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NEW QUESTION: 1
An IS auditor examining a biometric user authentication system establishes the existence of a control weakness that would allow an unauthorized individual to update the centralized database on the server that is used to store biometric templates. Ofthe following, which is the BEST control against this risk?
A. Kerberos
B. Before-image/after-image logging
C. Multimodal biometrics
D. Vitality detection
Answer: A
Explanation:
Explanation/Reference:
Explanation:
Kerberos is a network authentication protocol for client-server applications that can be used to restrict access to the database to authorized users. Choices B and C are incorrect because vitality detection and multimodal biometrics are controls against spoofing and mimicry attacks. Before-image/after-image logging of database transactions is a detective control, as opposed to Kerberos, which is a preventative control.
NEW QUESTION: 2
A. Option A
B. Option D
C. Option C
D. Option B
Answer: C
Explanation:
Explanation
Shim types are one of two technologies that the Microsoft Fakes Framework uses to let you easily isolate components under test from the environment. Shims divert calls to specific methods to code that you write as part of your test. Many methods return different results dependent on external conditions, but a shim is under the control of your test and can return consistent results at every call. This makes your tests much easier to write.
References:
http://msdn.microsoft.com/en-us/library/hh549176.aspx
NEW QUESTION: 3
Which of the following characteristics MOST separates a tablet from a laptop?
A. Lack of internal wireless
B. Lack of field serviceable parts
C. Lack of any storage medium
D. Lack of a touchscreen
Answer: B
NEW QUESTION: 4
The production server needs additional swap space due to lack of physical memory. The system has only one disk and is partitioned as follows:
Part Tag Flag Cylinders Size Blocks
0 root wm 1041 - 38491 18.00GB (37451/0/0) 37750608 1 swap wu 0 - 1040 512.37MB (1041/0/0) 1049328 2 backup wm 0 - 39532 19.00GB (39533/0/0) 39849264 3 unassigned wm 0 0 (0/0/0) 0 4 unassigned wm 0 0 (0/0/0) 0 5 unassigned wm 0 0 (0/0/0) 0 6 unassigned wm 0 0 (0/0/0) 0 7 unassigned wm 0 0 (0/0/0) 0
There is one swap slice configured and active:
# swap -l swapfile dev swaplo blocks free
/dev/dsk/c0t0d0s1 136,1 16 1049312 0
The root file system is a flat file system:
# df -k /
Filesystem kbytes used avail capacity Mounted on
/dev/dsk/c0t0d0s0 18575269 9549546 9025723 53% /
Which two methods can be used to temporarily add swap space? (Choose two.)
A. create a 512 megabyte swap file on the root disk:
# mkfile 512m /swapfile
# swap -a /swapfile
B. remove the existing swap and read it using swap compression: # swap -d
/dev/dsk/c0t0d0s3
# swap -a -c /dev/dsk/c0t0d0s3
C. repartition the disk so that slice 3 contains the unused cylinders 38492 through 39532,
and then add slice 3 as swap:
# swap -a /dev/dsk/c0t0d0s3
D. use NFS to mount a file system from the development server, and then create and use a
swap file on the remote system:
# mkdir /data
# mount -F nfs tokyo:/export/data /data
# mkfile /data/swapfile
# swap -a /data/swapfile
E. add slice 2 as swap:
# swap -a /dev/dsk/c0t0d0s2
Answer: A,C