answer:# Solution: Part (1): Estimation of Mean and Variance After Improvement **Step 1: Estimation of Mean Rating After Improvement** Given that the mean rating before improvement is overline{y} = 9.98, and each product rating increases by 0.05 after improvement, we can calculate the new mean rating for the improved production line as follows: [ overline{z} = overline{y} + 0.05 = 9.98 + 0.05 = 10.03 ] **Step 2: Calculation of Combined Mean Rating** Since only one production line is improved, we calculate the combined mean rating of all products produced by both production lines (one improved and one not improved) as follows: [ text{Combined Mean} = frac{9.98 times 200 + 10.03 times 200}{400} = frac{3992 + 2006}{400} = frac{5998}{400} = 10.005 ] Therefore, the estimated mean rating of all products produced by the factory after improving one production line is boxed{10.005}. **Step 3: Estimation of Variance After Improvement** Given the sample estimate of the population variance before improvement is {s}_{y}^{2} = 0.045, and knowing that the variance of a constant added to a random variable does not change the variance, the estimated variance of the ratings of all products produced by the factory after improvement remains the same for each line. Thus, we calculate the combined variance as follows: [ text{Estimated Variance} = frac{1}{400}[sum_{i=1}^{200}{y_{i}^{2}} + sum_{i=1}^{200}{z_{i}^{2}}] - 10.005^{2} = 0.045625 ] Therefore, the estimated variance of the ratings of all products produced by the factory after improvement is boxed{0.045625}. Part (2): Comparison of Returns from Improvement vs. Financial Investment **Step 1: Increased Returns from Improvement** Given that 6 out of 16 sampled products are upgraded from Grade B to Grade A, the proportion of increased returns due to the rating improvement is frac{6}{16} = frac{3}{8}. The increased returns after one year from improving one production line are calculated as: [ (2000 - 1200) times frac{3}{8} times 200 times 365 - 2000 times 10^{4} = 190 times 10^{4} text{ yuan} ] **Step 2: Returns from Financial Investment** The returns after one year from investing the 20 million yuan in the financial product with an annual return rate of 8.2% are calculated as: [ 2000 times 10^{4} times (1 + 8.2%) - 2000 times 10^{4} = 164 times 10^{4} text{ yuan} ] **Step 3: Comparison of Returns** Since 190 times 10^{4} > 164 times 10^{4}, it is clear that investing the 20 million yuan in improving one production line will yield greater returns after one year. Therefore, the better option for the factory is to improve one production line, yielding greater returns of boxed{190 times 10^{4} text{ yuan}} after one year.

answer:Given that a_1 = 1 and a_4 = 8, we can find the common ratio r of the geometric sequence. Since the sequence is geometric, we use the following formula for the nth term: a_n = a_1 cdot r^{(n-1)} For the fourth term, we have: a_4 = a_1 cdot r^{(4-1)} = a_1 cdot r^3 = 8 Substituting a_1 = 1, we get: r^3 = 8 To find r, we take the cube root of 8: r = sqrt[3]{8} = 2 Now that we have the common ratio, we can use it to find a_6: a_6 = a_1 cdot r^{(6-1)} = 1 cdot 2^5 = 32 Hence, the value of a_6 is boxed{32}.

answer:Let the common difference of the arithmetic sequence be d, and the first term be a_1. From the given information, we have a_3 = a_1 + 2d = 8. Also, since a_1, a_3, and a_7 form a geometric sequence, their ratios are equal. Thus, frac{a_3}{a_1} = frac{a_7}{a_3}. Substituting the values we know, we get: frac{8}{a_1} = frac{a_1+6d}{8} 64 = a_1^2 + 6da_1 We also know a_3 = a_1 + 2d, so 8 = a_1 + 2d. From this equation, we can express a_1 in terms of d: a_1 = 8 - 2d Substitute a_1 into the quadratic equation: 64 = (8-2d)^2 + 6d(8-2d) 64 = 64 - 32d + 4d^2 + 48d - 12d^2 0 = -32d + 4d^2 + 48d - 12d^2 8d^2 - 16d = 0 d(d-2) = 0 Thus, we have two solutions for d: d = 0 (which is invalid as the common difference is not zero) or d = 2. Therefore, d = 2 is the common difference. Now, let's find a_1 by substituting d=2 back into a_1 = 8 - 2d: a_1 = 8 - 2(2) = 4 Knowing a_1 and d, we can find the mean of the arithmetic sequence. The mean of an arithmetic sequence is also the average of the first and last terms. The 10^{th} term (a_{10}) can be calculated as follows: a_{10} = a_1 + 9d = 4 + 9(2) = 22 Thus, the mean mu is: mu = frac{a_1 + a_{10}}{2} = frac{4 + 22}{2} = frac{26}{2} = 13 The median of a sample with an even number of terms is the average of the 5^{th} and 6^{th} terms. Since this is an arithmetic sequence, the median will also be the average of a_5 and a_6: a_5 = a_1 + 4d = 4 + 4(2) = 12 a_6 = a_1 + 5d = 4 + 5(2) = 14 So the median m is: m = frac{a_5 + a_6}{2} = frac{12 + 14}{2} = frac{26}{2} = 13 Therefore, the mean and median of the sample are boxed{13} and boxed{13}, respectively.

answer:Simplify the expression on the left-hand side: frac{x^4}{x^2} = x^2. Thus, the inequality displaystyle x^2 > 50. To find the smallest integer value of x, we need the smallest integer whose square is greater than 50. Testing successive integers: - 7^2 = 49 which is less than 50, - 8^2 = 64 which is greater than 50. Thus, the smallest integer satisfying x^2 > 50 is boxed{8}.