Answer:
In curriculum, a vision of knowledge, the role of the educator and a concept of the process of education are all present. In this sense, Stenhouse suggests that the role of teachers and professors is fundamental in the elaboration and implementation of curriculum.
Answer:
C. A cold site is a leased facility that contains only electrical and communications wiring, air conditioning, plumbing, and raised flooring. No communications equipment, networking hardware, or computers are installed at a cold site until it is necessary to bring the site to full operation.
Explanation:
Answer:
Thank you
Explanation:
That made my day. I will make sure to pass it on.
Answer:
Spyware
Explanation:
Internet is a type of computer network that allow device communicate with each other world wide. So, it is not the correct option.
Worm: this is a standalone malware computer program that replicates itself in order to spread to other computers. It spreads copies of itself from computer to computer. A worm can replicate itself without any human interaction, and it does not need to attach itself to a software program in order to cause damage. The major feature of a worm is ability to replicate on it own. So, it is not the correct option.
Bot: this is a software application that runs automated tasks. So, it is not the correct option.
Middleware: this is software that lies between an operating system and the applications running on it, enabling communication and data management. It provides services to software applications beyond those available from the operating system. So, it is not the correct option.
Spyware: this is the correct answer. Spyware is similar to trojan horse because it hides itself in a system and a user may not know that it exist on the system. Spyware is a form of malware that hides on your device, monitors your activity, and steals sensitive information without knowledge of the user.
For a direct mapped cache the general rule is: first figure out the bits of the offset (the right-most bits of the address), then figure out the bits of the index (the next-to right-most address bits), and then the tag is everything left over (on the left side).
One way to think of a direct mapped cache is as a table with rows and columns. The index tells you what row to look at, then you compare the tag for that row, and if it matches, the offsettells you which column to use. (Note that the order you use the parts: index/tag/offset, is different than the order in which you figure out which bits are which: offset/index/tag.)
So in part (a) The block size is 1 word, so you need 0 offset bits (because <span><span><span>20</span>=1</span><span><span>20</span>=1</span></span>). You have 16 blocks, so you need 4 index bits to give 16 different indices (because <span><span><span>24</span>=16</span><span><span>24</span>=16</span></span>). That leaves you with the remaining 28 bits for the tag. You seem to have gotten this mostly right (except for the rows for "180" and "43" where you seem to have missed a few bits, and the row for "181" where you interchanged some bits when converting to binary, I think). You are correct that everything is a miss.
For part (b) The block size is 2 words, so you need 1 offset bit (because <span><span><span>21</span>=2</span><span><span>21</span>=2</span></span>). You have 8 blocks, so you need 3 index bits to give 8 different row indices (because <span><span><span>23</span>=8</span><span><span>23</span>=8</span></span>). That leaves you with the remaining 28 bits for the tag. Again you got it mostly right except for the rows for "180" and "43" and "181". (Which then will change some of the hits and misses.)
