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zmey [24]
2 years ago
6

Suppose that you have two different algorithms for solving a problem of size n. The first algorithm uses exactly n(log n) operat

ions and the second algorithm uses exactly n 3/2 operations. As n grows, determine which algorithm uses fewer operations?
Mathematics
1 answer:
Mumz [18]2 years ago
5 0

Answer:

Algorithm 1 uses:

n*log(n) operations.

While algorithm 2 uses:

n^(3/2) operations.

We want to see, as n grows, which algorithm uses fewer operations.

So we would want to first solve:

n*log(n) = n^(3/2)

This will give us the exact value of n such that the number of operations is the same in both algorithms.

dividing both sides by n we get:

log(n) = n^(3/2)/n = n^(3/2 - 1) = n^(1/2)

where we can use:

log(n) = ln(n)/ln(10)

ln(n) = ln(10)*n^(1/2)

This equation actually has no solutions.

This happens because the right side is always larger than the left side.

Then, the same thing happens for our two initial equations:

n^(3/2) is always larger than n*log(n), as you can see in the graph below, where n^(3/2)  is represented with the orange graph:

So we can conclude that the fist algorithm uses less operations as n grows.

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Natasha2012 [34]

Hello from MrBillDoesMath!

Answer:

x < -2

Discussion:

7 < 3 - 2x                       => Add 2x to both sides

7 + 2x < 3 - (2x - 2x)     =>  As 2x -2x = 0

7 + 2x < 3                      =>  Subtract  7 from both sides

(7-7) +  2x < 3 - 7           => As 7-7 = 0 and 3-7 = -4

2x < -4                           => Divide both sides by 2

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Thank you,

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Length of other board =\frac{4x+20}{x^{2} -4} metres

Step-by-step explanation:

Given: Length of board = \frac{7}{x-2} metres

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Let l = length of the other board

Length of the other board = Length of board - Length of one piece

So, l=\frac{7}{x-2}-\frac{3}{x+2}

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Mode is what is the most frequent number that occurred; 95 occurred the most from the data.

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