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What decimal numbers are represented by this binary number 101?
101 = 5
Is -5 a irrational number?
No, -5 is not an irrational number. Irrational numbers are numbers that cannot be represented as the quotient of two integers. Since -5 is already an integer, it is rational.
How many number of bits are required to permit selection of 1 out of 32 equi-probable events?
5 bits, since 2 to the power 5 equals 32.
How many natural numbers between 1 and 10 can be repesented as product of two or more prime factors?
Between 1 and 10, 2, 3, 5 and 7 are already prime numbers. 4 (2x2), 6 (2x3), 8 (2x2x2) and 9 (3x3) can be represented as a product of prime factors, i.e. four natural numbers between 1 and 10 can be represented as such.
How many combinations of 5 different numbers if there are 5 numbers?
120
How many bits to represent twenty-six?
23 can be represented in binary as 10111 and would therefore require 5 bits to represent.
What is the largest number that can be represented using 5 bits?
31 - it's binary equivalent is 11111
What decimal numbers are represented by this binary number 101?
101 = 5
If 88 MB is used out of 5 gb how many is left?
5 gb= 5*230 bits 88 mb= 88*220 bits Therefore if 88 mb is used out of 5 gb then = 5 gb - 88 mb bits are left = 5*2 30 bits - 88*220 bits = 220(5*210-88) bits = 220(5*1024-88) bits = 220*5032 bits = 5032 mb are left
How many bits of mask are required to provide 30 host addresses?
5 bits
How many bits or bytes required to store the robot?
40 bits or 5 byrtes
Is -5 a irrational number?
No, -5 is not an irrational number. Irrational numbers are numbers that cannot be represented as the quotient of two integers. Since -5 is already an integer, it is rational.
How many mega bits in 5 gig a bits?
http://www.google.co.uk/search?q=5+gigabits+in+megabits 5120 megabits in 5 gigabits
What are some rational numbers that are not integers?
1/2, 5/8, and 6/7 are some. Rational numbers are numbers that can be represented as fractions, so there are practically unlimited numbers of them.
What numbers equal 17 when one number is greater by 5?
If this refers to the SUM of two numbers then let the lesser number be represented by a. Then, a + (a + 5) = 17 : 2a + 5 = 17 : 2a = 12 : a = 6 So the two numbers are 6 and 11.
How many number of bits are required to permit selection of 1 out of 32 equi-probable events?
5 bits, since 2 to the power 5 equals 32.
How many distinct values can we represent with 5 bits?
32 values. 2^5=32
