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3 years ago

If the three operations were combined, O(logN) + O(N) * O(logN) + 1, the overall algorithm cost would be:________

Computers and Technology

1 answer:

KiRa [710]3 years ago

3 0

Answer:

D. O(NlogN)

Explanation:

The computation of the overall algorithm cost is as follows:

Given that

O(logN) + O(N) × O(logN) + 1

In the case of complexity we considered the high order that dominates the other terms

Thus, that term would be

O(N) × O(logN)

It could be rewrite as

O(NlogN)

Hence, the correct option is D.

All the other options are wrong

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Which marketing strategy is represented by advertisers paying for online display ads?

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Answer:

The answer is (c) paid media

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3 years ago

Joshua wants to be a lawyer. He found the following table on the Bureau of Labor Statistics’ website to find out about the emplo

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Int[] myArray1= new int[24]; int[] myArray2= new int[10]; float[] myArray3= new float[8]; int index1, index =0 for(int index1 =0

anastassius [24]

Answer:

The value of myArray2[index2] when index1 = 12 is 30

Explanation:

In the source code, the formula for myArray2[index2] is;

myArray2[index2] = index2 + index3 + myArray1[index1],

myArray1[index1] = index1 * 2,

index2 = index % 10 (equal to the remainder) and

index3 = index % 8

When index1 increases to 12 in the for-loop statement, the "myArray1[index1]" is equal to 24, index2 is equal to 2 and index3 is 4. The total sum is equal to 30 and assigned to "myArray2[index2]".

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3 years ago

A group of N stations share a 56-kbps pure ALOHA channel. Each station outputs a 1000-bit frame on average once every 100 sec, e

Sunny_sXe [5.5K]

Explanation:

Here, in the given statement, maximum flow capacity of the ALOHA channel = 18.4% = 0.184

Then, the stations N share 56 kbps

And, the channel rate = $0.184*56$ $= 10.304 kbps.$

Then, the outputs of the wach stations is $1000-bit frames/100 sec$

$\therefore$ , the output of the bits/ sec by each stations = $\frac{1000}{100} bits/sec = 10 bits/sec$

So, The output of the N stations is 10 bits/sec on the channel that having $10.304kb/sec$

Finally, N = $\frac{(10.304 \times 10^{3})}{10} = 1030 stations$

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3 years ago