Neither of the following are true about 1 bit, it can not represent decimal values 0 and 9 nor can it be used to represent one character in the lowercase English alphabet and one binary digit four binary. A true statement would be that 1 bit is represented by the decimal values 0 or 1.
The binary values is 10110101.
Binary- 01100111 Decimal Value- 103
Decimal: 75Binary: 1001011
It is 1111.
The standard written format for an IP address is as 4 bytes written as their decimal values separated by periods. Just convert each decimal value to a binary byte and append them to make a 32 bit number. Reverse that to convert a 32 bit number to 4 decimal bytes separated by periods.
Many non-integral values, such as decimal 0.2, have an infinite place-value representation in binary (.001100110011...) but have a finite place-value in binary-coded decimal (0.0010)[bcd]. Consequently a system based on binary-coded decimal representations of decimal fractions avoids errors representing and calculating such values. Rounding at a decimal digit boundary is simpler in BCD. Addition and subtraction in decimal does not require rounding.
Decimal (more formally, binary coded decimal) values store numeric information as digits encoded using the four bit binary equivalents: 0 (0000) to 9 (1001). That means a single byte can hold values between 0 and 99. But simply using the same byte to hold a binary value will yield values between 0 and 255 (or –128 and +127).
"Binary decimal" is a contradiction in terms. Decimal has a base of 10, binary a base of 2 and hexadecimal a base of 16.The way I would do it is:If you have a value in binary then convert this to a decimal value. Then convert it to hexadecimal remembering that the number will now be comprised by the following (where x represents the digit):The first digit (from right to left) will equal x * 160, the next will equal x * 161 and so forth...An example:So in binary 11111 = (1 * 20) + (1 * 21) + (1 * 22) + (1 * 23) + (1 * 24) = 1 + 2 + 4 + 8 + 16 = 31 (in decimal).To write this in hexadecimal, 31 would be (15 * 160) + (1 * 161) = 1FNote: A tip - If you are using a Windows operating system, then if you go to the Start menu and choose search/run and type in "calc" or "calculator" then you will get a virtual calculator to use. If you choose "programmer" from the View menu and then choose the "Bin" button and type in a binary value and then choose the "Hex" button then the binary value will be converted to hexadecimal. (The above certainly applies for Windows 7).
Since a binary digit has only two possible values, each digit bears less information than in decimal, where each digit can have ten different values.
ASCII for K is 0x4b = 75 = 0100 1011
You divide percent values by 100 to get the decimal equivalent: 0.50% / 100 = 0.005
Binary to Decimal Working from the least-significant bit, the positional values are 1, 2, 4, 8, 16, 32, 64 and 128 (each more significant bit doubles the value of the preceding bit). If the corresponding bit is set, add that value to an accumulator initialised to zero. Decimal to Binary Repeatedly divide the decimal value by 2 and take the remainder (which can only be 0 or 1). Each division determines the value of the next most significant bit, starting with the least significant bit.
Advantage of binary over decimal: information can be recorded and stored in any dichotomous variable: magnetised or not magnetised (most electronic media), pit or no pit (optoelectronic media CDs/DVDs). For decimal it would be necessary to store as 10 different levels of magnetisation or depths of pits. Not so easy to make such a system error-free. Advantage of decimal over binary: fewer "digits" required. Every ten binary digits (1024 values) can be replaced by just a shade more than three decimal digits (1000 values). So the number of digits to be stored is less than a third.
Not sure what is meant by this. If 230 and 35 are decimal values then they are already in a high-level language: decimal. Computers only understand binary, so perhaps you are really asking how to convert these decimal values to their binary equivalent in a human-readable form. If so, divide each value by 2 repeatedly until the value is zero. At each division, take the remainder (the modus, which can only be a 0 or a 1) and fill a character array from right to left (least significant to most significant digit) with that value's character. To convert the remainder value to a character, add the remainder value to the character code '0' (which evaluates to '0' or '1'). Once the character array is created, you can display it.
Convert each decimal number to a binary number separately. You can use Windows calculator for this (in the menu you will find a command to switch between standard calculator and scientific calculator). Each group of bits must be completed to 8 bits - fill out with zeroes on the left, until you have 8 bits (so for the the second part, the calculator will give you 1101011 - add a zero to the left to get 01101011). If you want to do the conversion with pencil and paper, ask a separate question, or research existing questions, something like "how do you convert decimal to binary".
You might say so, but depending on the method used to assign decimal values, that may be a very weak encryption - in other words, easy to guess.
I presume you really mean how do you convert a value to a binary string for printing purposes. E.g., the decimal value 42 should print "0b101010" (the 0b prefix is optional, but helps to distinguish binary values from decimal values). Integers are reasonably simple to convert: int i = 42; System.out.println(Integer.toString(i) + " = 0b" + Integer.toBinaryString(i)); Doubles are a bit more complex: double d = Math.pow(2, 900); System.out.println(Double.toString(d) + " = 0b" + Long.toBinaryString(Double.doubleToRawLongBits(d))); For more complex types (such as class representations), treat the type as if it were an array of type char, then print the binary value of each char (padding with leading zeroes where necessary). Be aware that although Java uses big-endian binary representations, this may or may not match the representations used by the underlying architecture. This is important when converting values generated outside of Java (such as values stored in binary files generated from other programs). Before converting these values you first need to determine the underlying type of the value and then convert to a corresponding Java type. This ensures the bytes are in the correct order regardless of the physical representation. If you really want the physical representation then just treat the value as an array of type char.
A binary tree is simply a tree in which each node can have at most two children. A binary search tree is a binary tree in which the nodes are assigned values, with the following restrictions ; -No duplicate values. -The left subtree of a node can only have values less than the node -The right subtree of a node can only have values greater than the node and recursively defined; -The left subtree of a node is a binary search tree. -The right subtree of a node is a binary search tree.
Each hexadecimal digit can hold one of 16 values (0-F); 16 = 2^4, so exactly 4 bits (binary digits) can hold the same value as 1 hexadecimal digit. As a result the conversion from binary to hexadecimal is simply a matter of grouping the bits together in blocks of 4 (making nybbles) and converting each block into a single hexadecimal digit. Similarly for binary to octal but in this case as 8 = 2³ the bits are group into blocks of 3 which are then converted into octal digits. However, converting decimal to hexadecimal is not so "easy" as each decimal digit does not map to an exact number of binary digits. The only exception would be when using BCD (Binary Coded Decimal) where only the bit patterns for the decimal digits 0-9 are used in every 4 bits (wasting 6 possible digits) and where 0000 1001 (09) + 0000 0001 (01) = 0001 0000 (10). In this case the hexadecimal representation of the BCD is exactly the same as the decimal, but I have never seen it used as such (beyond the binary representation).
1. Binary computers do not understand decimal. The native language of a binary computer is binary language. Thus if we wish to avoid unnecessary computations converting to and from decimal encodings, we must represent all numbers using binary. 2. Decimal computers are complicated. While it is possible for a computer to differentiate between 10 possible states and thus work with native decimal values, logic circuits are much easier to implement when there are only two possible states to consider. Arithmetic is the application of pure logic, thus arithmetic circuitry is greatly simplified as a result.
This is the decimal value 15. A binary number uses exponents of 2 rather than 10, and the 8 digits shown represent 128, 64, 32, 16, 8, 4, 2, and 1 00001111 (or just 1111) = 0+0+0+0+8+4+2+1 = 15 Add the values of each exponential where there is a 1 value.
Usually it's converted to other values for human use. As far as the computer "sees" it, the number is always stored in binary.
Internally, the computer processes bits - ones and zeroes. Hexadecimal is a shorthand form of writing those values, or of showing them to humans - every hexadecimal digit corresponds to four bits. Conversion from binary to hexadecimal is fairly easy; converstion between binary and decimal is more complicated.
No. The metric system, since it is based on powers of ten, is better used with decimal representations of values.