💡Okay, so I have this problem here where I need to find the probability of drawing a number from a bag that's divisible by 11. The bag contains all natural numbers less than or equal to 999 whose digits sum to 6. Hmm, let me break this down step by step.First, I need to understand what numbers we're dealing with. Since we're talking about numbers less than or equal to 999, we're dealing with one-digit, two-digit, and three-digit numbers. But since all these numbers have digits that sum to 6, I can represent any number in the bag as a three-digit number where the digits can be zero. So, let's denote a number as ABC, where A, B, and C are digits (0-9), and A can be zero. The sum of the digits is A + B + C = 6.Now, the next part is about divisibility by 11. I remember that a number is divisible by 11 if the alternating sum of its digits is a multiple of 11. That is, (A - B + C) should be divisible by 11. So, for our number ABC, the condition is A - B + C = 0 or ±11 or ±22, etc. But since A, B, and C are single digits, the maximum value of A - B + C can be 9 - 0 + 9 = 18, and the minimum can be 0 - 9 + 0 = -9. So, the only possible multiples of 11 within this range are -11, 0, and 11. But wait, -11 is less than -9, so it's not possible. Similarly, 11 is more than 9, so the only feasible multiple is 0. Therefore, for the number to be divisible by 11, A - B + C must equal 0.So, we have two equations now:1. A + B + C = 62. A - B + C = 0Let me write these down:1. A + B + C = 62. A - B + C = 0If I subtract the second equation from the first, I get:(A + B + C) - (A - B + C) = 6 - 0Simplify:A + B + C - A + B - C = 6Which simplifies to:2B = 6So, B = 3.Alright, so the middle digit B must be 3. Now, plugging B = 3 back into the first equation:A + 3 + C = 6So, A + C = 3.Now, I need to find all possible non-negative integer solutions for A and C such that A + C = 3. Since A and C are digits, they can be 0 through 9, but their sum is 3, so the possible pairs (A, C) are:(0, 3), (1, 2), (2, 1), (3, 0)So, there are 4 possible combinations for A and C. Therefore, there are 4 numbers in the bag that are divisible by 11.Now, I need to find the total number of numbers in the bag, which is the total number of natural numbers less than or equal to 999 with digits summing to 6. This is a classic stars and bars problem. The number of non-negative integer solutions to A + B + C = 6 is C(6 + 3 - 1, 3 - 1) = C(8, 2) = 28. So, there are 28 numbers in total.Therefore, the probability is the number of favorable outcomes divided by the total number of outcomes, which is 4/28. Simplifying that, we get 1/7.Wait, let me double-check. Are there really 28 numbers? Let me think. For numbers with digits summing to 6, considering leading zeros, yes, it's 28. And the numbers divisible by 11 are 4, as we found. So, 4/28 is indeed 1/7.I think that's it. So, the probability is 1/7.

💡Alright, so I have this problem here about a normal distribution. Let me see if I can figure it out step by step. The random variable ξ follows a normal distribution N(1,1), which means it has a mean (μ) of 1 and a standard deviation (σ) of 1. The problem tells me that P(ξ < 3) is 0.977, and I need to find P(-1 < ξ < 3). The options are A: 0.683, B: 0.853, C: 0.954, and D: 0.977.Okay, first, I remember that in a normal distribution, the total area under the curve is 1, representing 100% probability. The distribution is symmetric around the mean, which in this case is 1. So, if I can find the probabilities on either side of the mean, I can combine them to find the probability between two points.Given that P(ξ < 3) is 0.977, that means 97.7% of the data lies to the left of 3. Since the mean is 1, 3 is two standard deviations above the mean because the standard deviation is 1. I recall that in a normal distribution, about 95% of the data lies within two standard deviations of the mean. But here, it's 97.7%, which is a bit more than 95%. Hmm, maybe I need to think about the exact values.Wait, actually, the 68-95-99.7 rule says that about 68% of the data is within one standard deviation, 95% within two, and 99.7% within three. So, 97.7% is close to 95%, but not exactly. Maybe it's a specific value from the standard normal distribution table.Since ξ follows N(1,1), I can standardize it to Z = (ξ - μ)/σ. So, Z = (ξ - 1)/1 = ξ - 1. Therefore, P(ξ < 3) = P(Z < 2) = 0.977. That makes sense because P(Z < 2) is approximately 0.9772, which rounds to 0.977.Now, I need to find P(-1 < ξ < 3). Let's break this down. First, I can find P(ξ < 3), which is given as 0.977. Then, I need to find P(ξ < -1) and subtract that from P(ξ < 3) to get the probability between -1 and 3.So, P(-1 < ξ < 3) = P(ξ < 3) - P(ξ < -1). I know P(ξ < 3) is 0.977, so I need to find P(ξ < -1).To find P(ξ < -1), I'll standardize it as well. Z = (-1 - 1)/1 = -2. So, P(ξ < -1) = P(Z < -2). From the standard normal distribution table, P(Z < -2) is approximately 0.0228, which is about 0.023.Now, subtracting that from P(ξ < 3): 0.977 - 0.023 = 0.954. So, P(-1 < ξ < 3) is approximately 0.954.Looking back at the options, C is 0.954, so that should be the correct answer.Wait, let me double-check. If I think about the normal distribution, the area between -1 and 3 is the area from the mean minus two standard deviations to the mean plus two standard deviations. Since the mean is 1, -1 is two standard deviations below, and 3 is two standard deviations above. According to the 68-95-99.7 rule, about 95% of the data lies within two standard deviations. But here, it's exactly 0.954, which is close to 95%. So that makes sense.I think I did everything correctly. I standardized the values, used the Z-table to find the probabilities, and subtracted appropriately. Yeah, I'm confident that the answer is 0.954.

💡First, I need to identify the 12 consecutive integers in list D, starting from -4.So, the list would be: -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7.Next, I'll extract the positive integers from this list, which are: 1, 2, 3, 4, 5, 6, 7.To find the range of these positive integers, I'll subtract the smallest positive integer from the largest one.The largest positive integer is 7, and the smallest is 1. So, the range is 7 - 1 = 6.

💡Alright, so I need to find the inverse of this matrix:[begin{pmatrix} 5 & 10 -15 & -30 end{pmatrix}]Hmm, okay. I remember that to find the inverse of a 2x2 matrix, there's a formula involving the determinant. Let me recall... I think the inverse of a matrix [begin{pmatrix} a & b c & d end{pmatrix}]is [frac{1}{ad - bc} begin{pmatrix} d & -b -c & a end{pmatrix}]provided that the determinant ( ad - bc ) is not zero. If the determinant is zero, then the matrix doesn't have an inverse, and I should return the zero matrix as per the problem's instructions.Alright, let's apply this to the given matrix. Here, ( a = 5 ), ( b = 10 ), ( c = -15 ), and ( d = -30 ).First, I need to calculate the determinant:[text{det} = ad - bc = (5)(-30) - (10)(-15)]Calculating each part:[(5)(-30) = -150][(10)(-15) = -150]So,[text{det} = -150 - (-150) = -150 + 150 = 0]Oh, the determinant is zero. That means the matrix doesn't have an inverse. According to the problem, if the inverse doesn't exist, I should enter the zero matrix.So, the zero matrix of the same dimensions (2x2) is:[begin{pmatrix} 0 & 0 0 & 0 end{pmatrix}]Wait, let me double-check my calculations to make sure I didn't make a mistake. Maybe I miscalculated the determinant.Calculating ( ad ) again: ( 5 times -30 = -150 ).Calculating ( bc ) again: ( 10 times -15 = -150 ).So, ( ad - bc = -150 - (-150) = 0 ). Yep, that's correct. The determinant is indeed zero.Therefore, the inverse doesn't exist, and I should return the zero matrix.Just to make sure, maybe I can think about why the determinant being zero means there's no inverse. I remember that a determinant of zero implies that the matrix is singular, meaning it doesn't have full rank, and thus it's not invertible. So, that aligns with what I found.Alright, I think I'm confident in my conclusion. The inverse doesn't exist, so the answer is the zero matrix.

💡Okay, so I've got this equation to solve: 16^x multiplied by itself three times equals 256 to the power of 4. Hmm, let me write that down to make sure I understand it correctly. It's 16^x * 16^x * 16^x = 256^4. Alright, first off, when I see the same base raised to a power and multiplied together, I remember that I can add the exponents. But wait, in this case, it's the same base, 16^x, multiplied three times. So, that should be 16^(x + x + x), right? Which simplifies to 16^(3x). So, the left side of the equation becomes 16^(3x).Now, the right side is 256^4. I need to see if I can express both sides with the same base so that I can set the exponents equal to each other. I know that 16 is a power of 2, specifically 2^4 because 2*2*2*2 is 16. Similarly, 256 is also a power of 2. Let me check: 2^8 is 256 because 2^10 is 1024, so 2^8 should be 256. Yep, that's right.So, if I rewrite both sides using base 2, the left side becomes (2^4)^(3x) and the right side becomes (2^8)^4. Now, when you raise a power to another power, you multiply the exponents. So, on the left side, that would be 2^(4*3x) which is 2^(12x). On the right side, it's 2^(8*4) which is 2^32.Now the equation is 2^(12x) = 2^32. Since the bases are the same, I can set the exponents equal to each other. So, 12x = 32. To solve for x, I divide both sides by 12. That gives me x = 32/12. Simplifying that fraction, both numerator and denominator are divisible by 4, so 32 divided by 4 is 8 and 12 divided by 4 is 3. So, x = 8/3.Wait, let me double-check that. Starting from the beginning: 16^x * 16^x * 16^x is indeed 16^(3x). 16 is 2^4, so (2^4)^(3x) is 2^(12x). 256 is 2^8, so (2^8)^4 is 2^32. So, 2^(12x) = 2^32 leads to 12x = 32, hence x = 32/12 = 8/3. Yep, that seems correct.I think I've got it. So, x is 8/3.

💡First, I need to find the roots of the original quadratic equation (2x^2 - 4x - 6 = 0). To simplify, I'll divide the entire equation by 2, resulting in (x^2 - 2x - 3 = 0).Next, I'll use the quadratic formula to find the roots of this simplified equation. The quadratic formula is (x = frac{-b pm sqrt{b^2 - 4ac}}{2a}), where (a = 1), (b = -2), and (c = -3). Plugging in these values, I get:[x = frac{-(-2) pm sqrt{(-2)^2 - 4(1)(-3)}}{2(1)} = frac{2 pm sqrt{4 + 12}}{2} = frac{2 pm sqrt{16}}{2} = frac{2 pm 4}{2}]This gives the roots (x = 3) and (x = -1).Now, I need to create a new quadratic equation where each root is three less than the original roots. Subtracting 3 from each root, the new roots become (3 - 3 = 0) and (-1 - 3 = -4).With the new roots (0) and (-4), I can form the quadratic equation using the factored form:[(x - 0)(x - (-4)) = x(x + 4) = x^2 + 4x]This equation is already in the form (x^2 + dx + e), where (d = 4) and (e = 0).Therefore, the value of (e) is (0).