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NEW QUESTION: 1
Your network contains an Active Directory domain named contoso.com. The domain has Remote Desktop Services (RDS) deployed. The deployment contains two Remote Desktop Web Access (RD Web Access) servers named Server1 and Server2.
You need to implement a solution to ensure that users can access RD Web Access if a single RD Web Access server fails.
Which solution should you implement?
More than one answer choice may achieve the goal. Select the BEST answer.
A. DNS round robin
B. Failover Clustering
C. Multipath I/O (MPIO)
D. Network Load Balancing (NLB)
Answer: D
Explanation:
http://blogs.technet.com/b/marobbin2/archive/2012/10/10/configuring-a-windows-2012-rdgnlb-cluster.aspx

NEW QUESTION: 2
As shown in the broadcast network, OSPF mns on four routers and in the same area and on the same network segment. OSPF automatically elects a OR Multiple BORs to achieve better backup results.
A. True
B. False
Answer: B

NEW QUESTION: 3
Which standard access control entry permits from odd-numbered hosts in the 10.0.0.0/24 subnet?
A. Permit 10.0.0.0.0.0.0.1
B. Permit 10.0.0.0.255.255.255.254
C. Permit 10.0.0.1.0.0.0.0
D. Permit 10.0.0.1.0.0.0.254
Answer: D
Explanation:
Explanation
Remember, for the wildcard mask, 1s are I DON'T CARE, and 0s are I CARE. So now let's analyze a simple ACL:
access-list 1 permit 172.23.16.0 0.0.15.255
Two first octets are all 0's meaning that we care about the network 172.23.x.x. The third octet of the wildcard mask, 15 (0000 1111 in binary), means that we care about first 4 bits but don't care about last 4 bits so we allow the third octet in the form of 0001xxxx (minimum:00010000 = 16; maximum: 0001111 = 31).

The fourth octet is 255 (all 1 bits) that means I don't care.
Therefore network 172.23.16.0 0.0.15.255 ranges from 172.23.16.0 to 172.23.31.255.
Now let's consider the wildcard mask of 0.0.0.254 (four octet: 254 = 1111 1110) which means we only care the last bit. Therefore if the last bit of the IP address is a "1" (0000 0001) then only odd numbers are allowed. If the last bit of the IP address is a "0" (0000 0000) then only even numbers are allowed.
Note: In binary, odd numbers are always end with a "1" while even numbers are always end with a "0".
Therefore in this question, only the statement "permit 10.0.0.1 0.0.0.254" will allow all oddnumbered hosts in the 10.0.0.0/24 subnet.