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
Hello from MrBillDoesMath!
Answer:
-10w - 20
Discussion:
5 ( -2w - 4 ) =
5(-2w) - 5(4) =
-10w - 20
Thank you,
MrB
Average speed = (distance final - distance initial)/(time final - time initial)
Average speed = (30-0)km /(100 -0)min = 3/10 km/min
3/10 km/min *60 min/1 h = 180/10 km/h= 18 km/h
Answer:
-2*3
Step-by-step explanation:
If she had to refund $2 for each cup (-2) and she refunded 3 cups (3) she would lose a balance of -2*3.