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abruzzese [7]
3 years ago
6

Green field country is planning to conduct a cricket match between two teams A and B. a large crowd is expected in the stadium a

s 10000 tickets have already been sold. To fulfill the viewer's requirements, the match administrators are willing to install large display screens inside the stadium
Which should display the match Live through several cameras installed on different angles in the stadium.
After carefully reading and analyzing the given scenario, you are required to suggest the most suitable serial transmission mode with two strong reasons.
Computers and Technology
1 answer:
AnnZ [28]3 years ago
5 0

Answer:

For this requirement, we first need the Network Interface Card which supports the serial transmission, and two such can be RS-485 and EIA RS-422A. And you need to remember that parallel transmission is speedier than serial, however, they are more prone to the noise. And in both of the above cases, we can have multiple receivers and one transmitter and all that over a single line. And this line can be as long as 6000 feet and the speed can be as high as 10 MBPS, which is quite awesome and certainly apt for the above requirement.

Explanation:

Please check the answer.

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________ is the actual speed of data transfer that is achieved between two nodes on a network and is always less than or equal t
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<u>Throughput </u> is the actual speed of data transfer that is achieved between two nodes on a network and is always less than or equal to the data transfer rate.

<h3>What is used to transfer data packets between two or more networks?</h3>

A router is a networking device that forwards data packets between computer networks. Routers perform the traffic directing functions on the Internet.

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7 0
2 years ago
. (a) Prove or disprove carefully and in detail: (i) Θ is transitive and (ii) ω is transitive. (b) Assume n is a positive intege
Sergio [31]

Answer:

The Following are the solution to this question:

Explanation:

In Option a:

In the point (i) \Omega is transitive, which means it converts one action to others object because if \Omega(f(n))=g(n) indicates c.g(n). It's true by definition, that becomes valid. But if \Omega(g(n))=h(n), which implies c.h(n). it's a very essential component. If c.h(n) < = g(n) = f(n) \. They  \Omega(f(n))   will also be h(n).  

In point (ii), The  value of \Theta is convergent since the \Theta(g(n))=f(n). It means they should be dual a and b constant variable, therefore a.g(n) could only be valid for the constant variable, that is  \frac{1}{a}\ \  and\ \ \frac{1}{b}.

In Option b:

In this algorithm, the input size value is equal to 1 object, and the value of  A is a polynomial-time complexity, which is similar to its outcome that is O(n^{2}). It is the outside there will be a loop(i) for n iterations, that is also encoded inside it, the for loop(j), which would be a loop(n^{2}). All internal loops operate on a total number of N^{2} generations and therefore the final time complexity is O(n^{2}).

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