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igomit [66]
3 years ago
10

It describes the physical and social elements common to this work. Note that common contexts are listed toward the top, and less

common contexts are listed toward the bottom. According to O*NET, what are common work contexts for Construction Carpenters? Check all that apply. Spend time sitting wear common protective or safety equipment face-to-face discussions exposed to hazardous equipment spend time using hands to handle, control, or feel objects, tools, or controls public speaking
Engineering
2 answers:
QveST [7]3 years ago
7 0

Answer:

BCDE

Explanation:

just look at the link, it tells you.

Dafna1 [17]3 years ago
3 0

Answer:

bcde!!

Explanation:

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In details and step-by-step, show how you apply the Bubble Sort algorithm on the following list of values. Your answer should sh
astraxan [27]

( 12 17 18 19 25 )

<u>Explanation:</u>

<u>First Pass:</u>

( 19 18 25 17 12 ) –> ( 18 19 25 17 12 ), Here, algorithm compares the first two elements, and swaps since 19 > 18.

( 18 19 25 17 12 ) –> ( 18 19 25 17 12 ), Now, since these elements are already in order (25 > 19), algorithm does not swap them.

( 18 19 25 17 12 ) –> ( 18 19 17 25 12 ), Swap since 25 > 17

( 18 19 17 25 12 ) –> ( 18 19 17 12 25 ), Swap since 25 > 12

<u>Second Pass:</u>

( 18 19 17 12 25 ) –> ( 18 19 17 12 25 )

( 18 19 17 12 25 ) –> ( 18 17 19 12 25 ), Swap since 19 > 17

( 18 17 19 12 25 ) –> ( 18 17 12 19 25 ), Swap since 19 > 12

( 18 17 12 19 25 ) –> ( 18 17 12 19 25 )

<u>Third Pass:</u>

( 18 17 12 19 25 ) –> ( 17 18 12 19 25 ), Swap since 18 > 17

( 17 18 12 19 25 ) –> ( 17 12 18 19 25 ), Swap since 18 > 12

( 17 12 18 19 25 ) –> ( 17 12 18 19 25 )

( 17 12 18 19 25 ) –> ( 17 12 18 19 25 )

<u>Fourth Pass:</u>

( 17 12 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 17 > 12

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 18 > 12

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted.

<u>Fifth Pass:</u>

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

6 0
3 years ago
The Eads Bridge, which crosses the Mississippi River near St Louis, Missouri, was one of the first all steel bridges built in th
MrMuchimi

Answer: At 520 feet between the piers, the center arch of Eads Bridge was the longest rigid span ever built at the time of its construction (only a few suspension bridges had longer spans).

Explanation:

5 0
2 years ago
Using the Distortion-Energy failure theory: 8. (5 pts) Calculate the hydrostatic and distortional components of the stress 9. (1
WITCHER [35]

Answer:

Detailed solution is given below:

7 0
3 years ago
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;

}

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

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
A seamless pipe 800mm diameter contains a fluid under a pressure of 2N/mm2. If the permissible tensile stress is 100N/mm2, find
Bad White [126]

Answer:

8 mm

Explanation:

Given:

Diameter, D = 800 mm

Pressure, P = 2 N/mm²

Permissible tensile stress, σ = 100 N/mm²

Now,

for the pipes, we have the relation as:

\sigma=\frac{\textup{PD}}{\textup{2t}}

where, t is the thickness

on substituting the respective values, we get

100=\frac{\textup{2\times800}}{\textup{2t}}

or

t = 8 mm

Hence, the minimum thickness of pipe is 8 mm

3 0
3 years ago
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