answer:Let (y = sqrt{x - 4}). Then (y^2 = x - 4), so (x = y^2 + 4). Substituting into the original expression gives: [ frac{x + 5}{sqrt{x - 4}} = frac{y^2 + 9}{y} = y + frac{9}{y}. ] Applying the AM-GM inequality: [ y + frac{9}{y} geq 2sqrt{y cdot frac{9}{y}} = 2sqrt{9} = 6. ] Equality occurs when (y = 3), so (x = y^2 + 4 = 3^2 + 4 = 13). Therefore, the minimum value is (boxed{6}).

answer:To determine the mass percentage of oxygen in the mixture, we first need to calculate the total mass of oxygen in both compounds and then divide it by the total mass of the mixture. The molar mass of Copper II carbonate (CuCO3) is: Cu: 63.55 g/mol C: 12.01 g/mol O: 16.00 g/mol x 3 = 48.00 g/mol Total = 63.55 + 12.01 + 48.00 = 123.56 g/mol The molar mass of Copper II sulfate (CuSO4) is: Cu: 63.55 g/mol S: 32.07 g/mol O: 16.00 g/mol x 4 = 64.00 g/mol Total = 63.55 + 32.07 + 64.00 = 159.62 g/mol Now, let's calculate the mass of oxygen in each compound: For CuCO3: The mass of oxygen in 12g of CuCO3 = (mass of CuCO3 / molar mass of CuCO3) x (molar mass of oxygen in CuCO3) = (12g / 123.56 g/mol) x (48.00 g/mol) = 0.0971 mol x 48.00 g/mol = 4.6608 g For CuSO4: The mass of oxygen in 8g of CuSO4 = (mass of CuSO4 / molar mass of CuSO4) x (molar mass of oxygen in CuSO4) = (8g / 159.62 g/mol) x (64.00 g/mol) = 0.0501 mol x 64.00 g/mol = 3.2064 g Now, let's add the mass of oxygen from both compounds: Total mass of oxygen = 4.6608 g + 3.2064 g = 7.8672 g The total mass of the mixture is the sum of the masses of CuCO3 and CuSO4: Total mass of mixture = 12g + 8g = 20g Finally, the mass percentage of oxygen in the mixture is: Mass percentage of O = (Total mass of oxygen / Total mass of mixture) x 100% = (7.8672 g / 20g) x 100% = 0.39336 x 100% = 39.336% Therefore, the mass percentage of oxygen in the mixture is approximately boxed{39.34%} .

answer:1. **Understanding the problem:** Sandra has 5 pairs of shoes, 10 shoes in total, with each pair being of a different color. Each day for 5 days, she draws two shoes randomly from the drawer and throws them out the window. If the two shoes drawn on a particular day are the same color, she does one practice problem. We need to find the expected value (average number) of practice problems she does over 5 days. 2. **Calculate the probability of drawing two shoes of the same color on a given day:** - We start by computing the probability that the two shoes drawn on any day are a matching pair. - The probability that the first shoe can be any shoe is 1. - When drawing the second shoe, there are 9 remaining shoes in the drawer. Out of these 9 shoes, only 1 will match the color of the first shoe drawn. - Hence, the probability that the second shoe matches the first shoe (i.e., both shoes are of the same color) is: [ P(text{same color}) = frac{1}{9} ] 3. **Expected number of practice problems per day:** - Since there is a ( frac{1}{9} ) probability that both shoes are of the same color, on any given day, the expected number of practice problems Sandra does is simply this probability: [ E(text{problems per day}) = frac{1}{9} ] 4. **Using the linearity of expectation to find the total expected problems over 5 days:** - The linearity of expectation states that the expected value of the sum of random variables is equal to the sum of their expected values. - So, the total expected number of practice problems Sandra does over 5 days is: [ E(text{total problems}) = 5 times E(text{problems per day}) = 5 times frac{1}{9} = frac{5}{9} ] # Conclusion: The expected value of the number of practice problems Sandra gets to do over 5 days is: [ boxed{frac{5}{9}} ]

answer:First, let's consider all possible combinations for selecting 3 questions out of 10. The number of combinations is given by the binomial coefficient C(10,3). This can be computed as: C(10,3) = frac{10!}{3!(10-3)!} = frac{10 times 9 times 8}{6} = 120 The student will pass if they answer correctly to exactly 2 questions or all 3 of the selected questions. The number of ways to select 1 question that the student cannot answer (from the 4 that they cannot answer correctly) and 2 questions they can answer (from the 6 that they can) is given by the product C(4,1) cdot C(6,2). This can be computed as: C(4,1) cdot C(6,2) = 4 times frac{6 times 5}{2} = 4 times 15 = 60 The number of ways to select 3 questions they can answer is given by C(6,3). This can be computed as: C(6,3) = frac{6!}{3!(6-3)!} = frac{6 times 5 times 4}{6} = 20 Adding these two numbers up gives us the total number of successful outcomes, which is 60 + 20 = 80. Hence, the probability of the student passing is the ratio of the number of successful outcomes to the total number of possible outcomes, which is: frac{80}{120} = frac{2}{3} So, the probability that the student passes the test is boxed{frac{2}{3}}.