Solution :
We have to provide an expression for the binary numbers. There can be binary fractions or integers. Whenever there is leading 0, it is not allowed unless the integer part is a 0.
Thus the expression is :
![$(\in +.(0+1)^*(0+1))+(0.(0+1)^*(0+1))]$](https://tex.z-dn.net/?f=%24%28%5Cin%20%2B.%280%2B1%29%5E%2A%280%2B1%29%29%2B%280.%280%2B1%29%5E%2A%280%2B1%29%29%5D%24)
Answer:
Explanation:
When you have a single copy, a large number of concurrent updates that are supposed to go to a file may result in the user obtaining incorrect information. This incorrect information obtained them leads to the file being left in an incorrect state. When you have a lot of or multiple copies, then storage waste exist and the various copies might happen not to be consistent with respect one other. In summary, what happens is that
a) Using one copy saves space, but also the change might have an effect on all the users.
b) Using multiple copies avoids eliminates the change problem, while creating its own problems, using more space.
Answer:
I don't think you can. You can try to make an email that has the name you want and use that.
Explanation:
Answer:
Normalization of storage is a typical method of storing the floating point number by shifting the decimal after the first figure of the number such as 1101.101 is normalized to 1.101101x23.
If the number that is in hovering point representation has 1 sign bit, 3-bit exponent with a 4-bit significant:
whenever the storage is normalized, then the biggest positive floating spot number in 2`s and the complement notation is 0.11112 x 23 = 111.12 =7.5
If the storage is returned to normal, then the minimum positive floating point number is 0.12 x 2-4 =0.000012 =1/32 = 0.03125.
Explanation:
Whenever the floating figure is keyed into the computer memory, then the first bit will be the sign bit, the next 8 bits are for exponent and 23 bits are used for storing significand. The array of exponents is from -127 to 128. While Exponent 127 stands for 0 and positive figures can be represented by values bigger than 127. The biggest floating point number will be represented as 0.111111.... 1111x 211111111
