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Mkey [24]
3 years ago
12

Binary numbers are based on __________.

Computers and Technology
2 answers:
Lena [83]3 years ago
8 0
D. Two digits (1s and 0s
yanalaym [24]3 years ago
5 0

Answer: 0 and 1s.

Explanation:

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Computer Networks - Queues
lyudmila [28]

Answer:

the average arrival rate \lambda in units of packets/second is 15.24 kbps

the average number of packets w waiting to be serviced in the buffer is 762 bits

Explanation:

Given that:

A single channel with a capacity of 64 kbps.

Average packet waiting time T_w in the buffer = 0.05 second

Average number of packets in residence = 1 packet

Average packet length r = 1000 bits

What are the average arrival rate \lambda in units of packets/second and the average number of packets w waiting to be serviced in the buffer?

The Conservation of Time and Messages ;

E(R) = E(W) + ρ

r = w + ρ

Using Little law ;

r = λ × T_r

w =  λ × T_w

r /  λ = w / λ  +  ρ / λ

T_r =T_w + 1 / μ

T_r = T_w +T_s

where ;

ρ = utilisation fraction of time facility

r = mean number of item in the system waiting to be served

w = mean number of packet waiting to be served

λ = mean number of arrival per second

T_r =mean time an item spent in the system

T_w = mean waiting time

μ = traffic intensity

T_s = mean service time for each arrival

the average arrival rate \lambda in units of packets/second; we have the following.

First let's determine the serving time T_s

the serving time T_s  = \dfrac{1000}{64*1000}

= 0.015625

now; the mean time an item spent in the system T_r = T_w +T_s

where;

T_w = 0.05    (i.e the average packet waiting time)

T_s = 0.015625

T_r =  0.05 + 0.015625

T_r =  0.065625

However; the  mean number of arrival per second λ is;

r = λ × T_r

λ = r /  T_r

λ = 1000 / 0.065625

λ = 15238.09524 bps

λ ≅ 15.24 kbps

Thus;  the average arrival rate \lambda in units of packets/second is 15.24 kbps

b) Determine the average number of packets w waiting to be serviced in the buffer.

mean number of packets  w waiting to be served is calculated using the formula

w =  λ × T_w

where;

T_w = 0.05

w = 15238.09524 × 0.05

w = 761.904762

w ≅ 762 bits

Thus; the average number of packets w waiting to be serviced in the buffer is 762 bits

4 0
3 years ago
Java uses ____ to implement method lookup.A) a jump tableB) an inheritance treeC) restricted accessD) polymorphism
sashaice [31]

Answer:

D) polymorphism

7 0
3 years ago
What is one effect the internet has had on the library
Shkiper50 [21]

Answer:

The internet provides access to an abundance of information from home, making it easier for people to get answers much faster. This efficiency has caused libraries to receive less business and less users, therefore making their service less popular.

7 0
3 years ago
Explain the types of computer on the basis of model<br>​
RoseWind [281]

XT(extra technology) computer: it cannot support GUI Based operating system. its processing speed is 477MHz

AT(advanced technology):it supports GUI Operating system. Its processing speed is 2GHz

PS/2:is a laptop computer which is rechargeable

and battery powered. it's operated with OS/2 operating system.

PLEASE MARK AS BRAINLIEST

6 0
3 years ago
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What do you use for soaking hands and holding soapy water​
sertanlavr [38]

Answer

???

Explanation:

Was this a school question?

5 0
3 years ago
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