Input a binary number: the decimal value is: the octal value is: the hex value is: Input a decimal value the binary value is: the octal value is: the hex value is: Input an octal value: the binary value is: the decimal value is: the hex value is: Input a hex value: the binary value is: the decimal value is: the octal value is:
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- binary = "010000000111111010111001" #convert to decimal from binarydecimal = int(binary,2)print("{:.5f}".format(decimal))print("\n") #Three digit choppingprint('%.3f'% decimal)print("\n") #Three digit roundingprint(round(decimal,3))print("\n") Please solve questions 4-6. My code for questions #1-3 is above. Code is written in Python.ARGUMENT TWO Valid ☐ Invalid ☐ Some X is Z Some Y is not Z ∴ Some Y is X ARGUMENT THREE Valid ☐ Invalid ☐ All C is G All C is L ∴ All L is G ARGUMENT FOUR Valid ☐ Invalid ☐ No Z is L All O is L ∴ No O is a Z ARGUMENT FIVE Valid ☐ ï Invalid ☐ No B is A Some R is not B ∴ Some R is not A ARGUMENT SIX Valid ☐ ï Invalid ☐ All F is M Some S is M ∴ Some S is FPut the following number in decimal notation 3.512 x 105 put the following number into decimal notation 6.82 x 104 put the following number into decimal notation 1.45 x 10-3 Put the following number into decimal notation 8.321 x 10-6
- What is the decimal equivalent of the code?Binary numbers need more places for counting because: A. 0's and 1's can be added in front of them B. 0's and 1's have to be properly placed C. They are always big numbers D. Binary base is small E. None of the aboveConverting a value from a number base other than decimal to a decimal value...use expanded notation, where the number base is raised to a power based on its position within the value and multiplied by the value at the particular position
- a has the binary value 01101001 and b has the binary value 00111011. Indicate the binary value returned by the expression: a & b1). Perform the following multiplications. Assume that all values are unsignednumbers. (No problem with unsigned (always positive) numbers, just use the samestandard techniques as in base 10.) 1010 * 0110 2) Perform the following multiplications. Assume that all values are signed 2’scomplement numbers. 11101 * 11101Use the format SEEMMMM. All digits are decimal. The exponent is stored excess-50. The implied decimal point is at the beginning of the mantissa. The sign is 1 for a positive number, 7 for a negative number. Hint: Note carefully the number of digits in the mantissa! a. Convert the decimal number 28643 to floating point.b. What is the range of numbers that can be stored in this format? c. What is the floating-point representation for: −28569? d. What is the floating-point representation for 0.0000019557?
- Binary-Decimal / Decimal Binary . Between 0 and 255 a number will be decided randomly by computer. Then asks to user 3 times a random digit of binary value of the that number. If user enters wrong number for a digit program will select another random number and ask random times random digit. Every time user enters input will be counted as one hour in that planet. For example: Computer selected number as 163 ( Which is 10100011) What is the digit 2 (question 1/3) :User enter 1 CorrectWhat is the digit 4 (question 2/3) : User enter 0 CorrectWhat is the digit 7 (question 3/3) : User enter 0 CorrectUser finished the quest with 3 input .:Binary-Decimal / Decimal Binary . Between 0 and 255 a number will be decided randomly by computer. Then asks to user 3 times a random digit of binary value of the that number. If user enters wrong number for a digit program will select another random number and ask random times random digit. For example: Computer selected number as 163 ( Which is 10100011) What is the digit 2 (question 1/3) :User enter 1 CorrectWhat is the digit 4 (question 2/3) : User enter 0 CorrectWhat is the digit 7 (question 3/3) : User enter 0 CorrectUser finished the quest with 3 input .If the computer has 32 kilobytes of memory and the storage location is the largest word (8 bytes), calculate the following: 1. The amount of storage. 2- The names of the first and last positions One binary position, or 3.