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Lady bird [3.3K]
3 years ago
8

Which ergonomic principle helps to maintain good posture?

Computers and Technology
2 answers:
Julli [10]3 years ago
6 0

Answer:

d is the answer i took the test

Explanation:

Sophie [7]3 years ago
5 0

Answer:

D. keep your hand and forearm in the same plane

Explanation:

The ergonomic principle explains how to maintain a good posture. And if hands and forearms are in the same plane then you will do least work, and hence not loss more energy, which you can lose if your elbows are close together while typing or your hand and forearm is not in the same plane, or your hands are crossed, or your neck is non aligned to your spines. And if you want to work more efficiently in your office, then you should study ergonomic principles certainly.

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Using ______ capabilities, network managers can connect VOIP phones directly into a VLAN switch and configure the switch to rese
aniked [119]

Using <u>QOS</u> capabilities, network managers can connect VOIP phones directly into a VLAN switch and configure the switch to reserve sufficient network capacity so that they will always be able to send and receive voice messages.

<u>Explanation:</u>

Basically VOIP means voice over the tcpip address. Each phone set is are connected with network socket and connected to VLAN switch. Advantage of VOIP no need to have separated land line connection.

End user can call either local or international by network call. Network administrator cans also general report on call logs and control the end user on local or international calls.

Due to VOIP an organization avoids telephone bills which are routed on network (internet).

All VOIP is service oriented by server. QOS Service is used to operation from windows server.

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3 years ago
How can I record Tv shows on the Android OTT TV BOX either on the Box or on Kodiak without buying a Tv Tuner?
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3 years ago
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Explain the importance of system software in the computer​
AveGali [126]

Answer:

System software is a type of computer program that is designed to run a computer's hardware and application programs. If we think of the computer system as a layered model, the system software is the interface between the hardware and user applications. The OS manages all the other programs in a computer.

Explanation:

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3 years ago
True or false? Software application packages function as the interface between the operating system and firmware.
olga2289 [7]

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8 0
4 years ago
Suppose a host has a 1-MB file that is to be sent to another host. The file takes 1 second of CPU time to compress 50%, or 2 sec
kozerog [31]

Answer: bandwidth = 0.10 MB/s

Explanation:

Given

Total Time = Compression Time + Transmission Time

Transmission Time = RTT + (1 / Bandwidth) xTransferSize

Transmission Time = RTT + (0.50 MB / Bandwidth)

Transfer Size = 0.50 MB

Total Time = Compression Time + RTT + (0.50 MB /Bandwidth)

Total Time = 1 s + RTT + (0.50 MB / Bandwidth)

Compression Time = 1 sec

Situation B:

Total Time = Compression Time + Transmission Time

Transmission Time = RTT + (1 / Bandwidth) xTransferSize

Transmission Time = RTT + (0.40 MB / Bandwidth)

Transfer Size = 0.40 MB

Total Time = Compression Time + RTT + (0.40 MB /Bandwidth)

Total Time = 2 s + RTT + (0.40 MB / Bandwidth)

Compression Time = 2 sec

Setting the total times equal:

1 s + RTT + (0.50 MB / Bandwidth) = 2 s + RTT + (0.40 MB /Bandwidth)

As the equation is simplified, the RTT term drops out(which will be discussed later):

1 s + (0.50 MB / Bandwidth) = 2 s + (0.40 MB /Bandwidth)

Like terms are collected:

(0.50 MB / Bandwidth) - (0.40 MB / Bandwidth) = 2 s - 1s

0.10 MB / Bandwidth = 1 s

Algebra is applied:

0.10 MB / 1 s = Bandwidth

Simplify:

0.10 MB/s = Bandwidth

The bandwidth, at which the two total times are equivalent, is 0.10 MB/s, or 800 kbps.

(2) . Assume the RTT for the network connection is 200 ms.

For situtation 1:  

Total Time = Compression Time + RTT + (1/Bandwidth) xTransferSize

Total Time = 1 sec + 0.200 sec + (1 / 0.10 MB/s) x 0.50 MB

Total Time = 1.2 sec + 5 sec

Total Time = 6.2 sec

For situation 2:

Total Time = Compression Time + RTT + (1/Bandwidth) xTransferSize

Total Time = 2 sec + 0.200 sec + (1 / 0.10 MB/s) x 0.40 MB

Total Time = 2.2 sec + 4 sec

Total Time = 6.2 sec

Thus, latency is not a factor.

5 0
3 years ago
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