answer:1. **Express p and q using the change-of-base formula and simplify:** [ p = log_4 400 = frac{log_2 400}{log_2 4} ] [ q = log_2 20 ] 2. **Simplify the expressions:** [ log_2 4 = 2 quad text{(since } 2^2 = 4text{)} ] Thus, [ p = frac{log_2 400}{2} ] 3. **Relate p and q:** [ frac{p}{q} = frac{frac{log_2 400}{2}}{log_2 20} = frac{log_2 400}{2 log_2 20} ] 4. **Use the properties of logarithms:** [ log_2 400 = log_2 (20 times 20) = 2 log_2 20 ] Substituting this into the expression for frac{p}{q}, we get: [ frac{p}{q} = frac{2 log_2 20}{2 log_2 20} = 1 ] 5. **Solve for p:** [ p = q ] 6. **Conclude with the correct answer:** [ q ] The final answer is boxed{B}

answer:Let the number of students with green eyes be y. Therefore, the number of students with red hair is 3y due to the given ratio. The students with both red hair and green eyes total 8, so the students with green eyes but not red hair are y - 8, and those with red hair but not green eyes are 3y - 8. The students with neither characteristic are 4. Representing all categories: - Green-eyed and red-haired students: 8 - Green-eyed only students: y - 8 - Red-haired only students: 3y - 8 - Neither: 4 The total number of students is given by the sum of these groups: [ (y - 8) + (3y - 8) + 8 + 4 = 40 ] [ 4y - 8 = 40 ] [ 4y = 48 ] [ y = 12 ] So, the number of students with green eyes is boxed{12}.

answer:Given the complex number z+2i-3=3-3i, we find |z|=|6-5i|=sqrt{6^{2}+(-5)^{2}}=sqrt{61}. Therefore, the correct option is boxed{D}. This problem involves simplifying the complex number and then calculating the modulus of the complex number. It tests the ability to calculate the modulus of a complex number.

answer:Since the random variable X follows a binomial distribution X sim B(n,p), and E(X) = 6, Var(X) = 3, we have the following two equations: 1. E(X) = 6 = np, (1) 2. Var(X) = 3 = np(1-p), (2) Divide equation (2) by equation (1) to find the ratio of (1-p) to p: frac{1-p}{p} = frac{3}{6} = frac{1}{2} Thus, 1-p = frac{1}{2}p. Solving this equation for p, we find: p = frac{1}{2}. Putting this value back into equation (1), we get: np = 6, therefore, n = frac{6}{p} = frac{6}{frac{1}{2}} = 12. Now, we have n = 12 and p = frac{1}{2}. To find P(X=1), we use the binomial probability formula: P(X=k) = binom{n}{k} p^k (1-p)^{n-k} Finally, calculate P(X=1) by plugging in the values of n, p, and k=1: P(X=1) = binom{12}{1} left(frac{1}{2}right)^1 left(1-frac{1}{2}right)^{11} = 12 cdot frac{1}{2} cdot frac{1}{2^{11}} = 12 cdot 2^{-12} = 3 cdot 2^{-10}. So the correct answer is boxed{3 cdot 2^{-10}}.