Answer:
For including a part of the text within the index, you need to make use of the Mark Entry option. And here you can fill the required information for curating your index, you need to just follow the instruction as required, and you will be fine.
Explanation:
Please check the answer.
Answer:
The reason is due to proprietary design of the Operating System (OS) which require a virtualization software to blanket or "disguise" the hardware (processor) borderlines of the computer onto which it is to be installed.
Explanation:
An Apple system that has the RISC processor and system architecture which has an operating system made entirely for the Apple system architecture
If the above Apple OS is to be installed on a windows computer, then the procedure to setup up the OS has to be the same as that used when on an Apple system, hence, due to the different processors and components of both systems, a virtualization will be be needed to be provided by a Virtual box, Parallels desktop or other virtualization software.
Answer:
Insert mode
Explanation:
In editing a document in text editors such as MS Word, there are two types of mode available.
i. Insert mode: This mode allows users to write texts without erasing anything. As new characters are typed, texts to the right of the insertion point are moved or displaced to the right thus preventing overwriting.
ii. Overtype mode: This mode allows users to write texts by erasing texts to the right of the insertion point.
Note: To enable either of this mode, press the "Insert" key on the keyboard. This key toggles between the <em>Insert</em> and the <em>Overtype</em> modes.
Answer:
1. 2588672 bits
2. 4308992 bits
3. The larger the data size of the cache, the larger the area of memory you will need to "search" making the access time and performance slower than the a cache with a smaller data size.
Explanation:
1. Number of bits in the first cache
Using the formula: (2^index bits) * (valid bits + tag bits + (data bits * 2^offset bits))
total bits = 2^15 (1+14+(32*2^1)) = 2588672 bits
2. Number of bits in the Cache with 16 word blocks
Using the formula: (2^index bits) * (valid bits + tag bits + (data bits * 2^offset bits))
total bits = 2^13(1 +13+(32*2^4)) = 4308992 bits
3. Caches are used to help achieve good performance with slow main memories. However, due to architectural limitations of cache, larger data size of cache are not as effective than the smaller data size. A larger cache will have a lower miss rate and a higher delay. The larger the data size of the cache, the larger the area of memory you will need to "search" making the access time and performance slower than the a cache with a smaller data size.
