Q2 Perform the following conversions: a) Converting (756.603)8 to hex b) Converting A0F9.0EB to decimel c) (1001010)2 to gray d) (163.875)10 to binary e)0110 BCD to excess -3
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Q2 Perform the following conversions:
a) Converting (756.603)8 to hex
b) Converting A0F9.0EB to decimel
c) (1001010)2 to gray
d) (163.875)10 to binary
e)0110 BCD to excess -3
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- Convert the decimal number 27.315 to binary. and Calculate the binary equivalent of 2/3 to eight places after the binary point. Then convert the binary equivalent back to decimal and see if you get the same result.BCD isA. Bit Coded digitB. Binary Coded DecimalC. Bit Code digitD. Binary Coded DigitConvert the following Convert hexadecimal value 22 to decimal Convert binary 111111110010 to hexadecimal Convert the following binary number to decimal 011112 Convert the binary number 1011.00102 to decimal. Convert 69.3210 to BCD.
- . Find the answers of the following questions?a) Find the excess-3 code of (175)10b) Convert binary to grey code 1101010.For the circuit shown below, Binary Number A = 1010 and Binary number B = 1110. What is the value at C 0D 0E 0F 1X 0please correct me if i am wrong3- How many 74LS00 NAND gate inputs can be driven by a 74LS00 NAND gate outputs Refer to data sheet of 74LS00, the maximum values of IOH = 0.4 mA, IOL = 8 mA, IH = 20 µA, and I = 0.4 mA.
- What type of electronic device would be used to convert binary numbers to decimal numbers? In this experiment a IDGH voltage (near + 5 V) stood for ( 0 , 1) The 7442 IC has active _____ (HIGH, LOW) inputs. The "bubbles" at the outputs of the 7442 decoder mean this IC has active _____ (HIGH, LOW) outputs. If the binary input to the 7442 decoder is 0110, then output _____ (decimal number) is activated. If the binary input to the 7442 decoder is 0111, then output pin number _____ of the IC will become active and go ____ (HlGH,LOW)Q) Divide a 32-bit number by a 16-bit number and round the reminder for any value.8086 assemblyQUESTION 5 When the MPU performs a memory write:(Note that RD* means RD is active low) IO/M*= 0 and WR* = 0 IO/M*= 0 and WR* = 1 IO/M*= 1 and WR* = 0 IO/M*= 1 and WR* = 1