Question

Ста­тья, на­бран­ная на ком­пью­те­ре, со­дер­жит 32 стра­ни­цы, на каж­дой стра­ни­це 32 стро­ки, в каж­дой стро­ке 25 сим­во­лов. Опре­де­ли­те ин­фор­ма­ци­он­ный объём ста­тьи в ко­ди­ров­ке Windows-1251, в ко­то­рой каж­дый сим­вол ко­ди­ру­ет­ся 8 би­та­ми.

Answer 1

32 * 32 * 25 * 8 = 204800 бит или 25600 байт

Answer 2

Informational Content of the Article in Windows-1251 Encoding

To determine the informational content of the article in the Windows-1251 encoding, we can calculate the total number of bytes required to store the entire article. Given that each symbol is encoded using 8 bits in Windows-1251, we can use the provided information to calculate the informational content.

The article contains 32 pages, with 32 lines per page and 25 symbols per line.

Calculation Steps:

1. Calculate the total number of symbols in the article: - Total symbols = (32 pages) * (32 lines per page) * (25 symbols per line) = 32 * 32 * 25 = 25,600 symbols.

2. Calculate the total number of bytes required to store the article in Windows-1251 encoding: - Total bytes = (Total symbols) * (8 bits per symbol) = 25,600 * 8 = 204,800 bits = 25,600 bytes.

3. Convert the total bytes to kilobytes (KB): - Total KB = (Total bytes) / 1024 = 25,600 / 1024 = 25 KB.

Therefore, the informational content of the article in Windows-1251 encoding is 25 kilobytes.