💡First, I identify the number of students who like each individual sport: Basketball (B) = 7, Cricket (C) = 10, and Soccer (S) = 8.Next, I note the overlaps between the sports:- Both Basketball and Cricket (BC) = 5- Both Basketball and Soccer (BS) = 4- Both Cricket and Soccer (CS) = 3Additionally, there are 2 students who like all three sports (BCS).To apply the inclusion-exclusion principle, I adjust the overlaps to exclude those who like all three sports:- Adjusted BC = 5 - 2 = 3- Adjusted BS = 4 - 2 = 2- Adjusted CS = 3 - 2 = 1Finally, I calculate the total number of students who like at least one sport by adding the individual sport preferences, subtracting the adjusted overlaps, and adding back the students who like all three sports:Total = B + C + S - (Adjusted BC + Adjusted BS + Adjusted CS) + BCSTotal = 7 + 10 + 8 - (3 + 2 + 1) + 2 = 21Therefore, 21 students like to play at least one of the sports.

💡Okay, so I need to find the largest number where all the digits are different, and when you add up all the digits, the total is 20. Hmm, that sounds like a fun puzzle. Let me think about how to approach this.First, I guess I need to remember that to make the largest possible number, I should use the largest digits possible. But they all have to be different, and their sum has to be exactly 20. So I can't just use a bunch of 9s because they would add up to more than 20, and also, they wouldn't be unique.Maybe I should start by trying to use the largest digit, which is 9. If I use 9, then I have 20 - 9 = 11 left to reach the total of 20. Now, I need to find other unique digits that add up to 11. The next largest digit is 8. If I add 8 to 9, that's 17, and I have 20 - 17 = 3 left. So I need a digit that's 3. That works because 3 is unique and hasn't been used yet.So, putting it all together, the digits would be 9, 8, and 3. To make the largest number possible, I should arrange these digits in descending order. That would give me 983. Let me check: 9 + 8 + 3 = 20, and all digits are unique. That seems to work.But wait, is there a way to get a larger number? Maybe by using more digits? Let's see. If I try to use four digits, I need to find four unique digits that add up to 20. Starting with 9 again, then 7, which would leave me with 20 - 9 - 7 = 4. Then I need two more digits that add up to 4, but they have to be unique and not used yet. The digits available would be 4, but I already used 4, so that doesn't work. Maybe 3 and 1? So the digits would be 9, 7, 3, and 1. Adding them up: 9 + 7 + 3 + 1 = 20. Arranged in descending order, that would be 9731. Hmm, that's a four-digit number, which is larger than 983. So maybe 9731 is a better answer.But wait, can I get even larger? Let's try with five digits. Starting with 9, then 8, which leaves 20 - 9 - 8 = 3. Then I need three more digits that add up to 3, but they have to be unique and not used yet. The digits available would be 3, but I already used 3, so that doesn't work. Maybe 2 and 1? So the digits would be 9, 8, 2, and 1. But that's only four digits, and I need five. Maybe 9, 7, 2, 1, and 1? But I can't use 1 twice. Hmm, this is tricky.Maybe I should try a different approach. Let's list all possible combinations of unique digits that add up to 20 and see which one gives the largest number. Starting with the largest digit, 9, and then trying to find the next largest digits that fit.9 + 8 + 3 = 20. That's three digits, giving 983.9 + 7 + 4 = 20. That's also three digits, giving 974.9 + 6 + 5 = 20. That's three digits, giving 965.Now, with four digits:9 + 8 + 2 + 1 = 20. That's four digits, giving 9821.9 + 7 + 3 + 1 = 20. That's four digits, giving 9731.9 + 6 + 4 + 1 = 20. That's four digits, giving 9641.9 + 5 + 4 + 2 = 20. That's four digits, giving 9542.8 + 7 + 5 = 20. That's three digits, giving 875.8 + 6 + 5 + 1 = 20. That's four digits, giving 8651.7 + 6 + 5 + 2 = 20. That's four digits, giving 7652.Okay, so looking at all these combinations, the largest number seems to be 9821, which is a four-digit number. But earlier I thought 9731 was larger than 983. Wait, 9821 is actually larger than 9731 because the second digit is 8 versus 7. So 9821 is larger.But let me double-check if there's a five-digit number that's larger. From earlier attempts, it seems difficult because the digits have to be unique and add up to 20. Let's try again.Starting with 9, then 7, then 2, then 1, and then... wait, 9 + 7 + 2 + 1 = 19, so I need one more digit that's 1, but I already used 1. That doesn't work. Maybe 9 + 6 + 3 + 1 + 1? No, can't use 1 twice.How about 8 + 7 + 4 + 1 = 20. That's four digits, giving 8741.Or 8 + 6 + 5 + 1 = 20, which is 8651.But none of these five-digit combinations seem to work because either I run out of unique digits or the sum doesn't reach 20.So, the largest number I can get is a four-digit number, and among those, 9821 is the largest. Let me verify:9 + 8 + 2 + 1 = 20, and all digits are unique. Arranged in descending order, it's 9821.Wait, but earlier I thought 9731 was a valid four-digit number. Let me check that:9 + 7 + 3 + 1 = 20, yes, and arranged as 9731.Comparing 9821 and 9731, 9821 is larger because the second digit is 8 versus 7.Is there a way to get a four-digit number larger than 9821? Let's see.What if I use 9, 8, 3, and 0? That would be 9 + 8 + 3 + 0 = 20. Arranged as 9830. But 0 at the end doesn't make it larger; it's still 9830, which is less than 9821 because the third digit is 3 versus 2.Wait, no, 9830 is actually larger than 9821 because the third digit is 3 versus 2. So 9830 is larger. But does 0 count as a digit? Yes, but it can't be the leading digit. So 9830 is a valid four-digit number.But wait, 9 + 8 + 3 + 0 = 20, yes. So 9830 is a valid number. Is there a way to get a larger number than 9830?What if I use 9, 8, 4, and -1? No, negative digits aren't allowed.How about 9, 7, 4, and 0? That would be 9 + 7 + 4 + 0 = 20, arranged as 9740, which is less than 9830.Alternatively, 9, 8, 2, and 1 is 9821, which is less than 9830.So 9830 seems to be the largest four-digit number. But wait, can I use more digits to make it larger? Like five digits?Let's try. Starting with 9, then 8, then 3, then 0, and then... I need one more digit, but I've already used 9, 8, 3, and 0. The remaining digits are 1, 2, 4, 5, 6, 7. But I need a digit that, when added to 9 + 8 + 3 + 0, which is 20, so I can't add any more digits without exceeding the sum of 20. So five digits won't work because I can't add another digit without making the total sum more than 20.Therefore, the largest number is 9830. But wait, is 9830 the largest? Let me think again.If I use 9, 8, 3, and 0, that's 9830. But what if I rearrange the digits to get a larger number? Like 9830 versus 9803. No, 9830 is larger because the third digit is 3 versus 0.Alternatively, 9380, but that's smaller because the second digit is 3 versus 8.So 9830 seems to be the largest four-digit number. But earlier, I thought 9821 was a candidate, but 9830 is larger.Wait, but 9830 uses a 0, which is allowed as long as it's not the leading digit. So 9830 is valid.But let me check if there's a way to get a larger number without using 0. For example, using 9, 8, 2, and 1, which gives 9821. That's less than 9830.Alternatively, 9, 7, 4, and 0 gives 9740, which is less than 9830.So, 9830 seems to be the largest number with unique digits adding up to 20.But wait, is there a way to get a five-digit number larger than 9830? For example, 9830 is a four-digit number, but if I can make a five-digit number that's larger, that would be better. But as I thought earlier, adding more digits would require the sum to stay at 20, which is difficult because each additional digit would have to be smaller, but we already used the largest digits.Wait, let's try. Starting with 9, then 8, then 2, then 1, and then... I need one more digit that's 0, but I already used 0. So 9 + 8 + 2 + 1 + 0 = 20, which is 98210. But that's a five-digit number, and it's larger than 9830 because it has an extra digit. But wait, 98210 is actually larger than 9830 because it's a five-digit number versus a four-digit number.But does that work? Let's check the sum: 9 + 8 + 2 + 1 + 0 = 20, yes. And all digits are unique. So 98210 is a valid number, and it's larger than 9830.But wait, can I make an even larger five-digit number? Let's see.If I use 9, 8, 3, 0, and... I need one more digit. 9 + 8 + 3 + 0 = 20, so I can't add another digit without exceeding the sum. So I can't make a five-digit number with 9, 8, 3, 0, and something else because that would require the sum to be more than 20.Alternatively, if I use smaller digits, like 9, 7, 4, 0, and... 9 + 7 + 4 + 0 = 20, so I can't add another digit. Similarly, 9, 6, 5, 0, and... 9 + 6 + 5 + 0 = 20, so no room for another digit.So the only five-digit number I can make is 98210. Is there a way to make a five-digit number larger than 98210?Let's see. If I rearrange the digits, 98210 is already the largest possible arrangement because the digits are in descending order. So 98210 is the largest five-digit number with unique digits adding up to 20.Wait, but earlier I thought 9830 was the largest four-digit number, but now I see that 98210 is a five-digit number, which is larger. So 98210 is the answer.But let me double-check. 9 + 8 + 2 + 1 + 0 = 20, yes. All digits are unique, and arranged in descending order, it's 98210.Is there a way to make a six-digit number? Let's try. Starting with 9, 8, 2, 1, 0, and... I need one more digit, but I've already used 9, 8, 2, 1, 0. The remaining digits are 3, 4, 5, 6, 7. But adding any of those would make the sum exceed 20. For example, 9 + 8 + 2 + 1 + 0 + 3 = 23, which is too much. So six digits won't work.Therefore, the largest number is 98210.But wait, is there a way to rearrange the digits to get a larger number? For example, 98210 versus 98201. No, 98210 is larger because the third digit is 2 versus 2, but the fourth digit is 1 versus 0, so 98210 is larger.Alternatively, 98120, but that's smaller because the third digit is 1 versus 2.So, 98210 is indeed the largest number with unique digits adding up to 20.But wait, earlier I thought 9830 was the answer, but now I see that 98210 is larger. So I need to correct myself. The largest number is 98210.But let me think again. Is there a way to make a larger number by using different digits? For example, using 9, 7, 4, 0, and... 9 + 7 + 4 + 0 = 20, so I can't add another digit. So 9740 is a four-digit number, which is less than 98210.Alternatively, 9, 6, 5, 0, and... 9 + 6 + 5 + 0 = 20, so I can't add another digit. So 9650 is a four-digit number, less than 98210.So, yes, 98210 is the largest number.But wait, can I make a five-digit number larger than 98210? For example, 98310. Let's check the sum: 9 + 8 + 3 + 1 + 0 = 21, which is too much. So that doesn't work.What about 98301? Sum is 9 + 8 + 3 + 0 + 1 = 21, still too much.How about 98201? Sum is 9 + 8 + 2 + 0 + 1 = 20, which works. But 98201 is less than 98210 because the fourth digit is 0 versus 1.So 98210 is still the largest.Alternatively, 98120. Sum is 9 + 8 + 1 + 2 + 0 = 20, but it's less than 98210.So, after considering all possibilities, I think 98210 is the largest number with unique digits adding up to 20.

💡Alright, so I'm looking at this problem where a player's free-throw shooting results are recorded in a table. The table shows the number of shots taken, the number of makes, and the make frequency, which is just the number of makes divided by the number of shots. The question is asking me to estimate the probability that this player will make a shot at the free-throw line based on the data provided.First, I need to understand what the table is telling me. There are seven different instances where the player took a certain number of shots: 50, 100, 150, 200, 300, 400, and 500 shots. For each of these, there's a corresponding number of makes and the make frequency, which is just the makes divided by the shots.Looking at the make frequencies: 0.56, 0.49, 0.52, 0.51, 0.51, 0.52, 0.51. So, these are the probabilities of making a shot in each of these instances. The first one is 0.56, which is pretty high, but then it drops to 0.49, then goes up to 0.52, and so on.I notice that as the number of shots increases, the make frequency seems to stabilize around a certain value. The first few make frequencies are a bit all over the place—0.56, 0.49, 0.52—but then from 150 shots onward, it's 0.51, 0.51, 0.52, 0.51. It seems like after a certain point, the make frequency doesn't vary much anymore.So, if I want to estimate the player's probability of making a shot, I should probably look at the make frequencies when the number of shots is large because those are more reliable. The law of large numbers says that as the number of trials increases, the observed frequency should get closer to the true probability.In this case, the largest number of shots is 500, and the make frequency there is 0.51. The previous ones at 300, 400, and 500 are all around 0.51 or 0.52. So, it looks like the make frequency is converging to around 0.51.But just to be thorough, I should check if there's any trend or pattern in the make frequencies. Let's list them out again:- 50 shots: 0.56- 100 shots: 0.49- 150 shots: 0.52- 200 shots: 0.51- 300 shots: 0.51- 400 shots: 0.52- 500 shots: 0.51Looking at this sequence: 0.56, 0.49, 0.52, 0.51, 0.51, 0.52, 0.51. It seems like after the first two, it's hovering around 0.51 or 0.52. There's no clear upward or downward trend; it's just fluctuating slightly around that 0.51 mark.So, based on this, I think it's reasonable to estimate the player's probability of making a shot as 0.51. That seems to be the value that the make frequency is stabilizing around as the number of shots increases.But wait, let me double-check. Maybe I should calculate the overall make frequency across all shots to see if that gives me a slightly different estimate. So, total shots are 50 + 100 + 150 + 200 + 300 + 400 + 500. Let's add those up:50 + 100 = 150150 + 150 = 300300 + 200 = 500500 + 300 = 800800 + 400 = 12001200 + 500 = 1700So, total shots are 1700.Now, total makes: 28 + 49 + 78 + 102 + 153 + 208 + 255.Let's add those up:28 + 49 = 7777 + 78 = 155155 + 102 = 257257 + 153 = 410410 + 208 = 618618 + 255 = 873So, total makes are 873.Therefore, overall make frequency is 873 / 1700.Let me calculate that:873 divided by 1700.Well, 1700 goes into 873 zero times. So, 0.But to get a decimal, I can think of it as 873 divided by 1700.Alternatively, I can simplify the fraction.Divide numerator and denominator by 17:873 ÷ 17 = 51.35291700 ÷ 17 = 100So, it's approximately 51.3529 / 100 = 0.513529.So, approximately 0.5135, which is roughly 0.514.That's about 0.514, which is very close to 0.51.So, that reinforces my earlier conclusion that the make frequency is stabilizing around 0.51.But, wait, is there any reason to think that the make frequency might not stabilize? Maybe the player is improving or getting tired?Looking at the data, the make frequency doesn't show a clear upward or downward trend as the number of shots increases. It's fluctuating around 0.51. So, there's no evidence of improvement or decline; it's just consistent around 0.51.Therefore, I think it's safe to estimate the probability as 0.51.But just to be extra cautious, maybe I should look at the make frequencies for the largest number of shots, which is 500, and that's 0.51. The one before that, 400 shots, is 0.52, and 300 shots is 0.51. So, again, it's consistent around 0.51.Therefore, I'm pretty confident that the probability of the player making a shot is approximately 0.51.

💡Alright, so I need to figure out the percentage of Margie's design that is black. The design consists of multiple circles, each with a radius increasing by 2 inches. The smallest circle has a radius of 2 inches. The Asymptote code shows that there are six circles, each alternating between black and white. First, I should probably visualize the design. The Asymptote code is a bit confusing, but it seems like there are six circles, each centered at points A0 to A5, with radii decreasing from 6 to 1. But wait, the problem states that the smallest circle has a radius of 2 inches, and each subsequent circle increases by 2 inches. Hmm, that seems contradictory. Let me double-check.Wait, the problem says the smallest circle has a radius of 2 inches, and each successive circle's radius increases by 2 inches. So, the radii should be 2, 4, 6, 8, 10, and 12 inches. That makes sense. So, the largest circle has a radius of 12 inches. Looking back at the Asymptote code, it seems like the circles are being drawn with radii 6, 5, 4, 3, 2, and 1. But the problem states the radii are 2, 4, 6, 8, 10, and 12. Maybe the Asymptote code is scaled down? Or perhaps it's a different representation. Maybe I should focus on the problem statement rather than the Asymptote code.So, the design has circles with radii 2, 4, 6, 8, 10, and 12 inches. Each circle is either black or white, alternating. The smallest circle (radius 2) is black, then the next (radius 4) is white, then black, and so on. So, the colors alternate starting with black for the smallest circle.To find the percentage of the design that is black, I need to calculate the total area of the black regions and divide it by the total area of the entire design (which is the area of the largest circle, radius 12 inches).First, let's calculate the area of each circle:1. Radius 2 inches: Area = π*(2)^2 = 4π2. Radius 4 inches: Area = π*(4)^2 = 16π3. Radius 6 inches: Area = π*(6)^2 = 36π4. Radius 8 inches: Area = π*(8)^2 = 64π5. Radius 10 inches: Area = π*(10)^2 = 100π6. Radius 12 inches: Area = π*(12)^2 = 144πNow, the black regions are the areas of the black circles minus the areas of the smaller white circles inside them. Since the circles alternate, the black regions are:- The smallest circle (radius 2) is entirely black.- The circle with radius 6 is black, but the area of the circle with radius 4 (white) is subtracted.- The circle with radius 10 is black, but the area of the circle with radius 8 (white) is subtracted.So, the total black area is:1. Area of radius 2 circle: 4π2. Area of radius 6 circle minus area of radius 4 circle: 36π - 16π = 20π3. Area of radius 10 circle minus area of radius 8 circle: 100π - 64π = 36πAdding these up: 4π + 20π + 36π = 60πThe total area of the design is the area of the largest circle, which is 144π.So, the percentage of the design that is black is:(60π / 144π) * 100% = (60/144) * 100% = (5/12) * 100% ≈ 41.67%Looking at the options, 42% is the closest. So, the answer should be A) 42.Wait, but let me double-check my calculations to make sure I didn't make a mistake.Total black area:- Radius 2: 4π- Radius 6 minus 4: 36π - 16π = 20π- Radius 10 minus 8: 100π - 64π = 36πTotal: 4π + 20π + 36π = 60πTotal area: 144π60π / 144π = 5/12 ≈ 0.4167, which is approximately 41.67%, so 42% is indeed the closest.I think that's correct. So, the answer is 42.

💡First, I need to determine the total costs Burt incurred. He spent 2.00 on basil seeds and 8.00 on potting soil, which adds up to 10.00 in total costs.Next, I know that Burt made a net profit of 90.00. Net profit is calculated by subtracting the total costs from the total revenue. So, if I let R represent the total revenue, the equation would be:90 = R - 10To find R, I add 10.00 to both sides of the equation:R = 90 + 10R = 100This means Burt's total revenue from selling the basil plants is 100.00.Since Burt has 20 basil plants, I can find the selling price per plant by dividing the total revenue by the number of plants:Selling price per plant = 100 / 20Selling price per plant = 5Therefore, Burt sells each basil plant for 5.00.

💡Okay, so Rich ran a 24-mile marathon and finished it in 3 hours and 36 minutes. I need to figure out how long, on average, it took him to run each mile. Hmm, let's see. First, I think I should convert the total time he took into minutes because the question asks for the average time per mile in minutes. Alright, so 3 hours. Since there are 60 minutes in an hour, I can multiply 3 by 60 to get the total minutes for the hours part. Let me do that: 3 times 60 is 180 minutes. Then, there's an additional 36 minutes. So, I need to add those 36 minutes to the 180 minutes from the hours. That would be 180 plus 36, which equals 216 minutes in total.Now, Rich ran 24 miles in 216 minutes. To find out how long it took him to run each mile on average, I think I need to divide the total time by the number of miles. So, I'll take 216 minutes and divide that by 24 miles. Let me calculate that: 216 divided by 24. Hmm, 24 times 9 is 216, so that means 216 divided by 24 is 9. Wait, so does that mean it took Rich 9 minutes to run each mile on average? That seems pretty fast for a marathon, but maybe he's a really good runner. Let me double-check my calculations to make sure I didn't make a mistake. First, converting 3 hours to minutes: 3 times 60 is indeed 180 minutes. Adding the 36 minutes gives me 216 minutes total. Then, dividing 216 by 24 miles: 24 times 9 is 216, so yes, that's correct. I guess 9 minutes per mile is the right answer. Maybe Rich is training for something or had a particularly good day. Anyway, based on the numbers, that's the average time per mile.