2.8*189] = [8.0*1e**9, rate = [(e.07, 8.08, e.09, e.10, e.11, e.12] multiplier = 1/100 for i in range(len(rate)): rate[i] = rate[i] 10.9 multiplier
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Type the following python code out. Include indentation if needed
![K = [8.8*1e**9, 10.9*18**9, 12.e*18**9]
rate = [0.07, 0.08, e.09, e.10, e.11, 8.12]
multiplier = 1/100
for i in range(len(rate)):
rate[i] = rate[1]
* multiplier](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1a179059-feed-4fa1-a3a6-37335513ae23%2Fb358d0fa-751f-4027-ab2a-021a94e979f0%2F7tuuih8_processed.gif&w=3840&q=75)
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- The e(x) function returns e^x. True Falsevoid baz(int a, int * p) { pushq %rbp movq %rsp,%rbp subq $16,%rsp int c; c = a-*p; 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 if (c) { int d = c*2; movl -4(%rbp),%eax addl %eax,%eax movl %eax,-8(%rbp) *p -= d; movl 24(%rbp),%rax movl (%rax),%eax subl -8(%rbp),%eax movl %eax,%edx movq 24(%rbp), %rax movl %edx, (%rax) jmp L4: } (continued on next page)#Solve it with C programing Suppose, you are working as a Data Entry Operator in a company. You have created a dataset that contains the names of N employees. Then your supervisor wants you to look for a particular employee by name. (Consider, every name in the data set is unique).Now your task is to write a program which will help you to find out the name which your supervisor is looking for.If found then print a message "Matched" otherwise print "Not Matched". (N.B: Without quotation) The first line of input will take an integer (N) which indicates the number of names in the dataset. Then the second line will take N space-separated strings. The third line of input will take another string as search value. Sample input:5rabbe sharif shazzad polash tutulsharif Sample output:Matched Sample input:4jahid nishat rasel shakibmunna Sample output:Not Matched
- function result = result = ''; 1 tokens_to_str_code(token_mat, time_unit) 3 for i =1:size(token_mat,1) 4 5 if(token_mat (i,1) if(token_mat(i,2) > 4*time_unit) result = strcat(result,"-"); elseif((token_mat(i,2) > time_unit) && (token_mat(i,2) 8*time_unit) result = strcat(result,"/"); elseif((token_mat(i,2) 11 == 0) 12 13 14 4*time_unit) && (token_mat(i,2) < 8*time_unit)) 15 result strcat(result," "); %3D 16 end 17 end 18 end 19 end 20 21 Check Test Expected Got tokens %3D [1 4; е 3; 1 3; е 1; 1 3; е 1; 1 1; ө 1; 1 1 ]; time_unit = .5; disp( tokens_to_str_code( tokens, time_unit )) tokens = [ 0 4; 1 1457; 0 463; 1 1457; 0 463; 1 497; 0 1423; 1 1457; ---/..--.. --.---/..--.. о 463; 1 1457; 0 463; 1 1457; ө 3343; 1 497; ө 463; 1 497; 0 463; 1 1457; 0 1423; 1 1457; 0 463; 1 497; 0 463; 1 497; e 379 ]; time_unit = 240; disp( tokens_to_str_code( tokens, time_unit ))Assume that: int x[5] = {2,4,5,7,8}; What are the values of x after calling: swap(x+2, x+4); swap(x+1, x+3); if the swap function is defined as follows void swap(int *a, int *b) { int t; t=*b; *b=*a; *a=t;} Select one: O a. {2,7,5,4,8} O b. {2,4,5,7,5} O c. {2,4,5,4,5} O d. {2,7,8,4,5}Let A = {a, b, c} and B = {u, v}. Write a. A × B b. B × A
- Solution Floating point representation: It is defined as the representation of floating numbers. It includes sign bit, exponent, and mantissa bits. Based on precision it has 2 types. 1. For IEEE 754 single-precision floating-point numbers, what is the exponent of a denormalized floating-point number in decimal? Solution: In IEEE 754 single-precision, exponent bits are 8. Therefore exponent = 2 ^(n-1) -1 = 2^(8-1) -1 = 127 OPTION D 2. For IEEE 754 single-precision floating-point numbers, how many bits for mantissa? Solution: In IEEE 754 single-precision, there are 23 bits for mantissa. sign = 1 bit exponent = 8 bits mantissa = 23 bits 3. For IEEE 754 single-precision floating-point numbers, which of the following is an example of NAN? Solution: In IEEE 754 single-precision, NAN is a special value where all exponents bits are 1's and the mantissa is non zero. a. 1 111 1 111 0000 0000 1101 0000 0000 0000 Exponent is not all 1's. Not a NAN b. 0 111 1 111 1000 0000…To perform pivoting The following function is supposed to accept an augmented matrix, ab, and a pivot row index, i and returns a matrix with all elements below the pivot element eliminated. There are 3 errors in this code... find all the errors and circle them, then write the corrected version of each line below the code. def forward_elim(ab_orig,i): n = len(ab_orig) ab = copy.deepcopy(ab_orig) | for k in range(i,n): for j in range(i,n): ab[k.j] = ab_orig[k,i]*ab_orig[i.j] return ab return ab should be return ab_orig for k in range(i,n) should be for k in range(i+1,n) O for j in range(i,n) n should be n+1 ab[kj] = ab_orig[k,i]*ab_orig[i,j] should be ab[kj] = ab_orig[j.i]*ab_orig[i,k] V There should be a line just above the return ab line that reads:def apply_gaussian_noise(X, sigma=0.1): """ adds noise from standard normal distribution with standard deviation sigma :param X: image tensor of shape [batch, 3, height, width] Returns X + noise. """ ### YOUR CODE HERE ### # noise tests theoretical_std = (X_train[:100].std() ** 2 + 0.5 ** 2) ** .5 our_std = apply_gaussian_noise(X_train[:100], sigma=0.5).std() assert abs(theoretical_std - our_std) < 0.01, \ "Standard deviation does not match it's required value. Make sure you use sigma as std." assert abs(apply_gaussian_noise(X_train[:100], sigma=0.5).mean() - X_train[:100].mean()) < 0.01, \ "Mean has changed. Please add zero-mean noise"
- A = {1,2,3} B = {1,2,3} C = {5,6,7} D = {0,1,2,3,4,5} Is A = C?Given the code:void c(int n) { if(n<2) { cout << n << " "; return; } c(n/2); cout << n << " ";}1. Trace the function when n is 12.497 Bytes Write a recursive function that returns the sum of the digits of a given integer.Input format :Integer NOutput format :Sum of digits of NConstraints :0 <= N <= 10^9Sample Input 1 :12345Sample Output 1 :15Sample Input 2 :9Sample Output 2 :9 Solution:///////////// public class solution { public static int sumOfDigits(int input){ int sum; if(input<10){ return input; } sum = (input % 10) + sumOfDigits(input / 10); return sum; }}. .