What are some things that AREN’T manufactured

Say you have a random, unordered list containing 4096 four-digit numbers. Describe the most efficient way to: sort the list and

Debora [2.8K]

Answer:

Answer explained below

Explanation:

It is given that numbers are four-digit so maximum value of a number in this list could be 9999.

So we need to sort a list of integers, where each integer lies between [0,9999].

For these given constraints we can use counting sort which will run in linear time i.e. O(n).

--------------------------------------------------------------------------------

Psuedo Code:

countSort(int numList[]) {

int count[10000];

count[i] = 0; for all i;

for(int num in numList){

count[num]+= 1;

}

return count;

}

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Searching in this count array will be just O(1).

E.g. Lets say we want to search if 3 was present in the original list.

Case 1: it was present in the original list:

Then the count[3] would have been incremented by our sorting algorithm. so in case element exists then count value of that element will be greater than 0.

Case 2: it was not present:

In this case count[3] will remain at 0. so in case element does not exist then count of that element will be 0.

So to search for an element, say x, we just need to check if count[x]>0.

So search is O(1).

Run times:

Sorting: O(n)

Search: O(1)

6 0

3 years ago

Find E[x] when x is sum of two fair dice?

Ksenya-84 [330]

Answer:

When two fair dice are rolled, 6×6=36 observations are obtained.

P(X=2)=P(1,1)=

36

1

P(X=3)=P(1,2)+P(2,1)=

36

2

=

18

1

P(X=4)=P(1,3)+P(2,2)+P(3,1)=

36

3

=

12

1

P(X=5)=P(1,4)+P(2,3)+P(3,2)+P(4,1)=

36

4

=

9

1

P(X=6)=P(1,5)+P(2,4)+P(3,3)+P(4,2)+P(5,1)=

36

5

P(X=7)=P(1,6)+P(2,5)+P(3,4)+P(4,3)+P(5,2)+P(6,1)=

36

6

=

6

1

P(X=8)=P(2,6)+P(3,5)+P(4,4)+P(5,3)+P(6,2)=

36

5

P(X=9)=P(3,6)+P(4,5)+P(5,4)+P(6,3)=

36

4

=

9

1

P(X=10)=P(4,6)+P(5,5)+P(6,4)=

36

3

=

12

1

P(X=11)=P(5,6)+P(6,5)=

36

2

=

18

1

P(X=12)=P(6,6)=

36

1

Therefore, the required probability distribution is as follows.

Then, E(X)=∑X

i

⋅P(X

i

)

=2×

36

1

+3×

18

1

+4×

12

1

+5×

9

1

+6×

36

5

+7×

6

1

+8×

36

5

+9×

9

1

+10×

12

1

+11×

18

1

+12×

36

1

=

18

1

+

6

1

+

3

1

+

9

5

+

6

5

+

6

7

+

9

10

+1+

6

5

+

18

11

+

3

1

=7

E(X

2

)=∑X

i

2

⋅P(X

i

)

=4×

36

1

+9×

18

1

+16×

12

1

+25×

9

1

+36×

36

5

+49×

6

1

+64×

36

5

+81×

9

1

+100×

12

1

+121×

18

1

+144×

36

1

=

9

1

+

2

1

+

3

4

+

9

25

+5+

6

49

+

9

80

+9+

3

25

+

18

121

+4

=

18

987

=

6

329

=54.833

Then, Var(X)=E(X

2

)−[E(X)]

2

=54.833−(7)

2

=54.833−49

=5.833

∴ Standard deviation =

Var(X)

=

5.833

=2.415

4 0

2 years ago

Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,

Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0

3 years ago

I want a problems and there solutions of The inception of cavitation?

Ugo [173]

Answer:

The overview of the given scenario is explained in explanation segment below.

Explanation:

The inception of cavitation, that further sets the restriction for high-pressure and high-free operation, has always been the matter of substantial experimental study over the last few generations.
Cavitation inception would be expected to vary on the segment where the local "PL" pressure mostly on segment keeps falling to that are below the "Pv" vapor pressure of the fluid and therefore could be anticipated from either the apportionment of the pressure.

⇒ A cavitation number is denoted by "σ" .

4 0

3 years ago

35 points an brainiest if correct

Jet001 [13]

I think the answer is B. 10D

6 0

3 years ago