Question

Помогите. Какой объём видеопамяти необходимо для хранения двух страниц изображения, если битовая глубина равна 4, а разрешающая способность дисплея- 800х600 пикселей?(ответ выразите в Кбайтах)

Answer 1

V = 800*600*4 = 1920000 бит = 240000 байт ≈ 234,38 Кб

Answer 2

Calculating Video Memory Required

To calculate the video memory required for storing two image pages with a bit depth of 4 and a display resolution of 800x600 pixels, we can use the following formula:

Video Memory (in kilobytes) = (Resolution Width x Resolution Height x Bit Depth x Number of Pages) / 8 / 1024

Substituting the given values: - Resolution Width = 800 pixels - Resolution Height = 600 pixels - Bit Depth = 4 - Number of Pages = 2

We can calculate the video memory required using the formula.

Calculation

Using the formula: Video Memory (in kilobytes) = (800 x 600 x 4 x 2) / 8 / 1024

Video Memory (in kilobytes) = (960000) / 8 / 1024

Video Memory (in kilobytes) = 120000 / 1024

Video Memory (in kilobytes) ≈ 117 KB

Therefore, approximately 117 kilobytes of video memory would be required to store two image pages with a bit depth of 4 and a display resolution of 800x600 pixels.