Answer:
System.out.println("i = " + i + " f = " + f);
Explanation:
The code above has been written using Java's syntax.
To print or display statements to the console, one of the methods to use is the System.out.println() method.
According to the question, the statement to output is i=value-of-i f=value-of-f which is a string of texts.
Note that value-of-i and value-of-f are already stored in variables i and f respectively. Therefore to print them as part of the string text statement, it is important to concatenate them using the "+" operator as follows:
"i = " + i + " f = " + f
If the value of i is 25 and the value of f is 12.34 then the above statement, after successful concatenation reads:
"i = 25 f = 12.34"
which can then be shown on the console using System.out.println() method as follows:
System.out.println("i = " + i + " f = " + f);
Hope this helps!
Answer:
-32 to +31
Explanation:
6 bit 2s complement representation will be of the form b1b2b3b4b5b6 where each bit is either 0 or 1.
The largest positive number that can be represented using this scheme is 011111
Translating this to decimal this is 1*2^4 + 1^2^3 + 1^2^2 + 1^2^1 + 1^2^0
=16 + 8 + 4 + 2 + 1 =31
The smallest negative number that can be represented using this scheme is 100000
Translating this to decimal = -1 * 2^5 = -32
So the range of decimal values that can be represented is -32 to +31.
Your answer would be:
<span>
In Microsoft Word you can access the </span><span>insert citation command from the mini toolbar.</span>
Answer:
X is in NP and Y is in NP-HARD ( A )
Explanation:
X is in NP and Y is in NP-HARD can be inferred from the previous statement made in the problem above because problem decision X can be in NP if it can BE reducible to a 3-SAT polynomial real time, if that can be achieved then 3SAT will be in NP since SAT is in NP as well.
also problem decision Y can be in NP-HARD if 3SAT can be reducible to it in polynomial time as well hence option A is the correct option
