answer：Okay, so I have this problem here: Little Ming's height is 1.3 meters, and that's 0.2 meters taller than Little Red. I need to find out Little Red's height. Hmm, let's see. Alright, first, I need to understand what the problem is saying. Little Ming is taller than Little Red by 0.2 meters. So, if I think about it, Little Ming's height is the same as Little Red's height plus 0.2 meters. That makes sense because if Little Ming is taller, you add that extra height to Little Red's height to get Little Ming's.So, mathematically, I can write this as:Little Ming's height = Little Red's height + 0.2 metersWe know Little Ming's height is 1.3 meters, so plugging that in:1.3 meters = Little Red's height + 0.2 metersNow, I need to solve for Little Red's height. To do that, I can subtract 0.2 meters from both sides of the equation to isolate Little Red's height. Let's do that:1.3 meters - 0.2 meters = Little Red's heightCalculating the left side:1.3 - 0.2 = 1.1So, Little Red's height is 1.1 meters.Wait, let me double-check that. If Little Red is 1.1 meters tall and Little Ming is 0.2 meters taller, then Little Ming's height should be 1.1 + 0.2 = 1.3 meters, which matches the given information. Okay, that seems right.I think I've got it. Little Red is 1.1 meters tall.

answer：First, I need to determine the total loss the shopkeeper incurred by selling 400 meters of cloth at a loss of 5 per meter. Next, I'll calculate the total cost price by adding the total loss to the total selling price of 18,000.Finally, to find the cost price per meter, I'll divide the total cost price by the total number of meters, which is 400.

answer：First, I need to understand the relationship between the sector of the circle and the cone it forms. When a sector is removed from a circle, the remaining part forms the lateral surface of a cone. The radius of the original circle becomes the slant height of the cone.Given that the sector removed is (300^circ), the remaining sector that forms the cone is (60^circ). The circumference of the base of the cone is proportional to the angle of the remaining sector.The circumference of the full circle is (2pi times 6 = 12pi) inches. Since the remaining sector is (60^circ), which is (frac{1}{6}) of the full circle, the circumference of the base of the cone is (frac{1}{6}) of the full circumference.Calculating this, (frac{1}{6} times 12pi = 2pi). Therefore, the circumference of the base of the cone is (2pi) inches.

answer：First, I need to determine the probability that one of the three cars (X, Y, or Z) will win the race. I know the individual probabilities for each car winning: X has a 1/7 chance, Y has a 1/3 chance, and Z has a 1/5 chance.Since these events are mutually exclusive (only one car can win the race), I can add their probabilities together to find the total probability that one of them will win.To add the fractions, I need a common denominator. The denominators are 7, 3, and 5. The least common multiple of these numbers is 105. I will convert each fraction to have a denominator of 105.After converting, I add the numerators together and keep the common denominator. This will give me the total probability that one of the cars X, Y, or Z will win the race.

answer：Okay, so I have this problem where Liam needs to save 30 files onto disks, each with 1.44 MB of space. The files are of different sizes: 5 files are 0.9 MB each, 10 files are 0.6 MB each, 10 files are 0.45 MB each, and 5 files are 0.3 MB each. The goal is to figure out the smallest number of disks needed to store all these files without splitting any file across disks.First, I need to understand the problem clearly. We have four types of files with different sizes, and we need to fit them onto disks that each have a maximum capacity of 1.44 MB. The challenge is to do this efficiently so that we use as few disks as possible.Let me list out the files:- 5 files of 0.9 MB- 10 files of 0.6 MB- 10 files of 0.45 MB- 5 files of 0.3 MBTotal files: 5 + 10 + 10 + 5 = 30 files.Total storage required: (5 * 0.9) + (10 * 0.6) + (10 * 0.45) + (5 * 0.3) = 4.5 + 6 + 4.5 + 1.5 = 16.5 MB.Each disk can hold up to 1.44 MB. So, if we divide the total storage by the disk capacity, we get 16.5 / 1.44 ≈ 11.45. Since we can't have a fraction of a disk, we'd need at least 12 disks. However, this is just a theoretical lower bound because we can't split files across disks. So, the actual number might be higher.Now, let's think about how to optimally pair the files to minimize the number of disks. The key is to pair larger files with smaller ones in a way that the total doesn't exceed 1.44 MB per disk.Starting with the largest files, the 0.9 MB files. Each of these is quite sizable, so we need to see if we can pair them with smaller files without exceeding the disk limit.0.9 + 0.3 = 1.2 MB. That's under 1.44, so we can pair each 0.9 MB file with a 0.3 MB file. Since there are 5 of each, this would use 5 disks, each holding a 0.9 and a 0.3 MB file.After pairing these, we have:- 0 files of 0.9 MB left- 10 files of 0.6 MB- 10 files of 0.45 MB- 0 files of 0.3 MB leftNow, we have 10 files of 0.6 MB and 10 files of 0.45 MB left. Let's see how we can pair these.First, let's consider pairing 0.6 and 0.45 MB files together. 0.6 + 0.45 = 1.05 MB. That's well under 1.44, so we can fit two such pairs on a single disk. Wait, no, because each disk can hold multiple files as long as the total doesn't exceed 1.44.Wait, actually, each disk can hold multiple files, but we need to make sure the total doesn't exceed 1.44. So, let's see:If we pair 0.6 and 0.45, that's 1.05. Then, we can add another 0.45 to make it 1.5, which is over 1.44. So, we can't add another 0.45. Alternatively, can we add a 0.3? But we've already used all the 0.3 MB files. So, perhaps pairing 0.6 and 0.45 is the best we can do for now.But wait, 1.05 is still under 1.44. Maybe we can fit another 0.45? Let me check: 0.6 + 0.45 + 0.45 = 1.5, which is over 1.44. So, no, we can't. Alternatively, maybe pair two 0.6 MB files together: 0.6 + 0.6 = 1.2, which is under 1.44. Then, we could add a 0.45 to make it 1.65, which is over. So, no.Alternatively, pair a 0.6 and two 0.45s: 0.6 + 0.45 + 0.45 = 1.5, which is over. So, that doesn't work.Alternatively, pair a 0.6 and a 0.45, which is 1.05, and then see if we can add another 0.45: 1.05 + 0.45 = 1.5, which is over. So, no.Alternatively, pair two 0.45s: 0.45 + 0.45 = 0.9. Then, we can add a 0.6 to make it 1.5, which is over. So, no.Alternatively, pair three 0.45s: 0.45 * 3 = 1.35, which is under 1.44. So, we can fit three 0.45s on a disk. That would use 10 / 3 ≈ 3.33 disks, but since we can't have a fraction, we'd need 4 disks for the 10 files, but let's see.Wait, 10 files of 0.45 MB. If we put three on each disk, we can do 3 disks with 3 files each (using 9 files) and then one disk with the remaining 1 file. But that would be 4 disks for the 0.45s. Alternatively, pairing them with 0.6s might be more efficient.Let me think again. If we pair 0.6 and 0.45, that's 1.05. Then, on the same disk, can we add another 0.45? 1.05 + 0.45 = 1.5, which is over. So, no. Alternatively, can we add a 0.3? But we've already used all the 0.3s.Alternatively, pair two 0.6s: 0.6 + 0.6 = 1.2. Then, we can add a 0.45: 1.2 + 0.45 = 1.65, which is over. So, no.Alternatively, pair a 0.6 and two 0.45s: 0.6 + 0.45 + 0.45 = 1.5, over.Alternatively, pair a 0.6 and one 0.45, which is 1.05, and then see if we can add another 0.45: no, as before.Alternatively, pair a 0.6 and a 0.45, and then see if we can add another 0.45 on the same disk: no, as that would exceed.Alternatively, pair a 0.6 and a 0.45, and then see if we can add another 0.45 on the same disk: no.Alternatively, pair a 0.6 and a 0.45, and then see if we can add another 0.45 on the same disk: no.Wait, maybe I'm overcomplicating this. Let's try a different approach.We have 10 files of 0.6 and 10 files of 0.45. Let's see how many disks we can fill by pairing 0.6 and 0.45.Each such pair takes 1.05 MB. So, how many such pairs can we fit on a disk? Since 1.44 / 1.05 ≈ 1.37, so we can only fit one pair per disk. So, each disk will have one 0.6 and one 0.45, totaling 10 pairs, which would require 10 disks.But wait, that's 10 disks for 10 pairs, but we have 10 of each, so that would use up all the 0.6 and 0.45 files. So, that would be 10 disks.But wait, that would mean we have 5 disks from the first pairing (0.9 and 0.3) and 10 disks from the second pairing (0.6 and 0.45), totaling 15 disks. But the answer choices go up to 16, with the highest being 16, but 15 is an option (D).But wait, is there a way to do better? Maybe by combining more files on a single disk.Let me think again. After pairing the 0.9 and 0.3, we have 10 of 0.6 and 10 of 0.45 left.Is there a way to fit more than one pair on a disk? For example, can we fit two 0.6 and one 0.45 on a disk? Let's check:0.6 + 0.6 + 0.45 = 1.65, which is over 1.44. So, no.Alternatively, can we fit one 0.6 and two 0.45s? 0.6 + 0.45 + 0.45 = 1.5, which is over.Alternatively, can we fit two 0.6s? 0.6 + 0.6 = 1.2, which leaves 0.24 MB free. Maybe we can add a 0.45? 1.2 + 0.45 = 1.65, over. So, no.Alternatively, can we fit one 0.6 and one 0.45, which is 1.05, and then see if we can add another 0.45? 1.05 + 0.45 = 1.5, over.Alternatively, can we fit three 0.45s? 0.45 * 3 = 1.35, which leaves 0.09 MB free. That's possible, but then we'd have to use more disks for the 0.6s.Wait, let's calculate:If we pair 0.6 and 0.45, that's 1.05 per disk, requiring 10 disks.Alternatively, if we pair two 0.6s per disk: 0.6 * 2 = 1.2 per disk. Then, we can fit 5 disks with two 0.6s each, using 10 files. Then, for the 0.45s, we can fit three per disk: 0.45 * 3 = 1.35 per disk, requiring 4 disks (since 10 / 3 ≈ 3.33, so 4 disks). So, total disks would be 5 (for 0.6s) + 4 (for 0.45s) = 9 disks. Plus the initial 5 disks for 0.9 and 0.3, totaling 14 disks.Wait, that's better than 15. So, is 14 possible?Let me check:- Pair 0.9 and 0.3: 5 disks.- Pair two 0.6s per disk: 5 disks (since 10 / 2 = 5).- Pair three 0.45s per disk: 4 disks (since 10 / 3 ≈ 3.33, so 4 disks).Total disks: 5 + 5 + 4 = 14.But wait, does that work? Let's see:- 5 disks with 0.9 and 0.3: 5 disks.- 5 disks with two 0.6s: 5 disks.- 4 disks with three 0.45s: 4 disks.Total: 14 disks.But wait, 0.45 * 3 = 1.35, which is under 1.44, so that's fine.But let me check if we can do even better. Maybe combine some 0.6 and 0.45s in a way that uses the space more efficiently.For example, if we have a disk with one 0.6 and two 0.45s: 0.6 + 0.45 + 0.45 = 1.5, which is over 1.44, so that's not possible.Alternatively, one 0.6 and one 0.45: 1.05, leaving 0.39 MB free. Maybe we can add another 0.45? 1.05 + 0.45 = 1.5, over.Alternatively, can we fit two 0.6s and one 0.45? 1.2 + 0.45 = 1.65, over.Alternatively, three 0.45s and one 0.6: same as above.Alternatively, can we fit four 0.45s? 0.45 * 4 = 1.8, which is way over.Alternatively, two 0.45s and one 0.6: 0.45 + 0.45 + 0.6 = 1.5, over.Alternatively, one 0.6 and one 0.45, and then see if we can add a 0.3, but we've already used all the 0.3s.Wait, maybe instead of pairing two 0.6s, we can pair one 0.6 and one 0.45, and then see if we can add another 0.45 on the same disk. But as we saw, that would exceed.Alternatively, maybe pair one 0.6 and two 0.45s on some disks and pair two 0.6s on others.Let me try:Suppose we have some disks with two 0.6s (1.2 MB) and some disks with one 0.6 and one 0.45 (1.05 MB). Let's see how many of each we need.Let x be the number of disks with two 0.6s.Let y be the number of disks with one 0.6 and one 0.45.Each x disk uses 2 * 0.6 = 1.2 MB.Each y disk uses 0.6 + 0.45 = 1.05 MB.We have 10 files of 0.6 and 10 files of 0.45.So, the total number of 0.6 files used is 2x + y = 10.The total number of 0.45 files used is y = 10.Wait, but y can't be 10 because we only have 10 0.45 files. So, y = 10.Then, 2x + 10 = 10 => 2x = 0 => x = 0.So, that means we can't have any disks with two 0.6s if we use all 10 0.45 files on y disks.Alternatively, if we don't use all 10 0.45 files on y disks, maybe we can have some disks with two 0.6s and some with one 0.6 and one 0.45.Let me set up the equations:Let x = number of disks with two 0.6s.Let y = number of disks with one 0.6 and one 0.45.Then:2x + y = 10 (total 0.6 files)y + z = 10 (total 0.45 files, where z is the number of disks with three 0.45s)But wait, if we have z disks with three 0.45s, then the total 0.45 files used would be 3z.So, y + 3z = 10.But we also have:Total disks = x + y + z.We want to minimize x + y + z.From 2x + y = 10 => y = 10 - 2x.From y + 3z = 10 => (10 - 2x) + 3z = 10 => -2x + 3z = 0 => 3z = 2x => z = (2/3)x.Since z must be an integer, x must be a multiple of 3.Let x = 3k, then z = 2k.Then, y = 10 - 2x = 10 - 6k.We need y >= 0 => 10 - 6k >= 0 => k <= 1.666. So, k can be 0 or 1.If k = 1:x = 3, z = 2, y = 10 - 6 = 4.Total disks = x + y + z = 3 + 4 + 2 = 9.If k = 0:x = 0, z = 0, y = 10.Total disks = 0 + 10 + 0 = 10.So, the minimal total disks for 0.6 and 0.45 files is 9 when k = 1.So, with x = 3 disks with two 0.6s, y = 4 disks with one 0.6 and one 0.45, and z = 2 disks with three 0.45s.Let me verify:- 3 disks with two 0.6s: 3 * 2 = 6 files of 0.6.- 4 disks with one 0.6 and one 0.45: 4 files of 0.6 and 4 files of 0.45.- 2 disks with three 0.45s: 6 files of 0.45.Total 0.6 files: 6 + 4 = 10.Total 0.45 files: 4 + 6 = 10.Perfect.So, total disks for 0.6 and 0.45 files: 3 + 4 + 2 = 9.Adding the initial 5 disks for 0.9 and 0.3 files, total disks: 5 + 9 = 14.So, the minimal number of disks needed is 14.But wait, the answer choices are 10, 12, 14, 15, 16. So, 14 is an option (C). But earlier, I thought 15 was the answer, but now I see that 14 is possible.Wait, let me double-check:- 5 disks: 0.9 + 0.3 = 1.2 each.- 3 disks: two 0.6s each: 1.2 each.- 4 disks: one 0.6 and one 0.45 each: 1.05 each.- 2 disks: three 0.45s each: 1.35 each.Total disks: 5 + 3 + 4 + 2 = 14.Yes, that seems correct.But wait, is there a way to reduce it further? Let's see.If we can find a way to combine some of the 0.6 and 0.45 files more efficiently.For example, if we have a disk with one 0.6 and two 0.45s: 0.6 + 0.45 + 0.45 = 1.5, which is over 1.44. So, no.Alternatively, can we fit two 0.6s and one 0.45? 1.2 + 0.45 = 1.65, over.Alternatively, can we fit one 0.6 and one 0.45 and one 0.3? But we've already used all the 0.3s.Alternatively, can we fit four 0.45s? 1.8, over.Alternatively, can we fit one 0.6 and three 0.45s? 0.6 + 1.35 = 1.95, over.Alternatively, can we fit two 0.6s and one 0.45? Over.Alternatively, can we fit one 0.6 and one 0.45 and one 0.3? But 0.3s are already used.So, it seems that the most efficient way is to have 3 disks with two 0.6s, 4 disks with one 0.6 and one 0.45, and 2 disks with three 0.45s, totaling 9 disks for the 0.6 and 0.45 files, plus 5 disks for the 0.9 and 0.3 files, totaling 14 disks.Therefore, the minimal number of disks needed is 14.

answer：First, identify the universal set ( U ) and the set ( M ):[U = {1, 2, 3, 4, 5, 6}][M = {1, 3, 5}]The complement of ( M ) in ( U ), denoted as ( complement_U M ), consists of all elements in ( U ) that are not in ( M ).Compare each element of ( U ) with the elements of ( M ):- ( 1 ) is in ( M )- ( 2 ) is not in ( M )- ( 3 ) is in ( M )- ( 4 ) is not in ( M )- ( 5 ) is in ( M )- ( 6 ) is not in ( M )Collect the elements that are not in ( M ):[complement_U M = {2, 4, 6}]Therefore, the correct answer is:[boxed{A}]