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Elis [28]
3 years ago
8

Michale spent 80 minutes at the libary he finished studying at 4:00 what time did michale start studying

Computers and Technology
1 answer:
Papessa [141]3 years ago
7 0

Answer:

He finished at 2:40 P.M.

Explanation:

This is because 80 minutes= 1 hour and 20 minutes. Add 1 hour and 20 minutes to 2:40, which equals 4:00.

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NAS - network attached storage
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A search engine displays a list of webpage names that contain the search text. what is the term for that list?
Minchanka [31]
<span>A search engine displays a list of webpage names that contain the search text. The term for that list is hit.
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3 years ago
What are hard ware and software requirments in multimedia computer system
marishachu [46]

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:

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4 0
3 years ago
What is the numerical ratio that is equivalent to 19.2 dB?
mrs_skeptik [129]

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

4 0
3 years ago
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For f
julia-pushkina [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:

  • Tw = K*512* bit time

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

5 0
3 years ago
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