Everything 1 4 1 1001 32 bit

× 10-1 Concept:32-bit floating-point representation of a binary number in IEEE- 754 is Sign (1 bit) Exponent (8 bit) Mantissa bit (23 bits) Calculation:Given binary number is00111110011011010000000000000000Here, sign bit is 0. So, number is positive. 0 01111100 11011010000000000000000 Exponent bits = E = 01111100 = 124 (in decimal)Mantissa bits M = 11011010000000000000000In IEEE-754 format, 32-bit (single precision) (-1)s × 1.M × 2E – 127 = (-1)0 × 1.1101101 × 2124 – 127= 1.1101101 × 2-3= (1 + 2-1 + 2-2 + 2-4 + 2-5 + 2-7) × 2-3= 0.231 = 2.31 × 10-1 ≈ 2.27 × 10-1 In IEEE floating point representation, the hexadecimal number 0xC0000000 corresponds to –3.0–1.0–4.0–2.0Answer (Detailed Solution Below) Option 4 : –2.0 Concept:32-bit floating-point representation of a binary number in IEEE- 754 is Sign (1 bit) Exponent (8 bit) Mantissa bit (23 bits) Calculation:Binary number is0xC0000000 = (11000000000000000000000000000000)2Here, the sign bit is 1. So, the number is negative. 1 10000000 00000000000000000000000 Exponent bits = E = 10000000 = 128 (in decimal)Mantissa bits M = 00000000000000000000000In IEEE-754 format, 32-bit (single precision)(-1)s × 1.M × 2E – 127= (-1)1 × 1. 0 × 2128 – 127= -1 × 1.0 × 2= -2In IEEE floating-point representation, the hexadecimal number 0xC0000000 corresponds to -2. Let R1 and R2 be two 4-bit registers that store numbers in 2’s complement form. For the operation R1 + R2, which one of the following values of R1 and R2 gives an arithmetic overflow? R1 = 1011 and R2 = 1110R1 = 1100 and R2 = 1010R1 = 0011 and R2 = 0100R1 = 1001 and R2 = 1111Answer (Detailed Solution Below) Option 2 : R1 = 1100 and R2 = 1010 The correct answer is option 2.Concept:Stored numbers in registers R1 and R2 are in 2's complement form. Register size is 4 bits. The range of numbers in 2's complement form is -8 to +7. If R1 + R2, the result is out of the above range, then it is overflow.The given data,Given two four-bit registers R1 and R2.Option 1: R1 = 1011 and R2 = 1110False, R1 = 1 0 1 1 = -(0101)= -5+ R2 = 1 1 1 0 = -(0010)= -2----------------------------------------------- 1 0 0 1 = = -7 Here No overflow occurred, because sign bit is same for (R1 + R2 ).Option 2: R1 = 1100 and R2 = 1010True,R1 = 1 1 0 0 = -(0100)= -4+ R2 = 1 0 1 0 = -(0110)= -6 -------------------------------------------- 0 1 1 0 = = -10 Here Overflow occurred because the sign bit is different for (R1 + R2 ).Option 3: R1 = 0011 and R2 = 0100False,R1 = 0 0 1 1 = +(0011)= +3+ R2 = 0 1 0

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Trying to follow this guide on including HDR10+ metadata in x265 encodes with StaxRip, but getting a muxing error every time it tries to encode the video. Not sure what I'm doing wrong.Everything works fine if I don't add the HDR Info File. I've tried with multiple videos and get the same result. Any help would be appreciated. Let me know if I can provide any more info that would be helpful------------------------- System Environment -------------------------StaxRip : v2.13.0Windows : Windows 10 Pro 2009Language : English (Canada)CPU : AMD Ryzen 9 5900X 12-Core ProcessorGPU : NVIDIA TITAN X (Pascal)Resolution : 3440 x 1440DPI : 96Code Page : 1252----------------------- Media Info Source File -----------------------D:\Usenet\Complete\Planes.Trains.and.Automobiles.1987\Planes Trains and Automobiles [1997].mkvGeneralComplete name : D:\Usenet\Complete\Planes.Trains.and.Automobiles.1987\Planes Trains and Automobiles [1997].mkvFormat : MatroskaFormat version : Version 4File size : 16.1 GiBDuration : 1 h 32 minOverall bit rate : 24.9 Mb/sWriting application : mkvmerge v67.0.0 ('Under Stars') 64-bitWriting library : libebml v1.4.2 + libmatroska v1.6.4VideoID : 1Format : HEVCFormat/Info : High Efficiency Video CodingFormat profile : Main 10@L5@HighHDR format : SMPTE ST 2094 App 4, Version 1, HDR10+ Profile B compatibleCodec ID : V_MPEGH/ISO/HEVCDuration : 1 h 32 minBit rate : 24.5 Mb/sWidth : 3 840 pixelsHeight : 2 076 pixelsDisplay aspect ratio : 1.85:1Frame rate mode : ConstantFrame rate : 23.976 (24000/1001) FPSColor space : YUVChroma subsampling : 4:2:0 (Type 2)Bit depth : 10 bitsBits/(Pixel*Frame) : 0.128Stream size : 15.9 GiB (98%)Default : YesForced : NoColor range : LimitedColor primaries : BT.2020Transfer characteristics : PQMatrix coefficients : BT.2020 non-constantMastering display color primaries : Display P3Mastering display luminance : min: 0.0050 cd/m2, max: 4000 cd/m2Maximum Content Light Level : 1080 cd/m2Maximum Frame-Average Light Level : 337 cd/m2AudioID : 2Format : AC-3Format/Info : Audio Coding 3Commercial name : Dolby DigitalCodec ID : A_AC3Duration : 1 h 32 minBit rate mode

0 = +(0100)= +4 -------------------------------------------- 0 1 1 1 = +7 Here No overflow occurred, because the sign bit is the same for (R1 + R2 ).Option 4: R1 = 1001 and R2 = 1111False, R1 = 1 0 0 1 = -(0111) = -7+ R2 = 1 1 1 1 = -(0001) = -1 -------------------------------------------- 1 0 0 0 = = -8Here No overflow occurred, because the sign bit is the same for (R1 + R2 ).Hence the correct answer is R1 = 1100 and R2 = 1010. Consider three floating-point numbers A, B and C stored in registers RA, RB and RC, respectively as per IEEE-754 single-precision floating point format. The 32-bit content stored in these registers (in hexadecimal form) are as follows. RA= 0xC1400000 RB = 0x42100000 RC = 0x41400000 Which one of the following is FALSE? A + C = 0C = A + BB = 3C(B - C) > 0Answer (Detailed Solution Below) Option 2 : C = A + B The correct answer is option 2.Concept:IEEE single-precision floating-point:IEEE single-precision floating-point computer numbering format is a binary computing format that takes up 4 bytes (32 bits) of memory. Binary32 is the official name for the 32-bit base 2 formats in IEEE 754-2008. IEEE 754-1985 referred to it as single.IEEE single-precision format:Explanation:The given data,Decimal value =(-1)s x 1.M x 2Base Exponent -BiasBias value in IEEE single-precision format is 127RA = 1100 0001 0100 0000 0000 0000 0000 0000RA sign= 1RA Base Exponent =100 0001 0 = 130RA Mantisa = 100 0000 0000 0000 0000 0000 = 1.100 0000 0000.....Decimal value = (-1)1 x1.1 x2130-127 =-1.1x23= -1100 = (-12)10A=-12RB = 0100 0010 0001 0000 0000 0000 0000 0000RA sign= 0RA Base Exponent =100 0010 0= 132RA Mantisa = 001 0000 0000 0000 0000 0000 = 1.001 000000.....Decimal value = (-1)0 x1.001 x2132-127 =+1.001x25= + 100100 = (+36)10B=+36RC = 0100 0001 0100 0000 0000 0000 0000 0000RA sign= 0RA Base Exponent =100 0001 0= 130RA Mantisa =100 0000 0000 0000 0000 0000= 1.100 0000.....Decimal value = (-1)0 x1.1 x2130-127 =+1.1x23= + 1100 = (+12)10C=+12Option 1: A + C = 0True, A+C= -12+12=0Hence it is true.Option 2: C = A + BFalse, A+B= -12+36=+24it not equal to C. Hence it is false.Option 3: B = 3CTrue, B=3C =3x+12 =36 =Bit equal to B. Hence it is true.Option 4: (B - C) > 0True, (B-C) >0=(36-12)=24>0Hence it is true.Hence the correct answer is C = A + B. Consider three registers R1, R2 and R3 that store numbers in IEEE-754 single precision floating point format. Assume that R1 and R2 contain the values (in hexadecimal notation) 0x42200000 and 0xC1200000, respectively.If R3 \(= \frac{{R1}}{{R2}},\) what is the value stored in R3?