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Naya [18.7K]
3 years ago
7

horseback riders, bicyclists, and skateboarders ____ the rules of right-of-way when they use the road ?

Computers and Technology
2 answers:
Sveta_85 [38]3 years ago
8 0

Answer:

must follow

Explanation:

The rules of right-of-way indicate that you should let another driver go first in a traffic situation to avoid an accident. When two drivers get to an intersection at the same time, the driver in the right should go first. The other driver give up the right of way to the one on the right. This means that all the drivers, horseback riders, bicyclists, and skateboarders must follow these rules to be safe and like that do everything possible to avoid a crash.

xz_007 [3.2K]3 years ago
4 0

Answer:

I do not know specifically what type of answer you are looking for, but horseback riders, bicyclists, and skateboarders all use the same rules of right-of-way when they are on the road, but cars yield to them as they are pedestrians.

Hope this helps!

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Recall the problem of finding the number of inversions. As in the text, we are given a sequence of n numbers a1, . . . , an, whi
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;

// to count the number of significant split inversions while(i <= length(B) and j <= length(C)) if(B[i] > 2*C[j]) z += length(B)-i+1; j += 1; else i += 1;

//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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Explanation:

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So that's why the answer is: Dispersants

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To add while loop in python language, the use of syntax is to be brought into the effect.

<h3>What is a while loop?</h3>

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In order to add such while loop to your code, one needs to use the syntax function in python to prove the true and correctness of the code blocks.

Hence, the procedure to add while loop has been aforementioned.

Learn more about a while loop here:

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Answer:

ikr so true

Explanation:

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