Why is this happening when I already uploaded files to my drive?

Does watching Beastars make me a furry?

jarptica [38.1K]

Answer:

Technically

Yes

Yes it does

8 0

3 years ago

Big data refers to huge collections of data that are difficult to process, analyze, and manage using conventional data tools. It

Helga [31]

Answer:

3rd;mobile;velocity;NoSQL;out;shards;dynamic;schema-less;value;column;Hadoop;MapReduce.

Explanation:

Big data refers to huge collections of data that are difficult to process, analyze, and manage using conventional data tools. It is a core component of the 3rd platform, which also includes cloud computing, mobile devices, and social networking. The five Vs of big data are high volume, high velocity, diversified variety, unknown veracity, and low-density value. Although SQL and relational databases can be used for big datasets, a collection of alternative tools referred to as NoSQL has become popular. These tools work well when databases scale out (horizontally) and when databases are broken into subsets called shards. Modern database tools also handle dynamic scaling as devices are added when additional capacity is required. NoSQL tools are sometimes said to create schema-less databases, but they usually have some type of structure, though it may be more flexible than the relational model. A key-value data model provides each data element with a key. A column-oriented data model makes it easy to access data stored in similar fields, rather than in individual records. Two very popular NoSQL tools include Hadoop, which is a big data file system, and MapReduce which sends processing logic to the data, rather than bringing the data to the computer that performs the processing.

4 0

3 years ago

Row array gameScores contains all player scores. Construct a row array highScores than contains all player scores greater than m

Sever21 [200]

Answer:

The solution is written using Python as it has a simple syntax.

def getHighScores(gameScores, minScore):
meetsThreshold = []
for score in gameScores:
if(score > minScore):
meetsThreshold.append(score)
return meetsThreshold
gameScores = [2, 5, 7, 6, 1, 9, 1]
minScore = 5
highScores = getHighScores(gameScores, minScore)
print(highScores)

Explanation:

Line 1-8

Create a function and name it as getHighScores which accepts two values, gameScores and minScore. (Line 1)
Create an empty list/array and assign it to variable meetsThreshold. (Line 2)
Create a for loop to iterate through each of the score in the gameScores (Line 4)
Set a condition if the current score is bigger than the minScore, add the score into the meetsThreshold list (Line 5-6)
Return meetsThreshold list as the output

Line 11-12

create a random list of gameScores (Line 11)
Set the minimum score to 5 (Line 12)

Line 13-14

Call the function getHighScores() and pass the gameScores and minScore as the arguments. The codes within the function getHighScores() will run and return the meetsThreshold list and assign it to highScores. (Line 13)
Display highScores using built-in function print().

8 0

3 years ago

32.

ra1l [238]

Answer:

B.

Explanation:

If you are trying to save an existing document you would use the save command and you will be prompted if you want to rename it to what ever you wish

Hope this helps :)

3 0

3 years ago

Write a C++ program to count even and odd numbers in array. The array size is 50. The array elements will be entered by the user

vlabodo [156]

Answer:

The program in C++ is as follows:

#include <iostream>

using namespace std;

int main(){

int numbers[50];

int evekount = 0, odkount = 0;

for(int i = 0; i<50;i++){

cin>>numbers[i];

if(numbers[i]%2==0){ evekount++; }

else{ odkount++; }

}

cout<<"Even Count: "<<evekount<<endl;

cout<<"Odd Count: "<<odkount<<endl;

return 0;

}

Explanation:

This declares the integer array of number

int numbers[50];

This initializes the even count and odd count to 0

int evekount = 0, odkount = 0;

This iterates from 1 to 50

for(int i = 0; i<50;i++){

This gets input for the array

cin>>numbers[i];

This checks for even

if(numbers[i]%2==0){ evekount++; }

This checks for odd

else{ odkount++; }

}

This prints the even count

cout<<"Even Count: "<<evekount<<endl;

This prints the odd count

cout<<"Odd Count: "<<odkount<<endl;

3 0

3 years ago