💡Okay, so I'm trying to figure out this problem about the diameters of rollers produced by an automated machine. The diameters, denoted as ( X ), are normally distributed with a mean ( a = 10 ) mm and a standard deviation ( sigma = 0.1 ) mm. The question is asking for the interval within which the diameters will fall with a probability of 0.9973.First, I remember that in a normal distribution, most of the data lies within a certain number of standard deviations from the mean. Specifically, I recall something about the 68-95-99.7 rule, which states that about 68% of the data falls within one standard deviation, 95% within two, and 99.7% within three standard deviations from the mean. Since 0.9973 is very close to 99.7%, I think this might be related to three standard deviations.So, if the probability is 0.9973, which is approximately 99.7%, that should correspond to three standard deviations from the mean. That means the interval should be from ( a - 3sigma ) to ( a + 3sigma ).Let me verify that. The formula for the interval in a normal distribution is:[P(a - zsigma < X < a + zsigma) = p]Where ( z ) is the z-score corresponding to the probability ( p ). For 0.9973, the z-score should be 3 because, as I mentioned earlier, 99.7% of the data lies within three standard deviations.So, substituting the values:[a - zsigma = 10 - 3 times 0.1 = 10 - 0.3 = 9.7 , text{mm}][a + zsigma = 10 + 3 times 0.1 = 10 + 0.3 = 10.3 , text{mm}]Therefore, the interval should be from 9.7 mm to 10.3 mm.Wait, let me double-check if 0.9973 is exactly 99.7%. Yes, 0.9973 is approximately 99.73%, which is almost the same as 99.7%. So, using three standard deviations is correct here.Is there another way to approach this? Maybe using the Z-table or the inverse normal function? Let's see.If I use the standard normal distribution table, I can find the z-score that corresponds to the cumulative probability of ( (1 + 0.9973)/2 = 0.99865 ). Looking up 0.99865 in the Z-table, I find that it corresponds to a z-score of approximately 3. So, that confirms my earlier conclusion.Alternatively, using a calculator or software to find the inverse of the standard normal distribution for 0.99865 should also give me a z-score of 3. So, both methods lead me to the same result.Therefore, the interval within which the diameters of the rollers will fall with a probability of 0.9973 is from 9.7 mm to 10.3 mm.I think I've covered all the steps and verified my answer using different methods. It seems consistent and correct.

💡Okay, so I have this problem where a half-sector of a circle with radius 6 inches is rolled up to form a right circular cone. I need to find the volume of the cone. Hmm, let me think through this step by step.First, I remember that when you roll a sector of a circle into a cone, the radius of the sector becomes the slant height of the cone. So, in this case, the slant height (let's call it ( l )) of the cone is 6 inches. That makes sense because the sector's radius is 6 inches.Next, the arc length of the sector becomes the circumference of the base of the cone. Since it's a half-sector, the arc length is half of the circumference of the full circle. The circumference of a full circle with radius 6 inches is ( 2pi times 6 = 12pi ) inches. So, half of that would be ( 6pi ) inches. This ( 6pi ) inches is the circumference of the base of the cone.Now, to find the radius of the base of the cone, I can use the formula for circumference, which is ( 2pi r ). Setting that equal to ( 6pi ), I get:[2pi r = 6pi]Dividing both sides by ( 2pi ), I find that ( r = 3 ) inches. So, the radius of the base of the cone is 3 inches.Now, I need to find the height of the cone to calculate the volume. I remember that in a right circular cone, the slant height, radius, and height form a right triangle, so I can use the Pythagorean theorem:[l^2 = r^2 + h^2]We know ( l = 6 ) inches and ( r = 3 ) inches, so plugging those in:[6^2 = 3^2 + h^2 36 = 9 + h^2 h^2 = 36 - 9 h^2 = 27 h = sqrt{27} h = 3sqrt{3} text{ inches}]Okay, so the height of the cone is ( 3sqrt{3} ) inches.Now, to find the volume of the cone, I use the formula:[V = frac{1}{3} pi r^2 h]Plugging in the values I have:[V = frac{1}{3} pi (3)^2 (3sqrt{3}) V = frac{1}{3} pi (9) (3sqrt{3}) V = frac{1}{3} pi (27sqrt{3}) V = 9pisqrt{3}]So, the volume of the cone is ( 9pisqrt{3} ) cubic inches. Let me double-check my steps to make sure I didn't make any mistakes.1. Slant height ( l = 6 ) inches.2. Arc length of half-sector: ( 6pi ) inches, which is the circumference of the base.3. Radius of base ( r = 3 ) inches.4. Height ( h = sqrt{6^2 - 3^2} = sqrt{27} = 3sqrt{3} ) inches.5. Volume ( V = frac{1}{3} pi (3)^2 (3sqrt{3}) = 9pisqrt{3} ).Everything seems to check out. I think that's the correct answer.**Final Answer**boxed{B}

💡First, I need to find the coordinates of point ( N ), which is symmetric to point ( M(-4, 3) ) with respect to the origin. To do this, I negate both the ( x ) and ( y ) coordinates of ( M ).Next, with the coordinates of ( N ) determined, I need to find the coordinates of point ( P ), which is symmetric to ( N ) with respect to the y-axis. This involves negating the ( x )-coordinate of ( N ) while keeping the ( y )-coordinate the same.By following these steps, I can accurately determine the coordinates of point ( P ).

💡Okay, so I need to find the number of lattice points in three-dimensional space that are exactly 3 units away from the origin. A lattice point is a point where all the coordinates are integers. So, basically, I'm looking for all integer solutions (x, y, z) such that the distance from the origin is 3.The distance formula in three dimensions is sqrt(x² + y² + z²). So, if the distance is 3, then sqrt(x² + y² + z²) = 3. If I square both sides, that gives me x² + y² + z² = 9. So, I need to find all integer triples (x, y, z) that satisfy this equation.Alright, so x, y, and z are integers, and their squares add up to 9. Let me think about what possible integer values x, y, and z can take. Since squares are always non-negative, each of x², y², and z² must be less than or equal to 9. So, the possible values for x, y, and z are integers from -3 to 3, because 3² is 9, and anything beyond that would make the square exceed 9.So, x, y, z ∈ {-3, -2, -1, 0, 1, 2, 3}. Now, I need to find all combinations of these numbers such that their squares add up to 9.Let me break this down. I can consider different cases based on how many of x, y, z are zero or non-zero.Case 1: All three coordinates are non-zero. So, x, y, z ≠ 0. Then, each of x², y², z² is at least 1. So, the minimum sum would be 1 + 1 + 1 = 3, and the maximum sum would be 9 + 9 + 9 = 27. But we need the sum to be exactly 9.Case 2: Two coordinates are zero, and one is non-zero. So, for example, x = 0, y = 0, z = ±3. In this case, z² = 9, so x² + y² + z² = 0 + 0 + 9 = 9. Similarly, this could be any of the three coordinates being ±3 and the other two being zero.Case 3: One coordinate is zero, and the other two are non-zero. So, for example, x = 0, y and z are non-zero. Then, y² + z² must equal 9. So, I need to find all pairs (y, z) where y² + z² = 9, with y and z being integers from -3 to 3.Case 4: All three coordinates are non-zero, and none are zero. So, x, y, z ≠ 0, and x² + y² + z² = 9.Wait, actually, I think I can combine Case 1 and Case 4 into one, since both involve all three coordinates being non-zero. Let me reorganize:Case 1: Exactly one coordinate is non-zero (i.e., two coordinates are zero). Then, the non-zero coordinate must be ±3 because 3² = 9.Case 2: Exactly two coordinates are non-zero. Then, their squares must add up to 9.Case 3: All three coordinates are non-zero. Then, their squares must add up to 9.Let me handle each case separately.Case 1: Exactly one coordinate is non-zero.As I thought, the non-zero coordinate must be either 3 or -3, and the other two coordinates are zero. So, for each axis (x, y, z), there are two points: (3, 0, 0) and (-3, 0, 0), (0, 3, 0) and (0, -3, 0), (0, 0, 3) and (0, 0, -3). So, that's 2 points per axis, and 3 axes, so 2 * 3 = 6 points.Okay, so Case 1 gives me 6 lattice points.Case 2: Exactly two coordinates are non-zero.So, I need to find all pairs (a, b) where a² + b² = 9, with a and b integers from -3 to 3.Let me list all possible pairs:- (0, 3): But wait, in this case, exactly two coordinates are non-zero, so neither a nor b can be zero. So, both a and b must be non-zero.So, possible pairs where both a and b are non-zero and a² + b² = 9.Let me think:- (1, 2√2): Wait, no, that's not integer.Wait, 1² + 2² = 1 + 4 = 5 ≠ 9.2² + 2² = 4 + 4 = 8 ≠ 9.3² + 0² = 9, but that's already covered in Case 1.Wait, so is there any pair of non-zero integers a and b such that a² + b² = 9?Let me check:- 1² + 2² = 1 + 4 = 5- 1² + 3² = 1 + 9 = 10- 2² + 2² = 4 + 4 = 8- 2² + 3² = 4 + 9 = 13- 3² + 3² = 9 + 9 = 18Hmm, none of these add up to 9. So, does that mean there are no solutions in Case 2? That seems odd.Wait, maybe I missed something. Let me think again.Wait, 0² + 3² = 9, but that's Case 1. So, in Case 2, both coordinates must be non-zero, but their squares add up to 9. But from the above, it seems there are no such pairs.Wait, but that can't be right because, for example, (1, 2√2) is not integer, but maybe there are other combinations.Wait, 3² + 0² = 9, but that's Case 1. So, in Case 2, both coordinates are non-zero, but their squares add up to 9. But from the above, it seems there are no such pairs.Wait, maybe I made a mistake in considering the possible pairs. Let me list all possible pairs where a and b are integers from -3 to 3, excluding zero, and check if a² + b² = 9.So, possible a and b values: ±1, ±2, ±3.Let's compute a² + b² for all combinations:- 1² + 1² = 2- 1² + 2² = 5- 1² + 3² = 10- 2² + 2² = 8- 2² + 3² = 13- 3² + 3² = 18None of these equal 9. So, indeed, there are no solutions in Case 2.Wait, that seems strange. So, in three dimensions, the only lattice points at distance 3 from the origin are those with exactly one coordinate being ±3 and the others zero, and the points where two coordinates are ±2 and one is ±1, but wait, that's actually in three dimensions, so maybe that's another case.Wait, hold on, maybe I need to consider three coordinates, not just two.Wait, no, in Case 2, I'm considering exactly two coordinates non-zero, so the third is zero. But as we saw, there are no such pairs where two non-zero integers squared add up to 9.Wait, but in three dimensions, maybe the third coordinate is non-zero as well, but in Case 2, I'm specifically considering exactly two non-zero coordinates. So, maybe there are no solutions in Case 2.Wait, but that can't be right because, for example, (2, 2, 1) squared is 4 + 4 + 1 = 9, but that's three non-zero coordinates, so that would be in Case 3.Wait, so maybe in Case 2, there are no solutions, and all solutions are in Case 1 and Case 3.So, moving on to Case 3: All three coordinates are non-zero, and their squares add up to 9.So, I need to find all triples (a, b, c) where a, b, c are integers from -3 to 3, none of them zero, and a² + b² + c² = 9.Let me think about possible combinations.Since each coordinate is at least 1 in absolute value, the minimum sum of squares is 1 + 1 + 1 = 3, and the maximum is 9 + 9 + 9 = 27. We need the sum to be exactly 9.So, let's think about possible combinations of squares that add up to 9.Possible square numbers we can use: 1, 4, 9.So, we need three numbers from {1, 4, 9} that add up to 9.Let me list all possible combinations:1. 1 + 1 + 7: But 7 isn't a square.2. 1 + 4 + 4: 1 + 4 + 4 = 9.3. 9 + 0 + 0: But we already covered that in Case 1.4. 1 + 1 + 7: Again, 7 isn't a square.5. 1 + 8 + 0: 8 isn't a square.6. 4 + 5 + 0: 5 isn't a square.Wait, so the only combination is 1 + 4 + 4 = 9.So, that means one coordinate has a square of 1, and the other two have squares of 4.So, in terms of absolute values, that means one coordinate is ±1, and the other two are ±2.So, the possible triples are permutations of (±1, ±2, ±2).Now, let's count how many such points there are.First, how many ways can we arrange the 1, 2, 2 in terms of positions? There are 3 positions, and we need to choose which position is the 1. So, there are 3 choices.For each such choice, the 1 can be positive or negative, so that's 2 options.Similarly, each of the two 2s can be positive or negative, so that's 2 options for each 2.Therefore, for each arrangement, the number of points is 2 (for the 1) * 2 (for the first 2) * 2 (for the second 2) = 8.But wait, since the two 2s are indistinct in terms of their positions, do we need to adjust for overcounting?Wait, no, because even though the two 2s are in different positions, each can independently be positive or negative. So, for each arrangement, we have 2 * 2 * 2 = 8 points.But wait, actually, the two 2s are in different positions, so their signs are independent. So, yes, 8 points per arrangement.But wait, let me think again. If I fix the position of the 1, say, in the x-coordinate, then y and z are both 2 or -2. So, for each fixed position of the 1, there are 2 choices for the sign of the 1, and for each of the two 2s, 2 choices for their signs. So, 2 * 2 * 2 = 8 points per position.Since there are 3 positions for the 1, total points are 3 * 8 = 24.Wait, but let me verify that.For example, if the 1 is in the x-coordinate, then the points are (1, 2, 2), (1, 2, -2), (1, -2, 2), (1, -2, -2), (-1, 2, 2), (-1, 2, -2), (-1, -2, 2), (-1, -2, -2). That's 8 points.Similarly, if the 1 is in the y-coordinate, we have (2, 1, 2), (2, 1, -2), (2, -1, 2), (2, -1, -2), (-2, 1, 2), (-2, 1, -2), (-2, -1, 2), (-2, -1, -2). Another 8 points.And if the 1 is in the z-coordinate, we have (2, 2, 1), (2, 2, -1), (2, -2, 1), (2, -2, -1), (-2, 2, 1), (-2, 2, -1), (-2, -2, 1), (-2, -2, -1). Another 8 points.So, total of 24 points.So, Case 3 gives me 24 points.Wait, but earlier I thought that in Case 2, there are no points, and in Case 1, there are 6 points, and in Case 3, 24 points. So, total points would be 6 + 24 = 30.But wait, let me make sure I didn't miss any other combinations in Case 3.Is there another way to get three non-zero integers whose squares add up to 9?We considered 1, 2, 2. Is there another combination?Let's see:- 3, 0, 0: Covered in Case 1.- 2, 2, 1: Covered in Case 3.- 1, 1, sqrt(7): Not integer.- 3, 1, 1: 9 + 1 + 1 = 11 ≠ 9.- 2, 1, 2: Same as 1, 2, 2.- 3, 2, 0: 9 + 4 + 0 = 13 ≠ 9.- 2, 2, 2: 4 + 4 + 4 = 12 ≠ 9.- 3, 3, 3: 9 + 9 + 9 = 27 ≠ 9.- 1, 1, 1: 3 ≠ 9.So, no, I think 1, 2, 2 is the only combination where all three coordinates are non-zero and their squares add up to 9.Therefore, total lattice points are 6 (from Case 1) + 24 (from Case 3) = 30.Wait, but let me double-check. Maybe I missed some other combinations.For example, could we have coordinates like (1, 1, sqrt(7))? But sqrt(7) isn't integer, so no.Or (2, 2, 1), which we've already considered.Or (3, 0, 0), which is Case 1.I think that's all.So, the total number of lattice points at distance 3 from the origin in three-dimensional space is 30.

💡Alright, so I've got this problem about a belt drive system with three wheels: K, L, and M. They all rotate without any slippage, which probably means that the distance each wheel covers in one turn is related to their perimeters. The problem gives me some information about how many turns each wheel makes relative to the others, and it also tells me the perimeter of wheel M. I need to find the perimeter of wheel K.First, let me try to understand the relationships between the wheels. It says that wheel L makes 4 full turns when wheel K makes 5 full turns. So, if K turns 5 times, L turns 4 times. That probably means that the ratio of their turns is 5:4. Similarly, it says that wheel L makes 6 full turns when wheel M makes 7 full turns. So, when M turns 7 times, L turns 6 times, giving a ratio of 7:6 for M to L.Since all the wheels are connected by a belt without slippage, the distance each wheel covers in one turn must be the same. That distance is essentially the perimeter of the wheel. So, if wheel K makes 5 turns, it covers 5 times its perimeter. Similarly, wheel L covers 4 times its perimeter in the same time. Since the distances must be equal, 5 times the perimeter of K equals 4 times the perimeter of L. Mathematically, that would be:5 * P_K = 4 * P_LSimilarly, for wheels L and M, when M makes 7 turns and L makes 6 turns, the distance covered should be the same. So:7 * P_M = 6 * P_LI know the perimeter of M is 30 cm, so I can plug that into the second equation:7 * 30 = 6 * P_LLet me calculate that:210 = 6 * P_LSo, P_L = 210 / 6 = 35 cm.Now that I have the perimeter of L, I can use the first equation to find P_K.5 * P_K = 4 * 35Calculating the right side:4 * 35 = 140So, 5 * P_K = 140Therefore, P_K = 140 / 5 = 28 cm.Wait, let me double-check my steps to make sure I didn't make a mistake.1. I started by noting the ratios of turns between K and L, and between L and M.2. I translated those turn ratios into perimeter relationships because the distance covered must be the same.3. I solved for P_L using the perimeter of M, which was given.4. Then I used P_L to find P_K.Everything seems to add up. I don't see any errors in my calculations. So, the perimeter of wheel K should be 28 cm.

💡First, I need to find the probability that at least one person goes to the place. To do this, I'll first calculate the probabilities of the complementary events, which are the probabilities that each person does not go to the place.For person A, the probability of not going to the place is:[1 - frac{1}{4} = frac{3}{4}]For person B, the probability of not going to the place is:[1 - frac{1}{5} = frac{4}{5}]Since the actions of person A and person B are independent, the probability that neither person goes to the place is the product of their individual probabilities of not going:[frac{3}{4} times frac{4}{5} = frac{12}{20} = frac{3}{5}]Finally, the probability that at least one person goes to the place is the complement of the probability that neither goes:[1 - frac{3}{5} = frac{2}{5}]