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NEW QUESTION: 1
How many G.711 calls (20 ms sampling) can traverse an IP/PPP T1 link without packet loss?
A. 0
B. 1
C. 2
D. 3
Answer: B
NEW QUESTION: 2
You want to repair Windows by using the Windows 7 System Recovery Disc. Before you begin, which files should you back up to protect your personal data?
A. Everything in My Documents.
B. All files in the Program Files directory.
C. All files in the Windows directory.
D. All files in drive C.
Answer: A
NEW QUESTION: 3
Your enterprise structure has one ledger and two business units. Business unit one wants to enable budgetary control for Requisitioning only on Procure-to-Pay Business Functions and business unit two wants to enable budgetary control for Payable Invoicing only in Procure-to-Pay Business Functions. Which two statements are correct? (Choose two.)
A. Define budgetary control at ledger level and only encumbrance control at the business units
B. Define budgetary control at ledger level with Budgetary Control Exceptions for each business unit
C. While defining control for business unit one, disable control for Purchasing, Payable Invoicing, and Receiving
D. Define control for business unit two to disable control for Requisitioning, Purchasing, and Receiving
E. While defining control for business unit one, enable control at purchasing and define the exceptions to only include requisitioning
F. While defining control for business unit two, enable control at Requisitioning and define the exceptions to only include invoicing
Answer: C,F
Explanation:
Section: (none)
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.