7.2

import math
import numpy as np

x = [2.91, 2.94, 6.35, 6.58, 3.8, 6.43, 0.57, 5.96, 3.4, 4.55]
y = [82.16, 61.02, 44.56, 82.52, 99.17, 70.24, 63.23, 66.48, 48.35, 40.24]
n = len(x)
sum_x = 0.0
sum_y = 0.0
sum_y2 = 0.0
sum_x2 = 0.0
sum_x3 = 0.0
sum_x4 = 0.0
sum_xy = 0.0
sum_x2y = 0.0
for i in range(0, len(x)):
    sum_y += y[i]
    sum_x += x[i]
    sum_y2 += y[i] ** 2
    sum_x2 += x[i] ** 2
    sum_x3 += x[i] ** 3
    sum_x4 += x[i] ** 4
    sum_xy += x[i] * y[i]
    sum_x2y += x[i] ** 2 * y[i]
a1 = (sum_x * sum_y - n * sum_xy) / (sum_x ** 2 - n * sum_x2)
a = np.array([
    [sum_x2, sum_x, n],
    [sum_x3, sum_x2, sum_x],
    [sum_x4, sum_x3, sum_x2]
])
b = np.array([sum_y, sum_xy, sum_x2y])
temp_result = np.linalg.solve(a, b)
result = list(temp_result)
print(f"P2(x) = {result[0]}x^2 + ({result[1]})x + ({result[2]})")
g = 0.0
sum_g = 0.0
for i in range(0, n):
    sum_g = (y[i] - (result[0] * x[i] ** 2 + result[1] * x[i] + result[2])) ** 2
e = 0.0

for i in range(0, n):
    e += (y[i] - (result[0] * x[i] ** 2 + result[1] * x[i] + result[2])) ** 2

print("Средреквадратическое отклонение = ", math.sqrt(e / n))
y_f = [.0, .0, .0, .0, .0, .0, .0, .0, .0, .0]
minus_y = 0.0
minus_y2 = 0.0
for i in range(0, n):
    minus_y += (y[i] - (result[0] * x[i] ** 2 + result[1] * x[i] + result[2])) ** 2
    minus_y2 += (y[i] - (sum_y / n)) ** 2
print("Величина достоверности аппроксимации R^2 =", 1 - minus_y / minus_y2)