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2 years ago

In the NumPy function, the data preparation technique that is used to help machine learning algorithms is called _________.

1 answer:

3 0

In the NumPy function, the data preparation technique that is used to help machine learning algorithms is called the reshape technique or function

For better understanding, let us explain what the reshape function means

The numpy package helps to give the right tools for scientific and mathematical computations in python . it includes functions that cam be used to perform common linear algebra operations, fast Fourier transforms, and statistics The reshape function simply alter or change the row and column arrangement of data in numpy function and it is said to just give new shape to an array without the altering of its data.

from the above, we can therefore say that the answer In the NumPy function, the data preparation technique that is used to help machine learning algorithms is called the reshape technique or function is correct

learn more about reshape function from:

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3 years ago

Read 2 more answers

Write a recursive function that takes a non-negative integer as an argument and displays the same number in reverse order (i.e.

avanturin [10]

Answer:

Following are the program in C++ language

#include<iostream> // header file

using namespace std; // namespace std

int reverse(int n1); // function prototype

int main() // main function()

{

int num; // Variable declaration

cout << "Enter the number" << endl;

cin >> num; // Read the number bu the user

cout<<"After Reverse the digit:\n";

int res= reverse(num); // calling function reverse

cout<<res; // display

}

int reverse(int n1) // function definition of reverse

{

if (n1<10) // checking the base condition

{

return n1;

}

else

{

cout << n1 % 10; // Printed the last digit

return reverse(n1/10); // calling itsself

}

Output:

Enter the number:

76538

After Reverse the digit:

83567

Explanation:

Following are the description of the program

In the main function read the number by user in the "num" variable of int type.

Calling the reverse and pass that "num" variable into it.

After calling the control moves to the body of the reverse function.In this function we check the two condition

1 base condition

if (n1<10) // checking the base condition

{

return n1;

}

2 General condition

else

{

cout << n1 % 10; // Printed the last digit

return reverse(n1/10); // calling itsself

}

Finally return the reverse number and display them into the main function.

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4 years ago

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3 years ago

Which of these improved the ability to send information over the network in the 1970s?

frez [133]

Answer:

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Explanation:

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3 years ago

Given a floating-point formal with a k-bit exponent and an n-bit (fraction, write formulas for the exponent E, significant M, th

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Answer:

A) Describe the number 7.0 bit

The exponential value ( E ) = 2

while the significand value ( M ) = 1.112 ≈ 7/4

fractional value ( F ) = 0.112

And, numeric value of the quantity ( V ) = 7

The exponent bits will be represented as : 100----01.

while The fraction bits will be represented as : 1100---0.

B) The largest odd integer that can be represented exactly

The integer will have its exponential value ( E ) = n

hence the significand value ( M )

= 1.11------12 = 2 - 2-n

also the fractional value ( F ) =

0.11------12 = 1 – 2-n

Also, Value, V = 2n+1 – 1

The exponent bits will be represented as follows: n + 2k-1 – 1.

while The bit representation for the fraction will be as follows: 11---11.

C) The reciprocal of the smallest positive normalized value

The numerical value of the equity ( V ) = 22k-1-2

The exponential value ( E ) = 2k-1 – 2

While the significand value ( M ) = 1

also the fractional value ( F ) = 0

Hence The bit representation of the exponent will be represented as : 11---------101.

while The bit representation of the fraction will be represented as : 00-----00.

Explanation:

E = integer value of exponent

M = significand value

F = fractional value

V = numeric value of quantity

A) Describe the number 7.0 bit

The exponential value ( E ) = 2

while the significand value ( M ) = 1.112 ≈ 7/4

fractional value ( F ) = 0.112

And, numeric value of the quantity ( V ) = 7

The exponent bits will be represented as : 100----01.

while The fraction bits will be represented as : 1100---0.

B) The largest odd integer that can be represented exactly

The integer will have its exponential value ( E ) = n

hence the significand value ( M )

= 1.11------12 = 2 - 2-n

also the fractional value ( F ) =

0.11------12 = 1 – 2-n

Also, Value, V = 2n+1 – 1

The exponent bits will be represented as follows: n + 2k-1 – 1.

while The bit representation for the fraction will be as follows: 11---11.

C) The reciprocal of the smallest positive normalized value

The numerical value of the equity ( V ) = 22k-1-2

The exponential value ( E ) = 2k-1 – 2

While the significand value ( M ) = 1

also the fractional value ( F ) = 0

Hence The bit representation of the exponent will be represented as : 11---------101.

while The bit representation of the fraction will be represented as : 00-----00.

8 0

3 years ago