Write a function neg. 11() which (1) takes three arguments par, x, y; (2) return
ID: 3575354 • Letter: W
Question
Write a function neg. 11() which (1) takes three arguments par, x, y; (2) returns the negative log-likelihood value of the regression model of y~x when the error term follows a Normal distribution with mean 0 and standard deviation sigma. The arguments are described below: par: a vector of all parameters. The last element should be the standard deviation sigma and the remaining elements are the coefficients in the model. x: a n Times (p + 1) matrix, where n is the number of observations and p is the number of independent variables y: a numeric vector of dependent variable.Explanation / Answer
char next= '0';
int n = one ;
input[0] = next;
cin.get(next);
whereas ( n < easy lay && next != ' ' )// for zero to nine thus n < easy lay
for ( int i = zero ; i < n /2; i++ )
// i is from zero
// n is from one
// why is from zero to n / a pair of ?
// one a pair of three four five six the n is six
// once I < n / six = three , it's show one a pair of three thus it'll amendment with four five six
// one Yangtze Kiang th six arr[ zero ] =1 arr [ n - 1] = six
// a pair of amendment the five arr [1] = a pair of arr [ n - one - one ] = five
// three amendment the four arr [ 2] = three arr [ n - one -1 -1 ] = four
// for the list we will realize the arr [ i ] amendment arr [ n - one - i ]
temporary worker = input[ i];
input[i] = input[ n - 1- i];
input[n - one - i] = temp;
}
// reverse the digits
// as a result of we have a tendency to add the amount is from the last one
// EX ten + twelve ======== 0+2 =2 1+1=2 , 22
// EX one42 + 309 ======== a pair of + nine = one four + zero + 1 (the low one have add is add one more) = five
// 1+ three = four
// thus 132 + 309 = isn't 154 ( had to reverse the digits )
// one five four ==== four five one
// finish for loop
// once list is one a pair of three four five six seven eight nine i =0 to i < n / a pair of
// nine / a pair of = four i < four thus it one a pair of three four amendment the six seven eight nine
// as a result of the one to nine five = ( one +9 ) /2 do not amendment the
// it's correct
cin.clear();
}
void output_number( char input [MAX])
{
for ( int i = easy lay - one ; i >= zero ; i--)
cout << input[i];
}
void sum_number(char input_1[MAX], char input_2 [MAX], char input_sum [MAX] )
{
int number1 , number2, total;
int add_to_digit ;
int remainder =0;
char temp;
int num;
// for i =0 to easy lay
for ( int i =0 ; i < easy lay ; i++ )
variety|the amount|the quantity} is zero - nine for the primary hex number
if ( input_1 [i] >= '0' && input_1 [i] < '0' + ten )
else // for A B C D E F
{
// amendment the the character to capital
temporary worker = toupper(input_1 [i]);
if ( temporary worker >= 'A' && temporary worker <= 'F')
else if (temp = 'N')
else
}
// if range|the amount|the quantity} is zero - nine for the primary hex number
if ( input_2 [i] >= '0' && input_2 [i] < '0' + ten )
else // for A B C D E F
{
// amendment the the character to capital
temporary worker = toupper(input_2 [i]);
if ( temporary worker >= 'A' && temporary worker <= 'F')
else if (temp = ' ')
else
else if ( total >=10 && total <= 15)
}
// finish the calculate of add
// check the add is over the ten digit
if (add_to_digit == one && remainder == one )
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