answer:Let's denote the number of red velvet cakes Carter usually bakes in a week as R. In a regular week, Carter bakes: - 6 cheesecakes - 5 muffins - R red velvet cakes The total number of cakes he usually bakes in a week is 6 + 5 + R. For this week, he baked triple the amount of each type of cake, so he baked: - 3 * 6 = 18 cheesecakes - 3 * 5 = 15 muffins - 3 * R red velvet cakes The total number of cakes he baked this week is 18 + 15 + 3R. We are told that Carter was able to bake 38 more cakes this week than he usually does. So, the equation to represent this situation is: (6 + 5 + R) + 38 = 18 + 15 + 3R Simplifying both sides of the equation, we get: 11 + R + 38 = 33 + 3R Combining like terms, we get: 49 + R = 33 + 3R Subtracting R from both sides, we get: 49 = 33 + 2R Subtracting 33 from both sides, we get: 16 = 2R Dividing both sides by 2, we get: R = 8 So, Carter usually bakes boxed{8} red velvet cakes in a week.

answer:First, calculate the side length of square C. Since the perimeter of a square is four times one of its side lengths, we have: [ s_C = frac{32}{4} = 8 text{ cm} ] The area of square C is then: [ text{Area of C} = s_C^2 = 8^2 = 64 text{ square cm} ] Now, since square D's area is half of square C’s area: [ text{Area of D} = frac{64}{2} = 32 text{ square cm} ] To find the side length of square D, take the square root of its area: [ s_D = sqrt{32} = 4sqrt{2} text{ cm} ] The perimeter of square D is: [ 4 times s_D = 4 times 4sqrt{2} = 16sqrt{2} text{ cm} ] So, the perimeter of square D is boxed{16sqrt{2}} cm.

answer:Since the random variable X sim B(4, frac{1}{3}), we have E(X) = 4 times frac{1}{3} = boxed{frac{4}{3}}, and D(X) = 4 times frac{1}{3}(1 - frac{1}{3}) = frac{8}{9}. Now, to find D(3X+2), we use the property that the variance of a linear transformation of a random variable is given by D(aX+b) = a^2D(X). Hence, D(3X+2) = 3^2(D(X)) = 9 times frac{8}{9} = boxed{8}. So, the answers are frac{4}{3} and 8. In this problem, we directly apply the properties of the binomial distribution to determine the expected value and variance. This is a fundamental problem that tests your understanding of how to compute the expected value and variance of a discrete random variable. When solving this problem, be sure to carefully read the question and make appropriate use of the properties of the binomial distribution.

answer:Consider the expression for a_n, which can be rewritten as a_n = -2n^2 + 9n + 3 = -2left(n- frac {9}{4}right)^2+ frac {105}{8}. The expression -2left(n- frac {9}{4}right)^2 reaches its maximum value when n = frac {9}{4}, meaning that the square term is zero and the entire expression equals frac {105}{8}. However, since n must be an integer, we consider the integer values close to frac {9}{4}, which are n = 2 and n = 3. Now, we calculate a_2 and a_3 to determine which one is larger: a_2 = -2(2)^2 + 9(2) + 3 = -8 + 18 + 3 = 13, a_3 = -2(3)^2 + 9(3) + 3 = -18 + 27 + 3 = 12. Hence, the maximum value of a_n in the sequence {a_n} is a_2 = 13. Therefore, the correct answer is: [boxed{B: a_2 = 13}] This problem tests understanding of quadratic function monotonicity, maximum term in a sequence, and computational skills. It is a fundamental question.