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prisoha [69]
3 years ago
10

Which type of film would typically require the least amount of input from a screenwriter?

Computers and Technology
1 answer:
denpristay [2]3 years ago
8 0

Answer:

comedy

Explanation:

thats for the writers ig

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High School or when you get hired.

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Quality assurance testing plays a very crucial role in the process of software development. It saves money and time by spotting bugs and errors early in the software development life-cycle. Software that is newly developed may have inconsistencies, redundancies, and flaws; therefore, proper QA testing is mandatory.

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Which task can a company perform with intranets
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C.) store data securely would be my choice

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Answer the question ASAP please
RUDIKE [14]

Answer:

Shows the programming checking if num1 is greater than num2

Explanation:

So num1 and num2 are inputs

for you to code this you would need to put

num1=int(input("What is your first number? ))

and the same for num2 except change num1 for num 2 and first for second

When the input is completed, the computer will check if num 1 is greater than num2

it will do this by using a code something like:

if num1>num2:

    Print("Your first input was greater than your second")

But in this example if it greater it just ends

But if it was less than you would put

if num1>num2:

    Print("Your first input was greater than your second")

elif num1<num2:

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So basically this code shows the computer checking if one number is greater than the other or not

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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

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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

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