Answer:
Question was incomplete and continued the question
For each of the following scenarios, which of these choices would be best? Explain your answer.
BST
Sorted Array
Un-sorted Array
a) The records are guaranteed to arrive already sorted from lowest to highest (i.e., whenever a record is inserted, its key value will always be greater than that of the last record inserted). A total of 1000 inserts will be interspersed with 1000 searches.
b) The records arrive with values having a uniform random distribution (so the BST is likely to be well balanced). 1,000,000 insertions are performed, followed by 10 searches.
Explanation:
Answer for a: Un-sorted array or Un-sorted linked list : as mentioned in the question itself that the records are arriving in the sorted order and search will not be O(log n) and insert will be not be O(n).
Answer for b : Un-sorted array or Un-sorted linkedlist : Number of the items to be inserted is already known which is 1,000,000 but it is very high and at the same time search is low. Unsorted array or Unsorted linked list will be best option here.
Answer:
Distributed memory systems
Distributed memory systems use multiple computers to solve a common problem, with computation distributed among the connected computers (nodes) and using message-passing to communicate between the nodes.
Explanation:
Complete Question:
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?
Answer:
a) 51.2 msec. b) 5.12 msec
Explanation:
If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:
The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:
a) BW = 1 Mbps = 10⁶ bps
⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec
b) BW = 10 Mbps = 10⁷ bps
⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec
The answer would be an A/V Specialist, because Betsy is working with technical equipment on a TV set.
Answer:
Site speed and usability.
Explanation:
Web pages are developed using web development tools like HTML, CSS, JavaScript etc. A collection of these web pages are called a website. The web is a hosted for internet use in a web server.
All requests for a website in a web server is done in a web browser in a client device. For a website to be easy searched for , the search engine optimisation technique is used.
A hosted website on a server must have a unique and usable domain name and the page loading on a browser should be made faster by optimising the size of the scripting language.
