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Hatshy [7]
2 years ago
10

A student who used a regression model that included indicator variables was upset when receiving only the following output on th

e multiple regression printout: X TRANSPOSE X SINGULAR. What is a likely source of the difficulty
Computers and Technology
1 answer:
jeka57 [31]2 years ago
4 0

Answer:

This is a multicolinearity problem and the student should determine the variable(s) that cause(s) the problem and remove it.

Explanation:

This information means that there exists a linear combination between the independent variables. The problem might have developed due to multicolinearity producing almost perfectly linearly dependent columns.

This could also be as a results of single matrix created when the student use an incorrect indicator variables and included an additional indicator column which created linearly dependent columns.

You might be interested in
Write an expression that continues to bid until the user enters 'n'.
alisha [4.7K]

Answer:

1)

Add this while statement to the code:

while(keep_going!='n'):  

#the program keeps asking user to continue bidding until the user enter 'n' to stop.

2)

Here is the code for ACTIVITY 53.3. While loop: Insect growth:

num_insects = int(input())

while num_insects <= 100:

   print(num_insects, end=' ')

   num_insects = num_insects * 2

 

Explanation:

Here is the complete code for 1)

import random

random.seed (5)

keep_going='-'

next_bid = 0

while(keep_going!='n'):

   next_bid = next_bid + random.randint(1, 10)

   print('I\'ll bid $%d!' % (next_bid))

   print('continue bidding?', end='')

   keep_going = input()

Here the the variable keep_going is initialized to a character dash keep_going='-'

The statement keep_going = input() has an input() function which is used to take input from user and this input value entered by user is stored in keep_going variable. while(keep_going!='n'):  is a while loop that keeps iterating until keep_going is not equal to 'n'.    print('continue bidding?', end='')  statement prints the continue bidding? message on the output screen. For example the user enters 'y' when this message appears on screen. So keep_going = 'y' . So the operation in this statement next_bid = next_bid + random.randint(1, 10) is executed in which next_bid is added to some randomly generated integer within the range of 1 to 10 and print('I\'ll bid $%d!' % (next_bid))  prints the result on the screen. Then the user is prompted again to continue biffing. Lets say user enters 'n' this time. So keep_going = 'n'. Now the while loop breaks when user enters 'n' and the program ends.

2)

num_insects = int(input()) #the user is prompted to input an integer

while num_insects <= 100: #the loop keeps iterating until value of num_insects exceeds 100

   print(num_insects, end=' ') #prints the value of num_insects  

   num_insects = num_insects * 2 #the value of num_insects is doubled

For example user enters 8. So

num_insects = 8

Now the while loop checks if this value is less than or equal to 100. This is true because 8 is less than 100. So the body of while loop executes:

   print(num_insects, end=' ') statement prints the value of num_insects

So 8 is printed on the screen.

num_insects = num_insects * 2 statement doubles the value of num_insects So this becomes:

num_insects = 8 * 2

num_insects = 16

Now while loop checks if 16 is less than 100 which is again true so next 16 is printed on the output screen and doubled as:

num_insects = 16 * 2

num_insects = 32

Now while loop checks if 32 is less than 100 which is again true so next 32 is printed on the output screen and doubled as:

num_insects = 32 * 2

num_insects = 64

Now while loop checks if 64 is less than 100 which is again true so next 64 is printed on the output screen and doubled as:

num_insects = 64 * 2

num_insects = 128

Now while loop checks if 128 is less than 100 which is false so the program stops and the output is:

8 16 32 64  

The programs along with their output is attached.

7 0
2 years ago
Write a code segment that uses an enhanced for loop to print all elements of words that end with "ing". As an example, if words
nadezda [96]

Answer:

 public static void main(String[] args) {

   String ing[] = {"ten","fading","post","card","thunder","hinge","trailing","batting"};

   for (String i: ing){

     if (i.endsWith("ing")){

       System.out.println(i);

    }

   }

 }

Explanation:

The for-loop cycles through the entire list and the if-statement makes it so that the string is only printed if it ends with "ing"

6 0
2 years ago
In a fantasy world, your character must face hordes of demons. Each demon is vulnerable to a type of magical spell. This weaknes
Y_Kistochka [10]

I think it's imperfect information.

7 0
2 years ago
Read 2 more answers
#Write a function called is_composite. is_composite should #take as input one integer. It should return True if the #integer is
malfutka [58]

Answer:

// A optimized school method based C++ program to check  

// if a number is composite.  

#include <bits/stdc++.h>  

using namespace std;  

bool isComposite(int n)  

{  

// Corner cases  

if (n <= 1) return false;  

if (n <= 3) return false;  

// This is checked so that we can skip  

// middle five numbers in below loop  

if (n%2 == 0 || n%3 == 0) return true;  

for (int i=5; i*i<=n; i=i+6)  

 if (n%i == 0 || n%(i+2) == 0)  

 return true;  

return false;  

}  

// Driver Program to test above function  

int main()  

{  

isComposite(11)? cout << " true\n": cout << " false\n";  

isComposite(15)? cout << " true\n": cout << " false\n";  

return 0;  

}

Explanation:

3 0
2 years ago
An encryption system works by shifting the binary value for a letter one place to the left. "A" then becomes: 1 1 0 0 0 0 1 0 Th
Vika [28.1K]

Answer:

The hexadecimal equivalent of the encrypted A is C2

Explanation:

Given

Encrypted binary digit of A = 11000010

Required

Hexadecimal equivalent of the encrypted binary digit.

We start by grouping 11000010 in 4 bits

This is as follows;

1100 0010

The we write down the hexadecimal equivalent of each groupings

1100 is equivalent to 12 in hexadecimal

So, 1100 = 12 = C

0010 is represented by 2 in hexadecimal

So, 0010 = 2

Writing this result together; this gives

1100 0010 = C2

Going through the conversion process;

A is first converted to binary digits by shifting a point to the left

A => 11000010

11000010 is then converted to hexadecimal

11000010 = C2

Conclusively, the hexadecimal equivalent of the encrypted A is C2

8 0
2 years ago
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