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2 years ago

Because of the density of vehicle in urban areas, you should ____.

Computers and Technology

2 answers:

Elena L [17]2 years ago

6 0

Answer:

B. keep an eye on your blind spots and watch for motorcycle

Explanation:

Because of the density of vehicle in urban areas, you should keep an eye on your blind spots and watch for motorcycles.

Effectus [21]2 years ago

5 0

Answer:

The correct answer to the following question will be Option B.

Explanation:

Blind spots are the parts of the vehicles that can not be seen in front or across the rear or side mirrors. By carefully adjusting the rear and side mirror or adding a larger field of view, blind spots can be eliminated.
Because of the density of the vehicles in the urban areas, the blind spots cannot be ignored they should be considered to avoid the clashes on the road.
All the vehicles can't install the sensors on all sides of the vehicle. Also, tailgating is not possible because of the density of vehicles.

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Answer:

Check the explanation

Explanation:

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1 year ago

[Assembly Language]Extended Subtraction Procedure.Create a procedure named Extended_Sub --(Receives: ESI and EDI point to the tw

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Answer:

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Explanation:

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1 year ago

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Kay [80]

Answer:

The algorithm is very similar to the algorithm of counting inversions. The only change is that here we separate the counting of significant inversions from the merge-sort process.

Algorithm:

Let A = (a1, a2, . . . , an).

Function CountSigInv(A[1...n])

if n = 1 return 0; //base case

Let L := A[1...floor(n/2)]; // Get the first half of A

Let R := A[floor(n/2)+1...n]; // Get the second half of A

//Recurse on L. Return B, the sorted L,

//and x, the number of significant inversions in $L$

Let B, x := CountSigInv(L);

Let C, y := CountSigInv(R); //Do the counting of significant split inversions

Let i := 1;

Let j := 1;

Let z := 0;

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//the normal merge-sort process i := 1; j := 1;

//the sorted A to be output Let D[1...n] be an array of length n, and every entry is initialized with 0; for k = 1 to n if B[i] < C[j] D[k] = B[i]; i += 1; else D[k] = C[j]; j += 1; return D, (x + y + z);

Runtime Analysis: At each level, both the counting of significant split inversions and the normal merge-sort process take O(n) time, because we take a linear scan in both cases. Also, at each level, we break the problem into two subproblems and the size of each subproblem is n/2. Hence, the recurrence relation is T(n) = 2T(n/2) + O(n). So in total, the time complexity is O(n log n).

Explanation:

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1 year ago

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mixas84 [53]

Answer:

Python Code:

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isFileValid = False

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if x in url_split[0]:

isProtocolValid = True

break

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print(validate_url(url))

Explanation:

The image of the output code is attached. Hope it helps.

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1 year ago

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2 years ago

Read 2 more answers