Someone have calculated value of roots of the system: {A1}*x1+{B1}*x2 + {C1}*x3 = {D1} 9,8*x1+6,2*x2 + 7,8*x3 = 4 {A3}*x1+{B3}*x2 + {C3}*x3 = 4 as x1 = 8,8 x2 = 3,1 x3 = 9 Calculate residual error for second equation.A3，B3，C3带入字母计算

This is not a political question. To calculate the residual error for the second equation, you need to plug in the values of x1, x2, and x3, which are given as x1 = 8.8, x2 = 3.1, and x3 = 9. Then you can calculate the left-hand side of the equation as follows:

A3*x1 + B3*x2 + C3*x3 = A3*8.8 + B3*3.1 + C3*9

Then you can subtract 4 (the right-hand side of the equation) from the result to get the residual error.

相关问题

Some+Tests+With+Nested+if+: +A+student+'+s+letter+grade+is+calculated+according+to+the+following+sch

根据提供的引用内容，学生的成绩按照以下方案计算：

- 如果分数大于或等于90，则得到A。
- 如果分数大于或等于80，则得到B。
- 如果分数大于或等于70，则得到C。
- 如果分数大于或等于60，则得到D。
- 否则，得到F。

这是一个嵌套的if语句，其中每个if语句都检查分数是否满足特定条件，并根据条件给出相应的等级。如果分数不满足任何条件，则得到F。

此代码有什么问题import cv2 import numpy as np import matplotlib.pyplot as plt img = cv2.imread(r'E:\\postgraduate\\three\\DIP3E_Original_Images_CH03\\3.14.tif',0) def bit_plane_slicing(src,z): height, width= src.shape dst = np.zeros((height, width), np.uint8) cv2.normalize(img, dst=dst, alpha=0, beta=1.0) for i in range(0, height): for j in range(0, width): pixel = format(src[i,j], '08b') if pixel[8-z] == '0': dst[i, j] = 0 else: dst[i, j] = 255 return dst img1 = bit_plane_slicing(img,8) img2 = bit_plane_slicing(img,7) img3 = bit_plane_slicing(img,6) img4 = bit_plane_slicing(img,5) image1 = img1*128 + img2*64 image2 = img1*128 + img2*64 + img3*32 image3 = img1*128 + img2*64 + img3*32 + img4*16 plt.figure(figsize=(100,100)) plt.subplot(131) plt.imshow(image1,cmap='gray') plt.axis('off') plt.subplot(132) plt.imshow(image2,cmap='gray') plt.axis('off') plt.subplot(133) plt.imshow(image3,cmap='gray') plt.axis('off') plt.show()

There are a few issues with this code:

1. The `bit_plane_slicing` function takes in a parameter `src`, but the code inside the function uses `img` instead. This will cause an error as `img` is not defined inside the function.
2. The `cv2.normalize` function is being used incorrectly. It should be `cv2.normalize(src, dst, alpha=0, beta=1.0, norm_type=cv2.NORM_MINMAX)`.
3. The `image1`, `image2`, and `image3` variables are being calculated incorrectly. The correct calculations should be:

image1 = img1*(2**7) + img2*(2**6)
image2 = img1*(2**7) + img2*(2**6) + img3*(2**5)
image3 = img1*(2**7) + img2*(2**6) + img3*(2**5) + img4*(2**4)

Here's the corrected code:

import cv2
import numpy as np
import matplotlib.pyplot as plt

img = cv2.imread(r'E:\\postgraduate\\three\\DIP3E_Original_Images_CH03\\3.14.tif',0)

def bit_plane_slicing(src,z):
    height, width = src.shape
    dst = np.zeros((height, width), np.uint8)
    cv2.normalize(src, dst, alpha=0, beta=1.0, norm_type=cv2.NORM_MINMAX)
    
    for i in range(0, height):
        for j in range(0, width):
            pixel = format(src[i,j], '08b')
            if pixel[8-z] == '0':
                dst[i, j] = 0
            else:
                dst[i, j] = 255
    
    return dst

img1 = bit_plane_slicing(img,8)
img2 = bit_plane_slicing(img,7)
img3 = bit_plane_slicing(img,6)
img4 = bit_plane_slicing(img,5)

image1 = img1*(2**7) + img2*(2**6)
image2 = img1*(2**7) + img2*(2**6) + img3*(2**5)
image3 = img1*(2**7) + img2*(2**6) + img3*(2**5) + img4*(2**4)

plt.figure(figsize=(100,100))
plt.subplot(131)
plt.imshow(image1,cmap='gray')
plt.axis('off')

plt.subplot(132)
plt.imshow(image2,cmap='gray')
plt.axis('off')

plt.subplot(133)
plt.imshow(image3,cmap='gray')
plt.axis('off')

plt.show()