Answer:
Streaming movies may stop to buffer
Explanation:
microphones have a set record limit, and games downloaded wont generate errors unless something in the code is wrong, printers don't really need to connect to internet to work except for some, and streaming movies buffer because as you are watching them it downloads the next few minutes.
Answer:
1.) 25 ; 15 ; 15
2.) 50 ; 15 ; 50
Explanation:
In the first function written :
The variable val was initially decaled or assigned a value of 25 and that was what was printed first.
However, after the example function was written, the val variable was finally assiagned a value of 15 within the function. However, it was also declared that the global variable takes uonthe val value. Hence, the val variable initially assigned a value, of 25 changes to 15 globally.
For the second code :
From the top:
Val was assigned a value of 50 ;
Hence,
print(val) gives an output of 50
Within the function definition which prints the value of val that is assigned a value of 25 within the function.
Since tbe global variable isnt reset.
Printing Val again outputs 50;since ito is outside the function.
There is not really a key benifit, but it does help with little things.
Answer:
The correct answer to the following question will be "Peak capacity" and "Bandwidth starvation".
Explanation:
Peak capacity has been used to characterize the gradient aqueous phase separation efficiency or performance. It represents the overall conceptual number of operations or components which can be isolated consistently with something like a given set of analytical circumstances and column with
⇒ Rs =1 (Figure 1 and Equation 1)
Certain traffic competing at its policies for the available or unused bandwidth will theoretically enable classes with lower value rates to starve to bandwidth.
Due to these activities, Sharon is concerned about "Bandwidth starvation" and "Peak capacity".
Explanation:
Here, in the given statement, maximum flow capacity of the ALOHA channel = 18.4% = 0.184
Then, the stations N share 56 kbps
And, the channel rate = 
Then, the outputs of the wach stations is 
, the output of the bits/ sec by each stations = 
So, The output of the N stations is 10 bits/sec on the channel that having 
Finally, N = 
