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3 years ago

8

Annie needs to provide a form field that will allow users to select from a predefined list of options but without the ability to

add their own options.
Which option should she use?

Computers and Technology

1 answer:

Lana71 [14]3 years ago

8 0

Answer:

Drop down list

Explanation:

You have not listed what the possible answers are, however a drop down list would be an option to accomplish this.

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All of the following are examples of roadblocks to achieving a goal except

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Having to many goals

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3 years ago

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The application reliability team at your company this added a debug feature to their backend service to send all server events t

Paraphin [41]

Answer:

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

Explanation:

The given question is incomplete and the complete question will be:

The event records are at least 50 KB and at most 15 MB and are expected to peak at 3,000 events per second. You want to minimize data loss. Which process should you implement?

Append metadata to file the body.

Compress individual files.

Name files with a random prefix pattern.

Save files to one bucket

(B)

Batch every 10,000 events with a single manifest file for metadata.

Compress event files and manifest files into a single archive file.

Name files using serverName-EventSequence.

Create a new bucket if a bucket is older than 1 day and save the single archive file to the new bucket. Otherwise, save the single archive file to the existing bucket.

(C)

Compress individual files.

Name files with serverName-EventSequence.

Save files to one bucket

Set custom metadata headers for each object after saving.

(D)

Append metadata to file the body.

Compress individual files.

Name files with serverName-Timestamp.

Create a new bucket if the bucket is older than 1 hour and save individual files to the new bucket. Otherwise, save files to the existing bucket.

Therefore, in option A, a name file can be used with a prefix pattern randomly. So, it will be the right answer

The other three options such as C, B and D are incorrect because they are using name files with "serverName-EventSequence" or "serverName-Timestamp" which causes the distribution in the backend irregularly.

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4 years ago

A technician has connected a laptop to a television via HDMI cable. The entire display cannot be seen on the television. Which o

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Answer: B) Resolution

Explanation:

Resolution is the method that monitor the size and sharpness the image according to the display of the television.

By using resolution the number of individual pixels express according to the terms of horizontal and vertical axis.

The high resolution only depends upon the computer display and the physical ability in which the light focus in particular direction. There are basically two types of resolution that are :

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3 years ago

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Code example 5-1 import java.util.Scanner; import java.text.NumberFormat; public class WeightConverter { public static void main

Trava [24]

Answer:

It displays an error message from the getDouble() method

Explanation:

The above would be the result because of the following

1. Variable d is declared as a double variable and should be used as such.

2. The getDouble() method is also defined to handle double variables only.

When the program tries to accept at

weightInPounds = getDouble(sc, prompt);

The getDouble method is called immediately and since the value

"two hundred" entered in string, it can't handle this data type and it (the getDouble method) will display an error message

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4 years ago

Data Structure in C++

agasfer [191]

The code .cpp is available bellow

#include<iostream>

using namespace std;

//declaring variables

void merge(int* ip, int sz, int* opt, bool opt_asc); //merging

int* mergesort(int* ip, int sz);

void mergesort(int *ip, int sz, int* opt, bool opt_asc);

void merge(int* ip, int sz, int* opt, bool opt_asc)

{

int s1 = 0;

int mid_sz = sz / 2;

int s2 = mid_sz;

int e2 = sz;

int s3 = 0;

int end3 = sz;

int i, j;

if (opt_asc==true)

{

i = s1;

j = e2 - 1;

while (i < mid_sz && j >= s2)

{

if (*(ip + i) > *(ip + j))

{

*(opt + s3) = *(ip + j);

s3++;

j--;

}

else if (*(ip + i) <= *(ip + j))

{

*(opt + s3) = *(ip + i);

s3++;

i++;

}

}

if (i != mid_sz)

{

while (i < mid_sz)

{

*(opt + s3) = *(ip + i);

s3++;

i++;

}

}

if (j >= s2)

{

while (j >= s2)

{

*(opt + s3) = *(ip + j);

s3++;

j--;

}

}

}

else

{

i = mid_sz - 1;

j = s2;

while (i >= s1 && j <e2)

{

if (*(ip + i) > *(ip + j))

{

*(opt + s3) = *(ip + i);

s3++;

i--;

}

else if (*(ip + i) <= *(ip + j))

{

*(opt + s3) = *(ip + j);

s3++;

j++;

}

}

if (i >= s1)

{

while (i >= s1)

{

*(opt + s3) = *(ip + i);

s3++;

i--;

}

}

if (j != e2)

{

while (j < e2)

{

*(opt + s3) = *(ip + j);

s3++;

j++;

}

}

}

for (i = 0; i < sz; i++)

*(ip + i) = *(opt + i);

}

int* mergesort(int* ip, int sz)

{

int* opt = new int[sz];

mergesort(ip, sz, opt, true);

return opt;

}

void mergesort(int *ip, int sz, int* opt, bool opt_asc)

{

if (sz > 1)

{

int q = sz / 2;

mergesort(ip, sz / 2, opt, true);

mergesort(ip + sz / 2, sz - sz / 2, opt + sz / 2, false);

merge(ip, sz, opt, opt_asc);

}

}

int main()

{

int arr1[12] = { 5, 6, 9, 8,25,36, 3, 2, 5, 16, 87, 12 };

int arr2[14] = { 2, 3, 4, 5, 1, 20,15,30, 2, 3, 4, 6, 9,12 };

int arr3[10] = { 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 };

int *opt;

cout << "Arays after sorting:\n";

cout << "Array 1 : ";

opt = mergesort(arr1, 12);

for (int i = 0; i < 12; i++)

cout << opt[i] << " ";

cout << endl;

cout << "Array 2 : ";

opt = mergesort(arr2, 14);

for (int i = 0; i < 14; i++)

cout << opt[i] << " ";

cout << endl;

cout << "Array 3 : ";

opt = mergesort(arr3, 10);

for (int i = 0; i < 10; i++)

cout << opt[i] << " ";

cout << endl;

return 0;

}

4 0

4 years ago