Answer:
Street prices are affected by the extent of illegal commercial copying. The availability of inexpensive, high-quality illegal copies reduces the demand for legal copies to the extent that some users buy illegal copies instead of legal ones.
There are Mainly Five basic Components of a computer system.
Input Unit.
Output unit.
Memory Unit.
Control unit.
Arithmetic and Logic Unit.
Answer:
They had to unzip or extract the zipped/ compressed presentation to see it.
Explanation:
- Compressed file or a zip file contains a large file or more than one files that are packed or enclosed that large file or several files into a file or a folder.
- The compressed file or folder takes less space than the actual large file or set of files.
- Just as in this scenario the presentation file was larger than the size of what an email file should be. So Tikenya had to compress the file in order to email it to other members.
- The other members can open the PowerPoint presentation by following any of the below mentioned methods:
- In Windows to open a compressed file, right click on that file and select Extract All option.
- File can also be seen by opening that file or folder and dragging that file or folder from the zipped folder to a new location. In Windows this will automatically extract the files.
- This file can also be opened using a software like WinZip. Download WinZip. Then download the file which is sent in email and double click on that file which will open WinZip wizard dialogue box. Clicking that presentation file from the dialogue box will open another dialogue box confirming the opening of that file. By clicking on OK the presentation file will be opened and can be viewed now.
Answer:
The page field is 8-bit wide, then the page size is 256 bytes.
Using the subdivision above, the first level page table points to 1024 2nd level page tables, each pointing to 256 3rd page tables, each containing 64 pages. The program's address space consists of 1024 pages, thus we need we need 16 third-level page tables. Therefore we need 16 entries in a 2nd level page table, and one entry in the first level page table. Therefore the size is: 1024 entries for the first table, 256 entries for the 2nd level page table, and 16 3rd level page table containing 64 entries each. Assuming 2 bytes per entry, the space required is 1024 * 2 + 256 * 2 (one second-level paget table) + 16 * 64 * 2 (16 third-level page tables) = 4608 bytes.
First, the stack, data and code segments are at addresses that require having 3 page tables entries active in the first level page table. For 64K, you need 256 pages, or 4 third-level page tables. For 600K, you need 2400 pages, or 38 third-level page tables and for 48K you need 192 pages or 3 third-level page tables. Assuming 2 bytes per entry, the space required is 1024 * 2 + 256 * 3 * 2 (3 second-level page tables) + 64 * (38+4+3)* 2 (38 third-level page tables for data segment, 4 for stack and 3 for code segment) = 9344 bytes.
Explanation:
16 E the answer