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3 years ago

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• Suppose that a particular algorithm has time complexity T(n) = 10 ∗ 2n, and that execution of the algorithm on a particular ma

chine takes T seconds for n inputs. Now, suppose you are presented with a machine that is 64 times as fast as your current machine. How many inputs can you process on you new machine in T seconds?

1 answer:

elena-s [515]3 years ago

8 0

Answer:

The number of inputs processed by the new machine is 64

Solution:

As per the question:

The time complexity is given by:

$T(n) = 10\times 2n$

where

n = number of inputs

T = Time taken by the machine for 'n' inputs

Also

The new machine is 65 times faster than the one currently in use.

Let us assume that the new machine takes the same time to solve k operations.

Then

T(k) = 64 T(n)

$\frac{T(k)}{T(n)} = 64$

$\frac{20k}{20n} = 64$

k = 64n

Thus the new machine will process 64 inputs in the time duration T

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Rosalind franklin<br> What was she famous for

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Answer:

She was known for her work on X-ray diffraction images of DNA, which led to the discovery of the DNA double helix for which Francis Crick, James Watson, and Maurice Wilkins shared the Nobel Prize in Physiology or Medicine in 1962.

Explanation:

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3 years ago

A reversible refrigeration cycle operates between cold and hot reservoirs at temperatures TC and TH, respectively. (a) If the co

podryga [215]

Answer:

a) $T_{H} = 1.967\,^{\circ}F$ , b) $COP_{R} = 9.105$ , c) $T_{H} = 115.934\,^{\circ}F$ , d) $COP_{R} = 6.995$ , e) $T_{H} = 25.129\,^{\circ}C$

Explanation:

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The temperature of the hot reservoir is:

$10\cdot T_{H} - 4196.7 = 419.67$

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b) The coefficient of performance is:

$COP_{R} = \frac{273.15\,K}{303.15\,K-273.15\,K}$

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$T_{H} = 575.604\,R$

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$COP_{R} = \frac{489.67\,R}{559.67\,R-489.67\,R}$

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$8.9 = \frac{268.15\,K}{T_{H}-268.15\,K}$

$8.9\cdot T_{H} - 2386.535 = 268.15$

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Answer:

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Answer:

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