Of the key reasons for creating organizational units which of the following is not one of them?

Computers and Technology

1 answer:

victus00 [196]2 years ago

5 0

There are no answrt choices it would be very helpful if you added them, then I could help you.

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The following method in Inventory is intended to count how many batteries are fully charged. What should replace /*Missing Code

OverLord2011 [107]

Trees are the most important part of our environment. They are having a direct impact on our lives. We can’t breathe except the existence of trees in the world. They leave the most important thing oxygen for us. But it is a matter of sorrow that day by day, the tendency of deforestation has been increased in the peak. It has been a matter of afraid now.

Due to the high tendency of deforestation, the world is lacking enough amount of oxygen. We need to plant more and more trees to make it normal. We have seen lots of examples that the world will face a hard time except for the existence of trees. Already, lots of species of trees have been extinct.

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4 0

2 years ago

. (a) Prove or disprove carefully and in detail: (i) Θ is transitive and (ii) ω is transitive. (b) Assume n is a positive intege

Sergio [31]

Answer:

The Following are the solution to this question:

Explanation:

In Option a:

In the point (i) $\Omega$ is transitive, which means it converts one action to others object because if $\Omega(f(n))=g(n)$ indicates $c.g(n)$ . It's true by definition, that becomes valid. But if $\Omega(g(n))=h(n)$ , which implies $c.h(n)$ . it's a very essential component. If $c.h(n) < = g(n) = f(n) \$ . They $\Omega(f(n))$ will also be $h(n)$ .

In point (ii), The value of $\Theta$ is convergent since the $\Theta(g(n))=f(n)$ . It means they should be dual a and b constant variable, therefore $a.g(n)$ could only be valid for the constant variable, that is $\frac{1}{a}\ \ and\ \ \frac{1}{b}$ .

In Option b:

In this algorithm, the input size value is equal to 1 object, and the value of A is a polynomial-time complexity, which is similar to its outcome that is $O(n^{2})$ . It is the outside there will be a loop(i) for n iterations, that is also encoded inside it, the for loop(j), which would be a loop $(n^{2})$ . All internal loops operate on a total number of $N^{2}$ generations and therefore the final time complexity is $O(n^{2})$ .