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NEW QUESTION: 1
Which two DHCP messages are always sent as broadcast? (Choose two.)
A. DHCPDISCOVER
B. DHCPRELEASE
C. DHCPDECLINE
D. DHCPREQUEST
E. DHCPOFFER
Answer: A,D
Explanation:
DHCP discovery
The client broadcasts messages DHCPDISCOVER on the network subnet using the destination
address 255.255.255.255 or the specific subnet broadcast address.
DHCP request
In response to the DHCP offer, the client replies with a DHCP request, broadcasts to the server,
requesting the offered address. A client can receive DHCP offers from multiple servers, but it will
accept only one DHCP offer.
Reference. http://en.wikipedia.org/wiki/Dynamic_Host_Configuration_Protocol

NEW QUESTION: 2
Which of the following can be a common objective of testing?
A. Providing information as part of the debugging activity
B. Fixing defects to improve the system's quality characteristics
C. Making sure the system performs as fast and as efficient as needed
D. Gaining confidence about the level of the system's quality
Answer: B

NEW QUESTION: 3
In an analytic view, you need two different logical joins from the data foundation to an attribute view. The analytic view already contains the attribute view but you have to use the attribute view again. To which of the following do you connect the second join?
A. To a shared attribute view
B. To a derived attribute view
C. To a new instance of the attribute view
D. To a copy of the attribute view
Answer: B,D

NEW QUESTION: 4
Drag and Drop Question
Drag and drop the IPv4 network subnets from the left onto the correct usable host ranges on the right Select and Place:

Answer:
Explanation:

Explanation:
This subnet question requires us to grasp how to subnet very well. To quickly find out the subnet range, we have to find out the increment and the network address of each subnet. Let's take an example with the subnet
172.28.228.144/18:
From the /18 (= 1100 0000 in the 3rd octet), we find out the increment is 64. Therefore the network address of this subnet must be the greatest multiple of the increment but not greater than the value in the 3rd octet (228). We can find out the 3rd octet of the network address is 192 (because 192 = 64 * 3 and 192 < 228) -> The network address is 172.28.192.0. So the first usable host should be 172.28.192.1 and it matches with the 5th answer on the right. In this case we don't need to calculate the broadcast address because we found the correct answer.
Let's take another example with subnet 172.28.228.144/23 -> The increment is 2 (as /23 = 1111 1110 in 3rd octet) -> The 3rd octet of the network address is 228 (because 228 is the multiply of 2 and equal to the 3rd octet) -> The network address is 172.28.228.0 -> The first usable host is 172.28.228.1. It is not necessary but if we want to find out the broadcast address of this subnet, we can find out the next network address, which is 172.28.(228 + the increment number).0 or 172.28.230.0 then reduce 1 bit -> 172.28.229.255 is the broadcast address of our subnet. Therefore the last usable host is 172.28.229.254.