made from the execution plan.
A. The EMP table is the driving table and the DEPT table us the driven table.
B. The DEPT table is driving table and the EMP table join is the driven table.
C. Rows from both the tables are sorted by the join key, but only rows from the DEPT table are hashed into memory.
D. Rows from the DEPT table are first hashed by the join key into memory and then joined to the EMP table on the join key.
E. The rows from the DEPT table are sorted first by the join key and then hashed into memory.
Answer: B,D
Explanation:
Note: *A hash join is performed by hashing one data set into memory based on join columns and reading the other one and probing the hash table for matches. The hash join is very low cost when the hash table can be held entirely in memory, with the total cost amounting to very little more than the cost of reading the data sets. The cost rises if the hash table has to
be spilled to disk in a one-pass sort, and rises considerably for a multipass sort.
You should note that hash joins can only be used for equi-joins, but merge joins are more flexible.
In general, if you are joining large amounts of data in an equi-join then a hash join is going to be a better bet.
*The 'driving' table is the table we will join FROM -- that is JOIN TO other tables. For example, lets say you have the query:
select * from emp, dept where emp.deptno = dept.deptno;
In this case the driving table might be DEPT, we would fetch rows from DEPT in a full scan and then find the rows in EMP that match. DEPT is the driving table.
NEW QUESTION: 4
A company has two production departments and two service departments (Maintenance and Stores). The overhead costs of each of the departments are as follows.
The following equations represent the reapportionment of each of the service department overheads to the other.
M = 4,700 + 0.1S
S = 5,800 + 0.2M
Where M = total Maintenance overhead after reapportionment from Stores
S = total Stores overhead after reapportionment from Maintenance
60% of the total Maintenance overhead and 50% of the total Stores overhead are to be apportioned to Production Department 1.
The total production overhead for Production Department 1 after reapportionment of the service departments' overhead costs is closest to:
A. $75,500
B. $70,720
C. $71,672
D. $71,821
Answer: B
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