Answer:
Lamda= 4 students/min, µ= 5 students/min
P= Lamda/µ= 4/5= 0.8
a.) Probability that system is empty= P0= 1-P= 1-0.8= 0.2
b.) Probability of more than 2 students in the system= ∑(n=3 to inf) P^n*P0= (1-P)*(1/(1-P) – (1-P) –(1-P)*P –(1-P)*P^2)= (.2)*(5- - .2 - (.8)*.2 – (.2)*.8^2))= 0.848
Probability of more than 3 students in the system= ∑(n=4 to inf) P^n*P0= (1-P)*(1/(1-P) – (1-P) –(1-P)*P –(1-P)*P^2 – (1-P)*P^3)= 0.768
c.) W(q)= Waiting time in Queue= lamda/µ(µ- lamda)= 4/5(1)= 0.8 minutes
d.) L(q)= lamda*W(q)= 4*.8= 3.2 students
e.) L(System)= lamda/(µ-lamda)= 4 students.
f.) If another server with same efficiency as the 1st one is added, then µ= 6 sec/student= 10 students/min.
P= 4/10= 0.4
Probability that system is empty= P0= 1-.4= 0.6
W(q)= 4/10(10-4)= 0.0667 minutes
L(q)= Lamda*W(q)= 4*.0667=0.2668
L(system)= Lamda/(µ-lamda)= 4/6= .667
Step-by-step explanation:
Answer:
3.9 x 107
Step-by-step explanation:
Answer:
6, 7, 8, 9
Step-by-step explanation:
you plug in the number in the X column into the variable, which you divide by 4 and add 7 to your number.
Answer:
18 cm.
Step-by-step explanation:
The circumference of a circle is found by calculating 2 * pi * r.
In this case, the circumference is 36 pi cm.
2 * pi * r = 36 * pi
2 * r = 36
r = 36 / 2
r = 18 cm.
Hope this helps!
<span>8.4 + (13.1-0.6) equals 20.9
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