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3 years ago

Time complexity of merge sort

Engineering

1 answer:

vovangra [49]3 years ago

6 0

Answer:

The correct answer is " $O (n\times Log n)$ ". A further explanation is given below.

Explanation:

Throughout all the three instances (worst, average as well as best), the time complexity including its Merge sort seems to be $O (n\times Log n)$ as the merge form often splits the array into two halves together tends to linear time to combine multiple halves.
As an unsorted array, it needs an equivalent amount of unnecessary capacity. Therefore, large unsorted arrays are not appropriate for having to search.

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The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8 What is the probability of observing

Viktor [21]

Answer:

0.14% probability of observing more than 4 errors in the carpet

Explanation:

When we only have the mean, we use the Poisson distribution.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8.

This means that $\mu = 0.8$

What is the probability of observing more than 4 errors in the carpet

Either we observe 4 or less errors, or we observe more than 4. The sum of the probabilities of these outcomes is 1. So

$P(X \leq 4) + P(X > 4) = 1$

We want P(X > 4). Then

$P(X > 4) = 1 - P(X \leq 4)$

In which

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493$

$P(X = 1) = \frac{e^{-0.8}*(0.8)^{1}}{(1)!} = 0.3595$

$P(X = 2) = \frac{e^{-0.8}*(0.8)^{2}}{(2)!} = 0.1438$

$P(X = 3) = \frac{e^{-0.8}*(0.8)^{3}}{(3)!} = 0.0383$

$P(X = 4) = \frac{e^{-0.8}*(0.8)^{4}}{(4)!} = 0.0077$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.4493 + 0.3595 + 0.1438 + 0.0383 + 0.0077 = 0.9986$

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.9986 = 0.0014$

0.14% probability of observing more than 4 errors in the carpet

5 0

3 years ago

The phrase "positive to positive, negative to ground" is correct when jump starting a car.

Artist 52 [7]

The correct answer is A; True.

Further Explanation:

This is a correct phrase that is important to learn when owning any type of vehicle. When a car battery is dead, it can usually be jump started by using another cars battery or a portable battery charger. It is extremely important to put the positive battery cable on the positive battery post. Then the negative cable will be placed on the negative car battery post and the negative ground wire can be anywhere on the car except on the battery.

The car needs to be connected properly for a few minutes before trying to start the car. This helps the car battery to get enough "juice" to start. If the battery cables are placed wrong this can cause sparks to come out of the cables/battery and cause bodily harm.

Learn more about car batteries at brainly.com/question/7734062

#LearnwithBrainly

7 0

3 years ago

Drag each tile to the correct box.

Trava [24]

Answer:

Bluray

DVD

Explanation:

Blu ray can hold 25gb per layer

Dvd can hold 4.7GB on a single layer

Cd can hold around 737 mb

Also, dvds can go up to 2 layers

Blu ray can go up to 4

6 0

3 years ago

How will the delay and active power per device change as you increase the doping density of both the N- and the P-MOSFET?

Murljashka [212]

Answer:

hello your question is incomplete attached below is the missing part of the question

Consider an inverter operating a power supply voltage VDD. Assume that matched condition for this inverter. Make the necessary assumptions to get to an answer for the following questions.

answer : Nd ∝ rt

Explanation:

Determine how the delay and active power per device will change as the doping density of N- and P-MOSFET increases

Pactive ( active power ) = Efs * F

Pactive = $\frac{q^2Nd^2*Xn^2}{6Eo} * f$