Answer:
dsdasd1.Discuss the differences and similarities between a peer-to peer network and a client server network.
1.Discuss the differences and similarities between a peer-to peer network and a client server network.
1.Discuss the differences and similarities between a peer-to peer network and a client server network.
1.Discuss the differences and similarities between a peer-to peer network and a client server network.
1.Discuss the differences and similarities between a peer-to peer network and a client server network.
1.Discuss the differences and similarities between a peer-to peer network and a client server network.
1.Discuss the differences and similarities between a peer-to peer network and a client server network.
Explanation:
jaiusfkybuetcrjnwxyefshrcxijkwuecvashcnzx
Answer:
C)
Explanation:
One principle that can improve the efficiency of I/O would be to move processing primitives into hardware. Primitives are a semantic value representing something else such as words or numbers within the programming language. By moving them into hardware they system is able to read them at a much faster speed making the I/O more efficient.
Answer:
The risk is a buffer overflow.
Explanation:
Whatever the user passes as a command line argument, will be copied into the buffer. If the user passes more than 499 characters, the end of the buffer will be overwritten.
To solve it, compare the string length of argv[1] to 500 before copying, or even better, start using the new strcpy_s( ) function.
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
Answer:
It allows companies to make definitive predictions about the future. It gives companies the ability to make informed decisions.
It helps Data Analysts shape an analytics problem from a business problem.
Explanation:
