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Sliva [168]
3 years ago
7

A ____________ variable is available throughout a program.

Computers and Technology
1 answer:
kvv77 [185]3 years ago
3 0

Answer:

global

Explanation:

Its the global variable, and its scope is the whole program. And hence, a global variable is available throughout the program. And you need to know that a local variable has its scope in the block where it has been declared. Its scope is that block or the function or the loop only. And if you want it to have a global scope, then you will have to declare it as a global variable.

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Is the process of modifying something to make it fit certain criteria.
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Design Process

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Write a program that prompts the user to enter a date, using integer values for the month, day, and year, and then prints out th
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Answer:

import datetime

user = input("Enter date in yyyy,m,d: ").split(",")

int_date = tuple([int(x) for x in user])

year, month, day =int_date

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print(mydate)

x = mydate.strftime("%B %d, %Y was a %A")

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Answer:

E

Explanation:

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7 0
2 years ago
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Select all that apply. Two physical things you can do to demonstrate you are paying attention to a speaker are _____ and _____.
omeli [17]
B and d ............
7 0
3 years ago
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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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