answer:First, calculate the number of ways to choose 4 days out of 6 where strawberry yogurt is packed. This is given by the binomial coefficient: [ binom{6}{4} = 15 ] Next, consider the successful scenarios. We need to choose which 4 days out of the 6 will have strawberry yogurt. Two cases arise: 1. All 4 chosen days include either 2 or 3 days from the first three days: - Case 1a: 2 days from the first three (1/2 chance each) and 2 days from the last three (3/4 chance each): [ binom{3}{2} cdot binom{3}{2} left(frac{1}{2}right)^2 left(frac{3}{4}right)^2 ] - Case 1b: 3 days from the first three (1/2 chance each) and 1 day from the last three (3/4 chance each): [ binom{3}{3} cdot binom{3}{1} left(frac{1}{2}right)^3 left(frac{3}{4}right)^1 ] Now, calculate each case: - Case 1a: ( binom{3}{2} = 3 ), ( binom{3}{2} = 3 ) [ 3 cdot 3 left(frac{1}{2}right)^2 left(frac{3}{4}right)^2 = 9 cdot frac{1}{4} cdot frac{9}{16} = frac{81}{64} ] - Case 1b: ( binom{3}{3} = 1 ), ( binom{3}{1} = 3 ) [ 1 cdot 3 left(frac{1}{2}right)^3 left(frac{3}{4}right)^1 = 3 cdot frac{1}{8} cdot frac{3}{4} = frac{9}{32} ] Adding both cases: [ frac{81}{64} + frac{9}{32} = frac{81}{64} + frac{18}{64} = frac{99}{64} ] Multiply by the 15 ways to choose the 4 successful days: [ 15 cdot frac{99}{64} = frac{1485}{64} = frac{1485}{64} = boxed{frac{1485}{64}} ] Conclusion: We found the probability for Emily’s scenario. Now let's verify.

answer:1. Start with the given equation and let lg(x+1) = frac{1}{2} log_3(x) = k. This implies that the logarithmic expressions are equal: [ lg(x+1) = k ] and [ frac{1}{2} log_3(x) = k ] 2. Rewrite these expressions using the definitions of logarithms: - The first equation gives us: [ x + 1 = 10^k quad text{(since base 10 logarithm: } lg(x+1) = k text{ means } x + 1 = 10^k) ] - The second equation can be rewritten by isolating the logarithm: [ log_3(x) = 2k quad text{(multiplying both sides by 2)} ] This implies: [ x = 3^{2k} quad text{(converting the logarithm into an exponential form)} ] 3. Substitute x = 3^{2k} into the equation x + 1 = 10^k: [ 3^{2k} + 1 = 10^k ] 4. To solve for (k), recognize that the equation 3^{2k} + 1 = 10^k is a functional relation. To find k that satisfies it, consider k = 1: [ 3^{2 cdot 1} + 1 = 10^1 ] [ 3^2 + 1 = 10 ] [ 9 + 1 = 10 ] [ 10 = 10 ] Since this equation holds true for k=1, we substitute back into the expression for x: [ x = 3^{2k} = 3^{2 cdot 1} = 3^2 = 9 ] 5. Verify that x = 9 satisfies both original logarithmic equations: - For lg(x+1): [ lg(9+1) = lg(10) = 1 ] - For frac{1}{2} log_3(x): [ frac{1}{2} log_3(9) = frac{1}{2} cdot 2 = 1 ] Both expressions equal k = 1, confirming our solution. # Conclusion: The value of x that satisfies the given logarithmic equation is: [ boxed{9} ]

answer:The derivative of y=x+1+ln x is y'=1+frac{1}{x}. The slope of the tangent line to the curve y=x+1+ln x at x=1 is k=2, thus, the equation of the tangent line to the curve y=x+1+ln x at x=1 is y-2=2(x-1), which simplifies to y=2x. Since the tangent line is also tangent to the curve y=ax^2+(a+2)x+1, by setting y=2x, we get ax^2+ax+1=0. Given aneq 0 and there is exactly one tangent point, we have Delta=a^2-4a=0, solving this gives a=4. Therefore, the correct choice is: boxed{C}. By finding the derivative of y=x+1+ln x to get the slope of the tangent line, and then deriving the equation of the tangent line, and since the tangent line is also tangent to the curve y=ax^2+(a+2)x+1 with exactly one tangent point, we can set up the equation of the tangent line and the curve together. By setting Delta=0, we find the value of a. This problem examines the application of derivatives: finding the equation of a tangent line, mainly exploring the geometric meaning of derivatives: the derivative of a function at a certain point is the slope of the curve at that point. Setting up the equation of the tangent line and using the property of two lines being tangent is key to solving the problem.

answer:Let's denote the sequence as 2, x, y, 20. Then, we have the system of equations: begin{cases} x^{2}=2y 2y=x+20 end{cases} Solving this system, we get: begin{cases} x=-4 y=8 end{cases} or begin{cases} x=5 y= dfrac {25}{2} end{cases} Therefore, the sum x+y=4 or 17 dfrac {1}{2}. Hence, the correct choice is boxed{B}. By setting up the sequence according to the problem statement and using the properties of geometric and arithmetic sequences to form a system of equations for x and y, we can find the sum of the two numbers. This problem tests the understanding of the properties of arithmetic and geometric sequences. The key to solving this problem lies in constructing equations for x and y based on the given conditions and the definitions of the mean of arithmetic and geometric sequences.