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4 years ago

Time complexity of merge sort

Engineering

1 answer:

vovangra [49]4 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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Rewrite the following nested if-else statement as a switch statement that accomplishes exactly the same thing. Assume that num i

seropon [69]

Answer:

Answer explained below

Explanation:

The following is the nested if-else statement:

% if-based statement

if num < -2 || num > 4

f1(num)

else

if num <=2

if num >= 0

f2(num)

else

f3(num)

end

else

f4(num)

end

<u>NOTE:</u> the num is an integer variable that has been initialized and that there are functions f1, f2, f3 and f4.

The nested if-else statement can be replaced by switch statement as shown below:

switch num

case(0, 1, 2)

f2(num)

case(-2, -1)

f3(num)

case(3, 4)

f4(num)

otherwise

f1(num)

In this case, the switch based code is easier to write and understand because the cases are single valued or a few discrete values (not ranges of values)

8 0

3 years ago

Which of the following approaches to lean engineering consists of teams of people from different departments sharing

AURORKA [14]

Answer:

PDSA, where you work as a group

Explanation:

8 0

3 years ago

Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t

kodGreya [7K]

Answer:

$\frac{T-20}{130-20}= e^{-14.28*0.05}$

And if we solve for T we got:

$T= 20 + 110e^{-14.28*0.05} = 73.86 C$

The answer for this case would be $T = 73.86 C$ at 5cm from the base of the fin.

Explanation:

Data given

For this case we have the following data given:

$h = 20 \frac{W}{m^2 K}$ represent the heat transfer coefficient.

$p$ represent the perimeter for this case and would be given by:

$p = 2*0.05m +2*0.001m= 0.102m$

$k = 200 \frac{W}{m C}$ represent the thermal conductivity

$w = 5cm =0.05 m$ represent the width

$h = 1mm =0.001m$ represent the thickness

$A= wh= 0.05m *0.001m = 0.00005 m^2$

Solution to the problem

For this case we assume that we have steady conditions, the temperature of the fins varies just in one direction, the heat transfer coefficient not changes with the time and the thermal properties of the fin not change.

We can determine the temperature if the fin at x=5 cm=0.05 m from the base with the following formula:

$\frac{T-T_{\infty}}{T_b -T_{\infty}} = e^{-mx}$

Where m is a coefficient given by:

$m = \sqrt{\frac{hp}{kA}}=\sqrt{\frac{20 W/m^2 C 0.102 m}{200 W/ mC 0.00005 m^2}}= 14.28 m^{-1}$

The value of x for this case represent the distance $x =5 cm =0.05m$

$T_b =130 C$ represent the base temperature

$T_{\infty}= 20$ represent the temperature of the sorroundings or the ambient.

If we replace we have this:

$\frac{T-20}{130-20}= e^{-14.28*0.05}$

And if we solve for T we got:

$T= 20 + 110e^{-14.28*0.05} = 73.86 C$

The answer for this case would be $T = 73.86 C$ at 5cm from the base of the fin.

3 0

3 years ago

Select the correct answer. Why should engineers keep themselves updated on the technological developments in their field? OA. to

Nataly_w [17]

OD is the correct answer

5 0

3 years ago

Write a program to input 6 numbers. After each number is input, print the biggest of the numbers entered so far.

likoan [24]

Answer:

P<u>rogram:</u>

# Enter Numbers #

number1 = int(input("Enter number: " ))

print("Largest: " + string(number1))

#for num 2 #

number2 = int(input("Enter a number: "))

if number2 > number1:

print("Largest: " + string(number2))

else:

print("Largest: " + string(num1))

#for num 3 #

number3 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3)))

#for num 4 #

number4 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4)))

#for num 5 #

number5 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4, number5)))

#for num 6 #

number6 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4, number5, number6)))

# END #

4 0

3 years ago