Question

Скорость передачи данных через аналоговый модем равна 128000бит/с. Через данное соединение передают файл размером 375 Килобайт. Определите время передачи файла в секундах.

Answer 1

375 Киболайт = 375 * 1024 * 8 = 3072000 бит

3072000 : 128000 = 24 с

Ответ. Понадобится 24 с.

Answer 2

Понадобится 24 секунды.

Answer 3

Calculation of Transmission Time

To determine the time it takes to transmit a file over an analog modem connection, we need to consider the transmission speed and the file size.

Given: - Transmission speed: 128,000 bits per second (128,000 bps) - File size: 375 kilobytes (375 KB)

To calculate the transmission time, we can use the following formula:

Transmission Time = File Size / Transmission Speed

Let's plug in the values:

File Size = 375 KB = 375,000 bytes = 3,000,000 bits Transmission Speed = 128,000 bps

Now we can calculate the transmission time:

Transmission Time = 3,000,000 bits / 128,000 bps

Using this formula, we can determine the time it takes to transmit the file in seconds.

Calculation:

Transmission Time = 3,000,000 bits / 128,000 bps

The transmission time is approximately 23.4375 seconds.

Please note that this calculation assumes ideal conditions and does not take into account factors such as overhead, latency, or error correction. The actual transmission time may vary in real-world scenarios.