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NEW QUESTION: 1
Refer to the exhibit.

Why was this message received?
A. The login command has not been set on CON 0
B. No enable password has been set.
C. The login command has not been set on the VTY ports.
D. No console password has been set.
E. No enable secret password has been set.
F. No VTY password has been set.
Answer: F
Explanation:
Your CCNA certification exam is likely going to contain questions about Telnet, an application-level protocol that
allows remote communication between two networking devices. With Telnet use being as common as it is, you had
better know the details of how to configure it in order to pass your CCNA exam and to work in real-world networks.
The basic concept is pretty simple - we want to configure R1, but we're at R2. If we telnet successfully to R1, we will
be able to configure R1 if we've been given the proper permission levels. In this CCNA case study, R2 has an IP address
of 172.12.123.2 and R1 an address of 172.12.123.1. Let's try to telnet from R2 to R1.
R2#telnet 172.12.123.1
Trying 172.12.123.1 ... Open
Password required, but none set
[Connection to 172.12.123.1 closed by foreign host]
This seems like a problem, but it's a problem we're happy to have. A Cisco router will not let any user telnet to it by
default. That's a good thing, because we don't want just anyone connecting to our router! The "password required"
message means that no password has been set on the VTY lines on R1. Let's do so now.
R1(config)#line vty 0 4
R1(config-line)#password baseball
A password of "baseball" has been set on the VTY lines, so we shouldn't have any trouble using Telnet to get from R2
to R1. Let's try that now.
R2#telnet 172.12.123.1
Trying 172.12.123.1 ... Open
User Access Verification
Password:
R1>
We're in, and placed into user exec mode.
Reference: http://www.mcmcse.com/cisco/guides/telnet_passwords_and_privilege_levels.shtml

NEW QUESTION: 2
You have the following C/AL code segment:
ItemLedgerEntry.SETCURRENTKEY("Item No.");
ItemLedgerEntry.FINDSET;
You have the following query, based on the C/AL code segment:
SELECT * FROM "CRONUS International Ltd_$Item Ledger Entry" ORDER BY "Item No_", "Posting Date", "Entry No_"
Which two statements are true based on the query? (Each correct answer presents a complete solution. Choose two.)
A. If the second key in the table is "Item No.", then it must be disabled. The key "Item No.", "Posting Date" will be used instead.
B. The SQLIndex value of the key "Item No.", "Posting Date" has been entered with the value "Item No.".
C. The key "Item No.", "Posting Date" is the first key in the list that starts with the field "Item No." in the Item Ledger Entry table.
D. The SQLIndex value of the key "Item No." has been entered with the value "Item No.", "Posting Date".
Answer: A,D
Explanation:
Ref: http://msdn.microsoft.com/en-us/library/hh168524(v=nav.70).aspx

NEW QUESTION: 3
The team establishes ground rules for considering business analysis performance improvements. They agree that speed and accuracy of executing analysis activities becomes very important at the following horizon:
A. Strategy horizon
B. Feedback horizon
C. Initiative horizon
D. Delivery Horizon
Answer: 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

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.