Well, honey, to represent months of the year, you need at least 4 bits because you've got 12 months in a year, and you need 4 bits to represent numbers from 0 to 15. So, technically, you could do it with just 4 bits, but if you want to be fancy, you could use 5 bits for a more efficient representation.
A bitmap is a series of bits which represents a rasterized graphic image, each pixel being represented as a group of bits.
200 = 1 1 0 0 1 0 0 0
signed magnitude
With 5 bits, you can represent (2^5) different numbers, which equals 32. This includes numbers ranging from 0 to 31 in unsigned binary representation. If using signed binary representation (like two's complement), the range would be from -16 to 15, still allowing for 32 distinct values.
This is an ambiguous question that needs more input. If you mean a 10-bit YUV format, then the lowest 6 bits (the rightmost bits on paper) are unused. Least significant means that they have the lowest value. The far right bit when a binary number is written on paper is only significant by one. The far left bit is the most significant and represents half of the maximum value. For instance, 8 bits can contain a value from 0 to 255, meaning there are only 256 possibilities. So the far left bit is significant by 128, the next one by 64, the next by 32, etc. The same works with standard decimal numbers. The far right digit is the one's place. The next digit to the left is the ten's place. The next is the hundred's place.
Well, honey, to represent months of the year, you need at least 4 bits because you've got 12 months in a year, and you need 4 bits to represent numbers from 0 to 15. So, technically, you could do it with just 4 bits, but if you want to be fancy, you could use 5 bits for a more efficient representation.
That's called a digital image.
41 in decimal is 0100 0001 in BCD (this is 8 bits not 6 bits)41 in decimal is 101001 in binary (this is 6 bits, but binary not BCD)There is no 6 bit BCD representation of the decimal number 41!
bits
A bitmap is a series of bits which represents a rasterized graphic image, each pixel being represented as a group of bits.
precision is the total number of bits or digits in the representation of a number.accuracy is the number of correct bits or digits in a number.Given a certain representation on a computer, all numbers stored in that representation will have the same precision; however the accuracy of different numbers will vary, depending on the source and on the calculations done on them.
255.255.255.0 - in decimal representation 11111111.11111111.11111111.00000000 - binary representation (3 bytes with all the bits equal to 1, the last byte with all the bits equal to 0) /24 - binary short representation (the number of bits equal to 1)
In a bitmap, information is stored as a grid of pixels, where each pixel represents a specific color. The bitmap uses a matrix of bits, with each pixel typically requiring multiple bits to define its color, depending on the color depth (e.g., 1-bit for black and white, 8-bit for 256 colors, or 24-bit for true color). The arrangement of these bits in the grid corresponds to the image's dimensions, allowing for the representation of detailed graphics. Additionally, bitmap files may include metadata, such as image size and color profiles, stored in a header.
1024bitsIf "Bitmap" refers to a specific entity, image or file: I do not know. But if "Bitmap" refers to a general image then it is 8 bytes or 64 bits per pixel. I just made 3 1*1 bitmap images at colordepths 2bits (monochrome), 8 bits (256 colors) and 24 bits (16 Million colors). The sizes of these images were the same! (surprized me too!) Then I made a 1*2 pixel image and it was 66 bytes (528 bits) so the "overhead Microsoft paint puts on a bitmap is 400 bits. This could be an effect of limitations inherent in Microsoft Paint.
30kb = ? pixel?
To calculate the time it would take to transmit the image, we first need to determine the total number of bits in the image. With 256 gray levels, we need 8 bits (1 byte) to represent each pixel. So, for a 512x512 image, we have 512 x 512 x 8 = 2,097,152 bits. At a baud rate of 300 baud, this means 300 bits are transmitted per second. Therefore, it would take 2,097,152 bits / 300 bits per second = 6,990.51 seconds to transmit the image. Converting this to minutes, it would take approximately 116.51 minutes.