Answer:
Option(c) is the correct answer.
Explanation:
Cross-site scripting is the type of security breach that are usually found in the software applications.The main objective of cross site scripting it is used by the hackers to exploit the data security.
- The cross site scripting is the collection of web pages that enables people to insert the text like comment, name stores it afterwards it save the data and then it appears to the other users.
- Others options are incorrect because they are not related to given scenario.
Solution:
Running computer programs and their data are stored in Rom.
ROM is "built-in" computer memory containing data that normally can only be read, not written to. ROM contains the programming that allows your computer to be "booted up" or regenerated each time you turn it on. Unlike a computer's random access memory (RAM), the data in ROM is not lost when the computer power is turned off. The ROM is sustained by a small long-life battery in your computer.
Answer: E. class, objects
Explanation:
<em>A </em><em><u>class</u></em><em> is the blueprint for </em><em><u>objects</u></em><em> having similar attributes.</em>
As much as classifying something means to define it based on the characteristics that is has that are similar to other things in that class, so also is a class here.
A class is the blueprint or rather template for making objects that have similar attributes which means that the class therefore gives the object its various attributes and its behavior.
Answer:
The VI in LabView indicates c-Virtual Instrument
Explanation:
The VI in LabView is a program-subroutine. The VI stands for Virtual Instrument. The VI is composed of a Block diagram, Connector panel and a Front Panel.
Complete Question:
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?
Answer:
a) 51.2 msec. b) 5.12 msec
Explanation:
If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:
The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:
a) BW = 1 Mbps = 10⁶ bps
⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec
b) BW = 10 Mbps = 10⁷ bps
⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec
