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Q- A desktop laser printer has a 300-pixel-per-inch resolution. If each pixel is stored in one bit of memory how many bytes of memory are required to store the complete image of one 8.5 X 11 inch of paper? A. About 8 KB B. About .5 MB C. About 1 MB D. About 4 MB E. About 8 MB SOLUTION: 8.5 x 11inch paper = (8.5 x 300) x (11 x 300) pixels on paper stated in the question. ... 300pixel-per-inch resolution is =2550 x 3300 pixels Total pixels on the sheet = 2550 x 3300 = 8415000 Each pixel is stored in one bit of memory, so 8415000 pixels are stored in 8415000 bits of memory 8bits make 1 byte So 8415000 bits = 8415000 / 8 byte = 1051875 bytes 1024 bytes make 1 kilobyte(1KB) Therefore, 1051875 bytes = 1051875/1024 = 1027.222 KB 1024 KB make up 1 MegaByte (MB) Therefore, 1027.22 / 1024 = 1.003MB OUR ANSWER IS OPTION C - ABOUT 1 MB. Q-Assuming nine-bit two's complement binary representation, convert 1B4 from hexadecimal to decimal. Remember to check the sign bit: SOLUTION: In 9-bit 2's-complement system you use 9 x 1's and 0's to represent the numbers from -256 to +255, with 0 in the middle. The left-most bit (the highest bit) is reserved to represent the sign of the number 0 means positive and 1 means negative. Which leaves you with only 7-bits left to represent the actual value of the number. Direct conversion of Hexadecimal 1B4 to binary4-bit Binary of 1 - 0001 4-bit Binary of B - 1011 4-bit Binary of 4 - 0100 I.e. 0001 1011 0100 - truncating the first three 0 bits, we get 110110100 .. This is our 9-bit 2’s complement form. First bit is 1, which indicates negative. Computing rest of the 8-bits from binary to decimal: 10110100 = [1 x (2^7)] + [1 x (2^5)] + [1 x (2^4)] + [1 x (2^2)] = 128 + 32 + 16 + 4 = 180 First bit indicated negative, so our value in decimal is: -180 Q- For IEEE 754 single precision floating point, what is the number, as written in , whose hexadecimal representation is: 0061 0000 Let us write our hexadecimal in the binary notation. 0061 0000 = 0000 0000 0110 0001 0000 0000 0000 0000 Sign bit : 0 Exponent: 0000 0000 Significand: 110 0001 0000 0000 0000 0000 to the base 2 = 610000 to base 16 We then add the implicit 24th bit to the significand 1110 0001 0000 0000 0000 0000 = E10000 base16 and decode the exponent value by subtracting 127 0 base 16 = 0 0 - 127 = -127 ... I AM SOLVING AFTER THIS. Example LINK: http://en.wikipedia.org/wiki/Single-precision_floating-point_format Q- For IEEE 754 single precision floating point, write the hexadecimal representation for -1.0. SOLUTION: We have -1.0 (1.0) base 10 = 1 base 10 + 0.0 base 10 =(0001) base 2 + (000) base 2 =(0001.000) base 2 =1.0 x 2^0 Exponent is 0 (and in the biased form it is therefore 127) Biased form > 0000 0000 (For 0) Reverse : 1111 1111 <- this is the biased form The fraction is 0 (looking to the right of the binary point in 1.0 is all 0 = 000...0) From these we can form the resulting 32 bit IEEE 754 binary32 format representation of real number 0 as: 1-01111111-00000000000000000000000 = bf800000 [TAKING bits 4 by 4] Our answer: bf800000