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NEW QUESTION: 1
An administrator is visiting a remote site that has on-net calls with headquarters and one voice gateway for PSTN calls. When using RTMT to monitor the bandwidth utilization of the remote site, the administrator notices the OutOfResources counter for the site in LBM has been increasing slowly in last two weeks, but no call failure reports have been sent from this site.
Which description about this issue is true?
A. AAR is routing some of the calls.
B. The bandwidth settings of the site are fulfilling on-net call volume.
C. The location-based CAC does not work properly.
D. The LBM service is malfunctioning.
Answer: A
NEW QUESTION: 2
The following data were extracted from the financial statements of a company for the year ended December 31.
A. US $54.000
B. US $69.000
C. US $17.000
D. US $11.000
Answer: D
Explanation:
Depreciation and amortization are noncash expenses and are added to profit A decrease in receivables indicates that cash collections exceed sales on an accrual basis, so it is added to profit. To account for the difference between cost of goods sold a deduction from profit) and cash paid to suppliers, a two-step adjustment of profit is necessary. The difference between CGS and purchases is the change in inventory. The difference between purchases s and the amount paid to suppliers is the change in accounts payable.
Accordingly, the conversion of .s to cash paid to suppliers requires deducting the inventory increase and adding the accounts payable increase. An increase in plant assets indicates an acquisition of plant assets, causing a decrease in cash, so it is deducted. An increase in share capital represents a cash inflow and is added to profit A decrease in short-term notes payable is deducted from profit because it reflects a cash outflow. Thus, cash increased by US $11.000$70,000 profit + $14,000 + $1.000 + $2,000 - $9,000 +$4.000 - $47.000 T $31.000 - $55,000}.
NEW QUESTION: 3
Given an IP address of 192.168.1.42 255.255.255.248, what is the subnet address?
A. 192.168.1.32/27
B. 192.168.1.16/28
C. 192.168.1.40/29
D. 192.168.1.48/29
E. 192.168.1.8/29
Answer: C
Explanation:
248 mask uses 5 bits (1111 1000) 42 IP in binary is (0010 1010) The base subnet therefore is the lowest binary value that can be written without changing the output of an AND operation of the subnet mask and IP ... 1111 1000 AND 0010 1010 equals 0010 1000 - which is .40 /24 is standard class C mask. adding the 5 bits from the .248 mask gives /29
NEW QUESTION: 4
When using vpn tu, which option must you choose if you only want to clear phase 2 for a specific IP (gateway)?
A. (7) Delete all IPsec+IKE SAs for a given peer (GW)
B. (5) Delete all IPsec SAs for a given peer (GW)
C. (8) Delete all IPsec+IKE SAs for a given User (Client)
D. (6) Delete all IPsec SAs for a given User (Client)
Answer: B