Suppose users share a 2 mbps link. also suppose each user transmits continuously at 1 mbps when transmitting, but each user transmits only 20 percent of the time. (consider statistical multiplexing discussed in lecture.) (a) (3 points) when circuit switching is used, how many users can be supported? (b) (3 points) for the reminder of this problem, suppose packet switching is used. why will there be essentially no queuing delay before the link if two or fewer users transmit at the same time? why will there be a queuing delay if three users transmit at the same time? (c) (3 points) find the probability that a given user is transmitting. (d) (6 points) suppose now there are three users. find the probability that at any given time, all three users are transmitting simultaneously.

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.

step-by-step explanation:

1

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.