<span>A search engine displays a list of webpage names that contain the search text. The term for that list is hit.
Basically, what this term refers to is the number of visits, or a number of downloads that happened on that particular webpage. It is a good tool to measure web traffic, or how much that page is visited.
</span>
Answer:
Some hardware requirement: Monitor, keyboard, mouse, sound card, memory, processor, graphics display card. Some software requirement: Windows XP/Vista, Video for Windows, Quicktime.
Explanation:
BRAINLEST
Answer: 83.17
Explanation:
By definition, the dB is an adimensional unit, used to simplify calculations when numbers are either too big or too small, specially in telecommunications.
It applies specifically to power, and it is defined as follows:
P (dB) = 10 log₁₀ P₁ / P₂
Usually P₂ is a reference, for instance, if P₂ = 1 mW, dB is called as dBm (dB referred to 1 mW), but it is always adimensional.
In our question, we know that we have a numerical ratio, that is expressed in dB as 19.2 dB.
Applying the dB definition, we can write the following:
10 log₁₀ X = 19.2 ⇒ log₁₀ X = 19.2 / 10 = 1.92
Solving the logarithmic equation, we can compute X as follows:
X = 10^1.92 = 83.17
X = 83.17
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
