answer:The value (x = g^{-1}(3/1240)) is the solution to (g(x) = 3/1240). This implies [ frac{x^7 - 1}{5} = frac{3}{1240}. ] Multiplying by 5 gives: [ x^7 - 1 = frac{15}{1240}. ] Adding 1 to both sides results in: [ x^7 = frac{15}{1240} + frac{1240}{1240} = frac{1255}{1240}. ] Calculating the 7th root, we find [ x = sqrt[7]{frac{1255}{1240}}. ] The calculation is as follows: [ x = left(frac{1255}{1240}right)^{frac{1}{7}}. ] Thus, the solution to ( g^{-1}(3/1240) ) is [ x = boxed{left(frac{1255}{1240}right)^{frac{1}{7}}}. ]

answer:Manuel wants to save 17 each week for the next 19 weeks. To find the total amount he wants to save over the 19 weeks, we multiply the weekly savings amount by the number of weeks: 17/week * 19 weeks = 323 So, Manuel wants to save a total of 323 over the 19 weeks. If we include his initial deposit of 177, the total amount in his savings account after 19 weeks would be: Initial deposit + amount saved over 19 weeks = 177 + 323 = 500 Therefore, the total amount Manuel wants to have in his savings account after 19 weeks is boxed{500} .

answer:Since g(x) = begin{cases} 1, & x in mathbb{Q} 0, & x notin mathbb{Q} end{cases} and π is an irrational number, we have g(π) = 0. Now, substituting g(π) = 0 into function f, we get: f(g(π)) = f(0) = [0 + frac{3}{2}] = [frac{3}{2}]. Since the greatest integer less than or equal to frac{3}{2} is 1, [frac{3}{2}] = 1. Therefore, f(g(π)) = 1. So the answer is boxed{A}.

answer:Let's denote the mean of the original data as overline{x} = frac{1}{5}(x_1 + x_2 + x_3 + x_4 + x_5). The mean of the transformed data 2x_1+1, 2x_2+1, 2x_3+1, 2x_4+1, 2x_5+1 is: overline{x'} = frac{1}{5}[(2x_1+1) + (2x_2+1) + (2x_3+1) + (2x_4+1) + (2x_5+1)] Simplifying the expression, we get: overline{x'} = 2 times frac{1}{5}(x_1 + x_2 + x_3 + x_4 + x_5) + 1 = 2overline{x} + 1 Now, let's compute the variance of the transformed data. Recall that the variance of the original data is given as: S^2 = frac{1}{5}[(x_1 - overline{x})^2 + (x_2 - overline{x})^2 + (x_3 - overline{x})^2 + (x_4 - overline{x})^2 + (x_5 - overline{x})^2] = 3 The variance of the transformed data 2x_1+1, 2x_2+1, 2x_3+1, 2x_4+1, 2x_5+1 is: S'^2 = frac{1}{5}[(2x_1+1-2overline{x}-1)^2 + (2x_2+1-2overline{x}-1)^2 + (2x_3+1-2overline{x}-1)^2 + (2x_4+1-2overline{x}-1)^2 + (2x_5+1-2overline{x}-1)^2] Simplifying the expression, we get: S'^2 = frac{1}{5}[(x_1 - overline{x})^2 + (x_2 - overline{x})^2 + (x_3 - overline{x})^2 + (x_4 - overline{x})^2 + (x_5 - overline{x})^2] times 4 = 3 times 4 = 12 Therefore, the variance of the transformed data is boxed{12}.