Answer:
The answer is below
Explanation:
Given that:
Frame transmission time (X) = 40 ms
Requests = 50 requests/sec, Therefore the arrival rate for frame (G) = 50 request * 40 ms = 2 request
a) Probability that there is success on the first attempt = 
but k = 0, therefore Probability that there is success on the first attempt = 
b) probability of exactly k collisions and then a success = P(collisions in k attempts) × P(success in k+1 attempt)
P(collisions in k attempts) = [1-Probability that there is success on the first attempt]^k = ![[1-e^{-G}]^k=[1-0.135]^k=0.865^k](https://tex.z-dn.net/?f=%5B1-e%5E%7B-G%7D%5D%5Ek%3D%5B1-0.135%5D%5Ek%3D0.865%5Ek)
P(success in k+1 attempt) =
Probability of exactly k collisions and then a success = 
c) Expected number of transmission attempts needed = probability of success in k transmission = 
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Answer:
The code is given below
Explanation:
The correct syntax would be to place appropriate parenthesis.
(month==1?"jan":(month==2?"feb":(month==3?"mar":(month==4?"apr":(month==5?"may":(month==6?"jun":(month==7?"jul":(month==8?"aug":(month==9?"sep":(month==10?"oct":(month==11?"nov":"dec")))))))))));
Similarly, you can also use the following code:
String[] months = { "jan", "feb", "mar", "apr", "may", "jun", "jul", "aug", "sep", "oct", "nov", "dec" };
int month = 1;
String monthDescription = months[month - 1];
To get the number ot decoder output:
2^n; where n is number of input
2^4
2 × 2 × 2 × 2 = 16 outputs
