Answer:
<u>B. Tip the bowl slightly, then spoon up the last bit</u>
Explanation:
Generally speaking none of the other answers made much of sense.. "tip the bowl from side to side" doesn't help you at all, neither does "Keep spooning as much as you can, then stop eating". Neither of these will help you spoon up the last little bit, <u>the most logical answer is B. "Tip the bowl slightly, then spoon up the last bit"</u>. If this is not the correct answer then it'd be D, but I don't believe/feel that it's "poor etiquette" to "leave the last little bit".
Answer:
a) AL will contains 0011 1100
Explanation:
In assembly language, shifting bits in registers is a common and important practice. One of the shifting operations is the SHR AL, x where the x specifies that the bits be shifted to the right by x places.
SHR AL, 2 therefore means that the bits contained in the AL should be shifted to the right by two (2) places.
For example, if the AL contains binary 1000 1111, the SHR AL, 2 operation will cause the following to happen
Original bit => | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |
Shift once to the right => | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | (0) |
Shift once to the right => | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | (0) | (0) |
Notice;
(i) that there are two shifts - one at a time.
(ii) that the bits in bold face are the bits in the AL after the shift. Those that in regular face are those in the carry flag.
(iii) that the new bits added to the AL after a shift are the ones in bracket. They are always set to 0.
Answer:
The 2 main parts to a VR experience is getting comfortable and interacting with people in the VR if you have friends :3
Explanation:
:3
Bot. A program that performs a repetitive task on a network. Cybercriminals install malicous bots on unprotected computers to create a botnet. Zombie army. (also calledBotnet) groups of bots
Answer:
See the attached pictures for detailed answer.
Explanation:
The cache contain 8K number of blocks.each block has 16 Byte worth of data. Each of these bytes are addressable by block offset. Now, each of these blocks are addressable and their address is provided by line offset or set offset. If block number is given in Decimal format then the block to which that block will be mapped is given by expression
mappedToBlock = (bNumber)mod213
TAG and valid but consititues something called cache controller. TAG holds the physical address information because the TAG in physical address get divided into TAG and line offset in cache address. Valid bit tells status of blocks. If it is 1 then data blocks are referred by CPI else not.
