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NEW QUESTION: 1
A company's Web development team has become aware of a certain type of security vulnerability in their Web software. To mitigate the possibility of this vulnerability being exploited, the team wants to modify the software requirements to disallow users from entering HTML as input into their Web application.
What kind of Web application vulnerability likely exists in their software?
A. Cross-site scripting vulnerability
B. Web site defacement vulnerability
C. Cross-site Request Forgery vulnerability
D. SQL injection vulnerability
Many operators of particular web applications (e.g. forums and webmail) allow users to utilize a limited subset of HTML markup. When accepting HTML input from users (say,
<b>very</b> large), output encoding (such as <b>very</b> large) will not suffice since the user input needs to be rendered as HTML by the browser (so it shows as "very large", instead of "<b>very</b> large"). Stopping an XSS attack when accepting HTML input from users is much more complex in this situation. Untrusted HTML input must be run through an HTML sanitization engine to ensure that it does not contain cross-site scripting code.
NEW QUESTION: 2
Which of these is the equation used to derive a 64 Kbps bit rate?
A. 8 kHz x 8-bit code words
B. 2 x 4 kHz x 8-bit code words
C. 2 x 8 kHz x 4-bit code words
D. 2 x 4-bit code words x 8 kHz
Explanation: While the human ear can sense sounds from 20 to 20, 000 Hz, and speech encompasses
sounds from about 200 to 9000 Hz, the telephone channel was designed to operate at about 300 to 3400
Hz. This economical range carries enough fidelity to allow callers to identify the party at the far end and
sense their mood. Nyquist decided to extend the digitization to 4000 Hz, to capture higher-frequency
sounds that the telephone channel may deliver. Therefore, the highest frequency for voice is 4000 Hz.
According to Nyquist theory, we must double the highest frequency, so 2x4kHz = 8kHz.
Each sample will be encoded into a 8-bit code. Therefore 8kHz x 8-bit code = 64 Kbps (notice about the
unit Kbps: 8kHz = 8000 samples per second so 8000 x 8-bit = 64000 bit per second = 64 Kilobit per
second = 64 Kbps)
"When sampling a signal (e.g., converting from an analog signal to digital), the sampling frequency must
be greater than twice the bandwidth of the input signal in order to be able to reconstruct the original
perfectly from the sampled version."