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GalinKa [24]
3 years ago
7

En el disco duro de mi PC (que es de 500 Gigabytes de espacio útil total), tengo ocupado un 50% de ese espacio. ¿Cuantos juegos

pesados como el Call of Duty Ghost (ocupa 40960 Megabytes) podre instalar en mi computadora en el espacio restante?. Plantea las operaciones que hiciste para llegar al resultado
Computers and Technology
1 answer:
liubo4ka [24]3 years ago
5 0

Answer:si

Explanation:

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Russian newspaper says U.S. journalism is conducting 'experiments' to introduce fast-growing artificial intelligence technology.
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Answer:

Russian newspaper says U.S. journalism is conducting 'experiments' to introduce fast-growing artificial intelligence technology.

Explanation:

Artificial intelligence (AI) is a wide-ranging tool that enables people to rethink how we integrate information, analyze data, and use the resulting insights to improve decision making—and already it is transforming every walk of life. In this report, Darrell West and John Allen discuss AI's application across a variety of sectors, address issues in its development, and offer recommendations for getting the most out of AI while still protecting important human values.

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Give a recursive version of the algorithm Insertion-Sort (refer to page 18 in the textbook) that works as follows: To sort A[1..
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Answer:

see explaination

Explanation:

void insertion( int e,int *x, int start, int end)

{

if (e >= x[end])

x[end+1] = e;

else if (start < end)

{

x[end+1] = x[end];

insertion(e, x, start, end-1);

}

else

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x[end+1] = x[end];

x[end] = e;

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}

void insertion_recurssion(int *b, int start, int end)

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if(start < end)

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insertion_sort_recur(b, start, end-1);

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3 years ago
________ are typically comprised of a mix of ________ and ________.
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Your answer is "Mutual funds; stocks; bonds".
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3 years ago
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We have said that the average number of comparisons need to find a target value in an n-element list using sequential search is
bija089 [108]

Answer:

Part a: If the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b: If the list contains n elements (where n is even) the middle terms are  at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c: The average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

Explanation:

Suppose the list is such that the starting index is 0.

Part a

If list has 15 elements, the middle item would be given at 7th index i.e.

there are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(8,9,10,11,12,13,14) above it. It will have to run 8 comparisons  to find the middle term.

If list has 17 elements, the middle item would be given at 8th index i.e.

there are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(9,10,11,12,13,14,15,16) above it.It will have to run 9 comparisons  to find the middle term.

If list has 21 elements, the middle item would be given at 10th index i.e.

there are 10 indices (0,1,2,3,4,5,6,7,8,9) below it and 10 indices (11,12,13,14,15,16,17,18,19,20) above it.It will have to run 11 comparisons  to find the middle term.

Now this indicates that if the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b

If list has 16 elements, there are two middle terms as  one at would be at 7th index and the one at 8th index .There are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(9,10,11,12,13,14,15) above it. It will have to run 9 comparisons  to find the middle terms.

If list has 18 elements, there are two middle terms as  one at would be at 8th index and the one at 9th index .There are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(10,11,12,13,14,15,16,17) above it. It will have to run 10 comparisons  to find the middle terms.

If list has 20 elements, there are two middle terms as  one at would be at 9th index and the one at 10th index .There are 9 indices(0,1,2,3,4,5,6,7,8) below it and 9 indices(11,12,13,14,15,16,17,18,19) above it. It will have to run 11 comparisons  to find the middle terms.

Now this indicates that if the list contains n elements (where n is even) the middle terms are  at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c

So the average number of comparisons is given as

((n+1)/2+(n+2)/2)/2=(2n+3)/4

So the average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

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