Answer: (B) Lossless compression
Explanation:
According to the question, the given technique is the example of the lossless compression. The lossless compression is one of the technique that decompress the data into the original data form without any type of losses.
The lossless compression is the technique that usually compress the data, text and the databases. This technique also improve the compression rate helps in reconstruct the original data. It is used in various types of encoding methods, GNU tool and zip file.
Answer:
B. False
Explanation:
Sarbanes-Oxley Act or SOX also known as the Public Company Accounting Reform and Investor Protection Act and Corporate and Auditing Accountability, Responsibility, and Transparency Act is a United State federal law that creates or modify requirements for U.S public company board, management and public accounting firm. some of its policies are meant for private companies as well.
This act does not restrict any electronic and paper data containing personally identifiable financial information.
Answer:
1.2124
2.1212 is the correct answer right
Answer:
The circular individually linked list is more efficient for time sharing process ,when multiple application are running on pc it is responsibility of an output system to put all process on a list and execute them all by giving them piece of time and make them wait when cpu is selected to other process.
It will be more suitable for output system to use circular list as when it reaches to last of list it will be manually reaches to starting node or process.
Singly circular linked list is used when we are concerned with the memory as only one process will be allocated memory at once and there are no chances of process to go never-ending waiting.
Explanation:
Answer:
a. Utilization = 0.00039
b. Throughput = 50Kbps
Explanation:
<u>Given Data:</u>
Packet Size = L = 1kb = 8000 bits
Transmission Rate = R = 1 Gbps = 1 x 10⁹ bps
RTT = 20 msec
<u>To Find </u>
a. Sender Utilization = ?
b. Throughput = ?
Solution
a. Sender Utilization
<u>As Given </u>
Packet Size = L = 8000 bits
Transmission Rate = R = 1 Gbps = 1 x 10⁹ bps
Transmission Time = L/R = 8000 bits / 1 x 10⁹ bps = 8 micro-sec
Utilization = Transmission Time / RTT + Transmission Time
= 8 micro-sec/ 20 msec + 8 micro-sec
= 0.008 sec/ 20.008 sec
Utilization = 0.00039
b. Throughput
<u>As Given </u>
Packet Size = 1kb
RTT = 20ms = 20/100 sec = 0.02 sec
So,
Throughput = Packet Size/RTT = 1kb /0.02 = 50 kbps
So, the system has 50 kbps throughput over 1 Gbps Link.
