void baz(int a, int * p) { int c; c = a-*p; if (c) { int d = c*2; *p -= d; (continued on next page) pushq %rbp movq %rsp,%rbp subq $16,%rsp movl %ecx,16(%rbp) movq %rdx, 24(%rbp) movq 24(%rbp), %rax movl (%rax),%eax movl 16(%rbp),%edx subl %eax,%edx movl %edx,%eax movl %eax,-4(%rbp) cmpl $0, -4(%rbp) je L2 movl 4(%rbp),%eax addl %eax,%eax movl %eax,-8(%rbp) movl 24(%rbp),%rax movl (%rax),%eax -8(%rbp),%eax %eax,%edx movq 24(%rbp), %rax movl %edx, (%rax) jmp L4: subl movl
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- X= (ACB)((A+C)+(BC)) Lalo dolo! 1 walhe Jlsl Laa X = AB · (A + C) + ĀB · A + B + C alo aolo! 1disp(" f(x) = 4x4 - 32x3 + 89x2 - 103x + 42 = 0 "); disp(" find the root between 0 and 1.2 "); disp(" ---------------------------------------- "); a=0; b=1.2; for i=1:10 c=(a+b)/2; fa = 4*a^4 - 32*a^3 + 89*a^2 - 103*a + 42; fc = 4*c^4 - 32*c^3 + 89*c^2 - 103*c + 42; if fa*fc < 0 a = a; b = c; else a = c; b = b; end; if fa*fc == 0 fprintf(" root is found \n"); end; fprintf(" root is %2.4f \n", c); end; (can you do it with c++)A = {1,2,3} B = {1,2,3} C = {5,6,7} D = {0,1,2,3,4,5} Is A = C?
- Select all statements that are equivalent to (V1) và g Ⓒpv (q→r) Op-(94r) ✓(15)^(947)import java.lang.System; 3. public class FibonacciComparison { 4. // Fibonacci Sequence: 0, 1, 1, 2, 3, 5, 8 .... /* 7 input cases 8. 1) 0 9 2) 3 10 3) -1 11 4) 9 12 output cases 13 1) 0 14 2) 2 15 3) 0 16 4) 34 17 */ // Note that you need to return 0 if the input is negative. // Please pay close attention to the fact that the first index in our fib sequence is 0. 18 19 20 // Recursive Fibonacci public static int fib(int n) { // Code this func. 21 22 23 24 return -1; 25 26 // Iterative Fibonacci 27 28 public static int fiblinear(int n) { // Code this func. 29 30 return -1; 31 32 33 public static void main(String[] args) { 34 // list of fibonacci sequence numbers int[] nlist w { 5,10, 15, 20, 25, 30, 35, 40, 45}; 35 36 37 // Two arrays (one for fibLinear, other for fibRecursive) to store time for each run. // There are a total of nlist.length inputs that we will test double[] timingsEF = new double[nlist.Length]; double[] timingsLF = new double[nlist.length]; 38 39 40 41 42 // Every…Q6/Full the following blanks (1101 1010)BCD(5211) = ( )BCD(3321)=( )EX-3=( )16 (1101 1010)BCD(5211) = (1110 0111)BCD(3321)=(1011001)EX-3= (56)16 O None of them (1101 1010)BCD(5211) = (1101 O 0111)BCD(3321)=(1011001)EX-3= (2B)16 (1101 1010)BCD(5211) = (1011 O 1110) BCD (3321)=(1000111)EX-3= (44)16 (1101 1010)BCD(5211) = (1110 O 1100) BCD (3321)=(1010110)EX-3= (A6)16
- Find the bound variables in the example below (Ay. y (Ax. yxxzy) x) abNORMALIZATION PROBLEM: Normalization is one of the most basic preprocessing techniques in data analytics. This involves centering and scaling process. Centering means subtracting the data from the mean and scaling means dividing with its standard deviation. Mathematically, normalization can be expressed as: X-X 0 In Python, element-wise mean and element-wise standard deviation can be obtained by using .mean() and .std() calls. In this problem, create a random 5 x 5 ndarray and store it to variable X. Normalize X. Save your normalized ndarray as X_normalized.npyCode in C Code in the file IO: /************************************************************* This program prints a degree-to-radian table using a for- loop structure. The results are printed to a file and the the screen. *************************************************************/ #include <stdio.h> #define PI 3.141593 #define FILENAME "tableD2R.dat" int main(void) { /* Declare variables. */ double radians; FILE *fileout; /* Open file. */ fileout = fopen(FILENAME,"w"); if (fileout == NULL) printf("Error opening input file. \n"); else { /* Print radians and degrees in a loop. */ printf("Degrees to Radians \n"); for (int degrees=0; degrees<=360; degrees+=10) { radians = degrees*PI/180; printf("%6i %9.6f \n",degrees,radians); fprintf(fileout,"%6i %9.6f \n",degrees,radians); } /* Exit program. */ }
- Suppose a8 = 8, a9 = 9, a10 = 10part (d) (e) (f)A good string is a string or a substring that starts with the character "g" and ends with the character "d". For a given string, write a program to find the number of "good substrings" of a given string. Note - A substring of a string is a contiguous subsequence of that string. The list of all non-empty substrings of the string "apple" would be "apple", "appl", "pple", "app", "ppl", "ple", "ap", "pp", "pl", "le", "a", "p", "p", "I", and "e" There will be multiple test cases running so the Input and Output should match exactly as provided. The base output variable result is set to a default value of -404 which can be modified. Additionally, you can add or remove these output variables. Input Format The first line contains the string S. Sample Input godk - denotes S Constraints 1 <= len|S| <= 100 Output Format The output contains a single integer denoting the number of good substrings of the given string. Sample Output 1 Explanation In the given string, only one substring "god" is good.…ftesting<-aov(DIABETES~BMI,data=data2017) #problem Error in ctrfn(levels(x), contrasts = contrasts) : orthogonal polynomials cannot be represented accurately enough for 266 degrees of freedom what is the meaning of the error? The BMI and DIABETES column number is 819. Is it because of the column number?