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NEW QUESTION: 1
Sales representatives use a custom report type for Account reports. New fields have been created on the Account Object.
What should a System Administrator do to report on the newly created fields?
A. From the Custom Report Type, Edit Object Relationships, then add the new fields to the report.
B. Create a new account report and add the new fields from the Report Builder.
C. Create a new account report folder, go to share, and then add the new fields.
D. From the Custom Report Type, Edit Layout, then add the new fields to the report. Calculator
Answer: D

NEW QUESTION: 2
You administer a Microsoft SQL Server 2012 database. The database contains a table named Employee.
Part of the Employee table is shown in the exhibit. (Click the Exhibit button.)

Unless stated above, no columns in the Employee table reference other tables.
Confidential information about the employees is stored in a separate table named EmployeeData. One record exists within EmployeeData for each record in the Employee table.
You need to assign the appropriate constraints and table properties to ensure data integrity and visibility.
On which column in the Employee table should you a create a self-reference foreign key constraint?
A. DateHired
B. FirstName
C. LastName
D. DepartmentID
E. EmployeelD
F. ReportsToID
G. EmployeeNum
H. MiddleName
I. JobTitle
Answer: F

NEW QUESTION: 3
What is a chain of custody?
A. Chain of custody refers to obtaining preemptive court order to restrict further damage of evidence in electronic seizures
B. A legal document that demonstrates the progression of evidence as it travels from the original evidence location to the forensic laboratory
C. It is a search warrant that is required for seizing evidence at a crime scene
D. It Is a document that lists chain of windows process events
Answer: B

NEW QUESTION: 4
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 10.0.0.1.0.0.0.0
C. Permit 10.0.0.0.255.255.255.254
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.