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Answer: 2The values are 0 or 1.
With 6 binary digits, you have 26 different possibilities. This is because there are two possibilities for each digit, and each digit is independent of the other digits - so you just multiply the possibilities for each digit together.
The binary system uses two digits, zero and one.
56 in binary is 111000. Unlike the decimal number system where we use the digits.
Binary form has 2 sections,Ternary form has 3 sections.
A 128-bit register can store 2 128th (over 3.40 × 10 38th) different values. The range of integer values that can be stored in 128 bits depends on the integer representation used.
Answer: 2The values are 0 or 1.
Answer: 2The values are 0 or 1.
64 or 123
65,536
It can have 0 to 1 It can have 0 to 1
4 these are 00,01,10 and 11...
24, or 16 (0 through 15) One binary digit (bit) can have 21 values (0 or 1). Two bits can have 22 values. Three bits can have 23 values. A five-bit number can have 25 values... and so on...
There are 14 Regiters in the computer where data is stored which is currently being process
0 o 1
Texas is too big, but you can fit in a Ohio, Maryland, and part of West Virginia. But seriously though...I am assuming that you mean "how many unique combinations of 1's and 0's can be stored in a 16 bit register." The answer is given as 2 (the number of different possibilities per digit) raised to the power of 16. The answer is 65536, made up of 0 through 65535. An 8 bit register can represent 256 different values, 0 through 255.
please tell me answer of this question. Suppose you are building an N node binary search tree with the values 1...N. how many structurally different binary trees is there that store those values? write a recursive function that, gives the number of distinct values, computes the number of structurally unique binary search trees that store those values. For example, countTrees(4) should return 14, since there are 14 structurally unique binary search trees that store 1,2,3 and 4. The base case us easy, and the recursion is short but dense. your code should not construct any actual trees; it's just a counting problem.