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2 years ago

Which of the following is true about scalability?

Computers and Technology

1 answer:

blsea [12.9K]2 years ago

4 0

Answer:

Option (i) is the correct answer to the following question.

Explanation:

Because to achieve the strong scaling then, we have required the Amdahl’s law and the scalability is the weak scaling that is easier to instate than the powerful scaling scalability because of only Amdahl’s law.

So, that's why the following option is correct.

Option (ii) is incorrect because usually, there no communication among the procedures.

Option (iii) is incorrect because the following option is not related to the following statement.

Option (iv) is incorrect because the dependency of the data or information is decreased.

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Is a collection of information stored under a single nam

ad-work [718]

Answer:

plz mark it as brainliest

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2 years ago

Read 2 more answers

Given the strings s1 and s2 that are of the same length, create a new string consisting of the first character of s1 followed by

LiRa [457]

Answer:

s1='abc'

s2='wxyz'

length=len(s1)

s3=''

for i in range(length):

s3+=s1[i]+s2[i]

print(s3)

3 0

3 years ago

To meet the requirement for the number of vdss on board, what must be true about pyrotechnic vdss?

asambeis [7]

Answer: They must be dated.

Explanation: If pyrotechnic VDSs are used, they must be dated. Expired VDSs may be carried on board, but a minimum of three unexpired VDSs must be carried in the vessel.

5 0

2 years ago

Suppose you have two arrays of ints, arr1 and arr2, each containing ints that are sorted in ascending order. Write a static meth

telo118 [61]

Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.

public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}

So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.

A quick explanation:

We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.

The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.

4 0

3 years ago

Electronically, or on paper draw a Binary Search Tree using the values below. Note: Insert each value into the tree in order fro

maria [59]

Answer:

A.) in order traversal - 9 32 35 14 27 25 2 7 29

B.) pre-order traversal - 25 14 32 9 35 27 7 2 29

C.) post-order traversal - 9 35 32 27 14 2 29 25

Explanation:

Given - 25, 14, 7, 32, 27, 2, 29,9,35

To find - On a separate electronic page or sheet of paper:

1. List the values using in order traversal

2. List the values using pre-order traversal

3. List the values using post-order traversal

Proof -

Given values are -

25, 14, 7, 32, 27, 2, 29,9,35

1.)

In order is -

9 32 35 14 27 25 2 7 29

2.)

Pre-order is - (Root left right)

List nodes of first time visit

25 14 32 9 35 27 7 2 29

3.)

Post-order is - (Left Right Root)

List nodes of third time visit

9 35 32 27 14 2 29 25

5 0

2 years ago