Recall that
d/dx sech(x) = - sech(x) tanh(x)
d/dx tan⁻¹(x) = 1/(1 + x²)
Then by the chain rule,
dy/dx = - sech(x) tanh(x) / (1 + x²)
Answer:
Explanation has been given below
Step-by-step explanation:
a) inter arrival times are exponentially distributed with mean 1/n , where n = rate = 1/sec.
probability distribution function is F(t)=n*exp(-n*t).
reference to any kth packet and the (k-1)th packet
the answer is = integration of F(t).dt with limits 0 to 2 = 1 - exp(-2*n) = 1 - exp(-2)
b) t=5 , P(q) = exp(-5)*(5)^q/factorial(q)
probability of fourth call within t=5 seconds is =
that is P(4) P(5) ...... = 1 - ( P(0) P(1) P(2) P(3) ) ; put the values and get the answer.
c) number of calls/rate = 4/n = 4 seconds
Answer: 9pi
Step-by-step explanation:
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