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Trava [24]
3 years ago
11

Hardware device can be either internal hardware devices or external hardware devices.

Computers and Technology
1 answer:
mariarad [96]3 years ago
6 0

Answer:

Yes

Explanation:

Because keyboard and mouse are external hardware and not inside the CPU cabinet but hardwares such as heat sink, sound card and graphic card are external hardware.

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Circuit-switched networks provide connection between two parties that: Group of answer choices is dedicated for use by the parti
loris [4]

Answer:

The answer is "The first choice"

Explanation:

The circuit switches are used to provide a network that connects among the two parties, which are dedicated for the use to the parties for the length of the connection. It is a familiar technique, that is used to create a network for communication, which is used on telecalls. It also enables the hardware and circuits to share among the users, and for every user, it has direct access to the circuit during the use of the network.

7 0
3 years ago
The one who will defeat me in this typing race I will mark the one brainliest:
fredd [130]

Answer:

yes

Explanation:

can you send the link....

5 0
3 years ago
Read 2 more answers
g What field in the IPv4 datagram header can be used to ensure that a packet is forwarded through no more than N routers? When a
ad-work [718]

Answer:

a) Time to live field

b) Destination

c) Yes, they have two ip addresses.

d) 128 bits

e) 32 hexadecimal digits

Explanation:

a) the time to live field (TTL) indicates how long a packet can survive in a network and whether the packet should be discarded. The TTL is filled to limit the number of packets passing through N routers.

b) When a large datagram is fragmented into multiple smaller datagrams, they are reassembled at the destination into a single large datagram before beung passed to the next layer.

c) Yes, each router has a unique IP address that can be used to identify it. Each router has two IP addresses, each assigned to the wide area network interface and the local area network interface.

d) IPv6 addresses are represented by eight our characters hexadecimal numbers. Each hexadecimal number have 16 bits making a total of 128 bits (8 × 16)  

e) IPv6 address has 32 hexadecimal digits with 4 bits/hex digit

4 0
3 years ago
In addition to regular watch features, which two features are often found on smart watches?
andriy [413]
Phone capabilities and fitness monitoring
5 0
3 years ago
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
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