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× 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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Everything 32 Everything 32 1.

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. ‎Ieee 754 Floating Point Number Representation Question 4: 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. Everything 1. (32-bit) Date released: (4 years ago) Download. Everything 1. (32-bit) Date released: (4 years ago) Download. Everything 1. (32-bit) Date released: (4 years ago) Download. Everything 1. (32-bit) Everything 1. (32-bit) Date released: (4 years ago) Download. Everything 1. (32-bit) Date released: (4 years ago) Download. Everything 1. (32-bit) Date released: (4

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Everything 1. Beta (32-bit) - FileHorse

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

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Titolo Originale: The Matrix Nazionalità: USA Anno: 1999 Genere: Fantascienza Regia: Lana Wachowski, Andy Wachowski Durata: 131 Min Cast: Keanu Reeves, Laurence Fishburne, Carrie-Anne Moss, Hugo Weaving Thomas A. Anderson is a man living two lives. By day he is an average computer programmer and by night a hacker known as Neo. Neo has always questioned his reality, but the truth is far beyond his imagination. Un hacker di nome Neo, grazie all'aiuto del misterioso Morpheus, scopre che quella per lui è la "realtà" non è altro che un facciata, un mondo virtuale creato dal super computer Matrix per controllare gli esseri umani. Ma Morpheus è convinto che Neo sia il prescelto di cui parla un'antica profezia, e che sarà in grado di guidare la rivoluzione contro Matrix.... Generale ID univoco : 133047713099731545923563987600861290555 (0x64180EB64B876CB395FA19624A3BE03 Nome completo : D:The.Matrix.1999.4K.HDR.DV.2160p.BDRemux Ita Eng x265-NAHOMThe.Matrix.1999.4K.HDR.DV.2160p.BDRemux Ita Eng x265-NAHOM.mkv Formato : Matroska Versione formato : Version 4 Dimensione : 48,0 GiB Durata : 2 o 16 min Bitrate totale : 50,4 Mb/s Film : The Matrix 4K Data codifica : UTC 2022-02-10 23:20:36 Creato con : mkvmerge v61.0.0 ('So') 32-bit Compressore : libebml v1.4.2 + libmatroska v1.6.4 Video #1 ID : 1 Formato : HEVC Formato/Informazioni : High Efficiency Video Coding Profilo formato : Main [email protected]@High HDR format : SMPTE ST 2086, HDR10 compatible ID codec : V_MPEGH/ISO/HEVC Durata : 2 o 16 min Bitrate : 48,8 Mb/s Larghezza : 3.840 pixel Altezza : 2.160 pixel Rapporto aspetto visualizzazione : 16:9 Modalità frame rate : Costante Frame rate : 23,976 (24000/1001) FPS Spazio colore : YUV Croma subsampling : 4:2:0 (Type 2) Profondità bit : 10 bit Bit/(pixel*frame) : 0.246 Dimensione della traccia : 46,5 GiB (97%) Default : Si Forced : No Color range : Limited Colori primari : BT.2020 Caratteristiche trasferimento : PQ Coefficienti matrici : BT.2020 non-constant Mastering display color primaries : BT.2020 Mastering display luminance : min: 0.0020 cd/m2, max: 1000 cd/m2 Maximum Content Light Level : 992 cd/m2 Maximum Frame-Average Light Level : 518 cd/m2 Video #2 ID : 2 Formato : HEVC Formato/Informazioni : High Efficiency Video Coding Profilo formato : Main [email protected]@High HDR format : SMPTE ST 2086, HDR10 compatible ID codec : V_MPEGH/ISO/HEVC Durata : 2 o 16 min Bitrate : 66,2 kb/s Larghezza : 1.920 pixel Altezza : 1.080 pixel Rapporto aspetto visualizzazione : 16:9 Modalità frame rate : Costante Frame rate : 23,976 (24000/1001) FPS Spazio colore : YUV Croma subsampling : 4:2:0 (Type 2) Profondità bit : 10 bit Bit/(pixel*frame) : 0.001 Dimensione della traccia : 64,5MiB (0%) Default : Si Forced : No Color range : Limited Colori primari : BT.2020 Caratteristiche trasferimento : PQ Coefficienti matrici : BT.2020 non-constant Mastering

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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?. Everything 1. (32-bit) Date released: (4 years ago) Download. Everything 1. (32-bit) Date released: (4 years ago) Download. Everything 1. (32-bit) Date released: (4 years ago) Download. Everything 1. (32-bit)

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Encode Screenshots (Click for full quality) Media InformationGeneral Format : Matroska Format version : Version 4 / Version 2 File size : 1.33 GiB Duration : 55 min 19 s Overall bit rate : 3 433 kb/s Movie name : Homeland, Season 1 Disc 1 Encoded date : UTC 2018-07-30 14:14:46 Writing application : mkvmerge v20.0.0 ('I Am The Sun') 64-bit Writing library : libebml v1.3.5 + libmatroska v1.4.8 Writing frontend : StaxRip v1.7.0.6Video ID : 1 Format : HEVC Format/Info : High Efficiency Video Coding Format profile : Main [email protected]@Main Codec ID : V_MPEGH/ISO/HEVC Duration : 55 min 19 s Bit rate : 3 042 kb/s Width : 1 920 pixels Height : 1 080 pixels Display aspect ratio : 16:9 Frame rate mode : Constant Frame rate : 23.976 (24000/1001) FPS Color space : YUV Chroma subsampling : 4:2:0 Bit depth : 10 bits Bits/(Pixel*Frame) : 0.061 Stream size : 1.18 GiB (89%) Writing library : x265 2.8+10-703eade86b53:[Windows][MSVC 1900][64 bit] 10bit Encoding settings : cpuid=1111039 / frame-threads=4 / numa-pools=16 / wpp / no-pmode / no-pme / no-psnr / no-ssim / log-level=2 / input-csp=1 / input-res=1920x1080 / interlace=0 / total-frames=79595 / level-idc=0 / high-tier=1 / uhd-bd=0 / ref=5 / no-allow-non-conformance / no-repeat-headers / annexb / no-aud / no-hrd / info / hash=0 / no-temporal-layers / open-gop / min-keyint=23 / keyint=250 / gop-lookahead=0 / bframes=8 / b-adapt=2 / b-pyramid / bframe-bias=0 / rc-lookahead=80 / lookahead-slices=4 / scenecut=40 / radl=0 / no-intra-refresh / ctu=64 / min-cu-size=8 / no-rect / no-amp / max-tu-size=32 / tu-inter-depth=1 / tu-intra-depth=1 / limit-tu=0 / rdoq-level=2 / dynamic-rd=0.00 / no-ssim-rd / signhide / no-tskip / nr-intra=0 / nr-inter=0 / no-constrained-intra / no-strong-intra-smoothing / max-merge=3 / limit-refs=3 / limit-modes / me=3 / subme=5 / merange=57 / temporal-mvp / weightp / weightb / no-analyze-src-pics / deblock=-3:-3 / no-sao

Download Everything Portable 1. Beta (32-bit

La función BIT.Y devuelve el valor Y booleano bit a bit de dos números. Más información Esta es la tabla de verdad de BIT.Y: A B BIT.Y(A;B) 0 0 0 0 1 0 1 0 0 1 1 1 Partes de la función BIT.YBIT.Y(valor1;valor2) Parte Descripción Notas valor1 Primer número. Debe ser la representación decimal del número. valor2 Segundo número. Debe ser la representación decimal del número. NotasSe puede usar BIT.Y junto con BIN.A.DEC de la siguiente manera: BIT.Y(BIN.A.DEC("1010"); BIN.A.DEC("1001")), que da como resultado "1000" en el sistema binario u 8 en el decimal.EjemplosEl número 10 en decimal es "1010" en binario; y el 9 en decimal es "1001" en binario. El resultado es "1000" en binario, que es 8 en decimal. A B 1 Fórmula Resultado 2 =BIT.Y(10;9) 8 El valor Y booleano bit a bit de los números binarios "1110" y "0100" es el número binario "0100", que es 4 en decimal. A B 1 Fórmula Resultado 2 =BIT.Y(BIN.A.DEC("1110"); BIN.A.DEC("0100")) 4 Funciones relacionadas BIT.O: devuelve el valor O booleano bit a bit de dos números. BIT.XO: devuelve el valor XO (O exclusivo) booleano bit a bit de dos números. BIT.DESPLIZQDA: La función BIT.DESPLIZQDA desplaza los bits del valor introducido cierto número de posiciones hacia la izquierda. Los bits de la derecha se rellenan con ceros (0). BIT.DESPLDCHA: desplaza los bits del valor introducido cierto número de posiciones hacia la derecha. Los bits de la derecha se rellenan con ceros (0). BIN.A.DEC: Convierte un número binario con signo a formato decimal. DEC.A.BIN: Convierte un número decimal a formato binario con signo.. Everything 1. (32-bit) Date released: (4 years ago) Download. Everything 1. (32-bit) Date released: (4 years ago) Download. Everything 1. (32-bit) Date released: (4 years ago) Download. Everything 1. (32-bit)

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I invalid, N Not found Network Next Hop Metric LocPrf Weight Path *>i Dest:3::/0-24,Source:4::/0-24 FEC0::1001 100 0 i Step 5 show bgp ipv6 flowspec detail This command displays the detailed information about IPv6 flowspec routes. Example:Device# show bgp ipv6 flowspec detailBGP routing table entry for Dest:3::/0-24,Source:4::/0-24, version 2 Paths: (1 available, best #1, table Global-Flowspecv6-Table) Advertised to update-groups: 2 Refresh Epoch 1 Local FEC0::1001 from FEC0::1001 (10.0.101.2) Origin IGP, localpref 100, valid, internal, best rx pathid: 0, tx pathid: 0x0 Step 6 show bgp ipv6 flowspec summary This command displays the IPv6 flowspec neighbors. Example:Device# show bgp ipv6 flowspec summary BGP router identifier 10.10.10.2, local AS number 239 BGP table version is 3, main routing table version 32 network entries using 16608 bytes of memory2 path entries using 152 bytes of memory2/2 BGP path/bestpath attribute entries using 304 bytes of memory1 BGP AS-PATH entries using 24 bytes of memory2 BGP extended community entries using 48 bytes of memory0 BGP route-map cache entries using 0 bytes of memory0 BGP filter-list cache entries using 0 bytes of memory BGP using 17136 total bytes of memory BGP activity 18/0prefixes, 18/0 paths, scan interval 15 secsNeighbor V AS MsgRcvd MsgSent TblVer InQ OutQ Up/DownState/PfxRcd10.0.101.1 4 239 70 24 3 0 0 00:10:58 210.0.101.2 4 239 0 0 1 0 0 neverIdle10.0.101.3 4 240 0 0 1 0 0 neverIdle10.10.10.1 4 239 19 23 3 0 0 00:10:53 Step 7 show bgp vpnv4 flowspec This command displays the VPNv4 flowspec neighbors. Example:Device# show bgp vpnv4 flowspec BGP table version is 2, local router ID is 10.10.10.2 Status codes: s suppressed, d damped, h history, * valid, > best, i - internal,r RIB-failure, S Stale, m multipath, b backup-path, f RT-Filter, x best-external, a additional-path, c RIB-compressed, Origin codes: i - IGP, e - EGP, ?