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3 years ago

Big Data often involves a form of distributed storage and processing using Hadoop and MapReduce.

Computers and Technology

1 answer:

olya-2409 [2.1K]3 years ago

5 0

One reason for this is: the processing power needed for the centralized model would overload a single computer.

<u>Explanation:</u>

Companies would be engrossed in achieving and examining the datasets because they can supplement significant value to the desicion making method. Such processing may include complicated workloads. Furthermore the difficulty is not simply to save and maintain the massive data, but also to investigate and extract a essential utilities from it.

Processing of Bigdata can consists of various operations depending on usage like culling, classification, indexing, highlighting, searching etc. MapReduce is a programming model used to process large dataset workloads. Hadoop is designed to scale up from single servers to thousands of machines, each offering local computation and storage.

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Programs for embedded devices are often written in assembly language. Some embedded processors have limited instructions, like M

oksano4ka [1.4K]

Answer:

#include <iostream>

using namespace std;

int main()

{

char address[2];

int tag, firstBit, secondBit, setNumber;

int cache[4][2]={{1,5}, {2,4}, {3,2}, {6,0}};

cout << "Enter the address as hex(in small letters: "<

cin >> address;

for (int i < 0; i < 8; i++){

if (address[0] == '0'){

tag = 0;

firstBit = 0;

} else if (address[0] == '1'){

tag = 0;

firstBit = 1;

} else if (address[0] == '2'){

tag =1;

firstBit = 0;

} else if (address[0] == '3'){

tag = 1;

firstBit = 1;

} else if (address[0] == '4'){

tag = 2;

firstBit = 0;

} else if (address[0] == '5'){

tag = 2;

firstBit = 1;

} else if (address[0] == '6'){

tag = 3;

firstBit = 0;

} else if (address[0] == '7'){

tag = 3;

firstBit = 1;

} else if (address[0] == '8'){

tag = 4;

firstBit = 0;

} else if (address[0] == '9'){

tag = 4;

firstBit = 1;

} else if (address[0] == 'A'){

tag = 5;

firstBit = 0;

} else if (address[0] == 'B'){

tag = 5;

firstBit = 1;

} else if (address[0] == 'C'){

tag = 6;

firstBit = 0;

} else if (address[0] == 'D'){

tag = 6;

firstBit = 1;

} else if (address[0] == 'E'){

tag = 7;

firstBit = 0;

} else if (address[0] == 'F'){

tag = 7;

firstBit = 1;

} else{

cout<<"The Hex number is not valid"<< endl;

}

if(address[1]>='0' && address[1]<'8'){

secondBit = 0;

} else if(address[1]=='8'|| address[1]=='9'||(address[1]>='a' && address[1]<='f')){

secondBit = 1;

} else{

cout<<"The Hex number is not valid"<< endl;

return 0;

}

setNumber = firstBit * 2 + secondBit;

if(cache[setNumber][0]==tag || cache[setNumber][1]==tag){

cout<<"There is a hit";

} else{

cout<< "There is a miss";

}

return 0;

}

Explanation:

The C++ source code prompts the user for an input for the address variable, then the nested if statement is used to assign the value of the firstBit value given the value in the first index in the address character array. Another if statement is used to assign the value for the secondBit and then the setNumber is calculated.

If the setNumber is equal to the tag bit, Then the hit message is printed but a miss message is printed if not.

4 0

3 years ago

Linux is: Select one: a. primarily concerned with the tasks of end users. b. designed for specific machines and specific micropr

ivann1987 [24]

Answer:

C. an example of open-source software.

Explanation:

open-source software is the type of software in which anyone can access, it can also be shared And modified by anyone simply because ita accessible to the public.

Hence and open source software's source code can be

inspected, enhanced and modified by anyone. A typical example is Linux.

7 0

3 years ago

Helen has no experience in security. She would like to earn a certification that demonstrates that she has the basic knowledge n

rewona [7]

Answer:

Option C. Security+ is the correct answer.

Explanation:

8 0

3 years ago

Can some one fix this <br> input ("Enter a number: ") <br> print (num * 8)

Kaylis [27]

If you save the input as num,
this will print the input 8 times.

num = input("Enter a number: ")
print(num * 8)

If you want to do actual math calculations,
then the input needs to be a number.

num = float(input("Enter a number: "))
print(num * 8)

This doesn't account for any errors in which the user doesn't input a number, but I don't think that's what you were looking for anyway :)

6 0

3 years ago

Find the root using bisection method with initials 1 and 2 for function 0.005(e^(2x))cos(x) in matlab and error 1e-10?

frutty [35]

Answer:

The root is:

$c=1.5708$

Explanation:

Use this script in Matlab:

-------------------------------------------------------------------------------------

function [c, err, yc] = bisect (f, a, b, delta)

% f the function introduce as n anonymous function

% - a y b are the initial and the final value respectively

% - delta is the tolerance or error.

% - c is the root

% - yc = f(c)

% - err is the stimated error for c

ya = feval(f, a);

yb = feval(f, b);

if ya*yb > 0, return, end

max1 = 1 + round((log(b-a) - log(delta)) / log(2));

for k = 1:max1

c = (a + b) / 2;

yc = feval(f, c);

if yc == 0

a = c;

b = c;

elseif yb*yc > 0

b = c;

yb = yc;

else

a = c;

ya = yc;

end

if b-a < delta, break, end

end

c = (a + b) / 2;

err = abs(b - a);

yc = feval(f, c);

-------------------------------------------------------------------------------------

Enter the function in matlab like this:

f= @(x) 0.005*(exp(2*x)*cos(x))

You should get this result:

f =

function_handle with value:

@(x)0.005*(exp(2*x)*cos(x))

Now run the code like this:

[c, err, yc] = bisect (f, 1, 2, 1e-10)

You should get this result:

c =

1.5708

err =

5.8208e-11

yc =

-3.0708e-12

In addition, you can use the plot function to verify your results:

fplot(f,[1,2])

grid on

5 0

3 years ago