💡Okay, so I need to find the probability that the sum of two fair dice is greater than seven. Hmm, let me think about how to approach this.First, I know that when you roll two dice, each die has six faces, numbered from 1 to 6. So, the total number of possible outcomes when rolling two dice is 6 times 6, which is 36. That makes sense because for each number on the first die, there are six possible numbers on the second die.Now, I need to figure out how many of these 36 outcomes result in a sum greater than seven. Maybe it would help to list out all the possible sums and count how many ways each sum can occur.Let's start by listing the possible sums and the number of ways to achieve each sum:- Sum of 2: Only one way, which is (1,1).- Sum of 3: Two ways: (1,2) and (2,1).- Sum of 4: Three ways: (1,3), (2,2), and (3,1).- Sum of 5: Four ways: (1,4), (2,3), (3,2), and (4,1).- Sum of 6: Five ways: (1,5), (2,4), (3,3), (4,2), and (5,1).- Sum of 7: Six ways: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1).- Sum of 8: Five ways: (2,6), (3,5), (4,4), (5,3), and (6,2).- Sum of 9: Four ways: (3,6), (4,5), (5,4), and (6,3).- Sum of 10: Three ways: (4,6), (5,5), and (6,4).- Sum of 11: Two ways: (5,6) and (6,5).- Sum of 12: One way: (6,6).Okay, so now I can see how many ways each sum can occur. The problem asks for the probability that the sum is greater than seven. So, I need to consider sums of 8, 9, 10, 11, and 12.Let me add up the number of ways for these sums:- Sum of 8: 5 ways- Sum of 9: 4 ways- Sum of 10: 3 ways- Sum of 11: 2 ways- Sum of 12: 1 wayAdding these together: 5 + 4 + 3 + 2 + 1 = 15.So, there are 15 favorable outcomes where the sum is greater than seven.Since there are 36 possible outcomes in total, the probability is the number of favorable outcomes divided by the total number of outcomes. That would be 15/36.Hmm, I can simplify this fraction. Both 15 and 36 are divisible by 3.15 ÷ 3 = 536 ÷ 3 = 12So, the simplified probability is 5/12.Wait, let me double-check to make sure I didn't miss any outcomes or make a counting error.Looking back at the sums greater than seven:- 8: 5 ways- 9: 4 ways- 10: 3 ways- 11: 2 ways- 12: 1 wayAdding them up: 5 + 4 is 9, plus 3 is 12, plus 2 is 14, plus 1 is 15. Yeah, that seems correct.And the total number of outcomes is definitely 36 because each die has 6 sides, and they're independent, so 6*6=36.So, 15/36 simplifies to 5/12. That seems right.Alternatively, I could have calculated the probability of the complementary event, which is the sum being less than or equal to seven, and then subtracting that probability from 1.Let's see if that gives the same result.Sum of 2: 1 waySum of 3: 2 waysSum of 4: 3 waysSum of 5: 4 waysSum of 6: 5 waysSum of 7: 6 waysAdding these up: 1 + 2 + 3 + 4 + 5 + 6 = 21.So, there are 21 outcomes where the sum is less than or equal to seven.Therefore, the probability of the sum being less than or equal to seven is 21/36.Simplifying that: 21 ÷ 3 = 7, and 36 ÷ 3 = 12, so 7/12.Then, the probability of the sum being greater than seven is 1 minus 7/12, which is 5/12.Okay, so both methods give me the same answer, 5/12. That makes me more confident that it's correct.I think I've covered all the possible sums and counted the outcomes accurately. I don't see any mistakes in my reasoning, so I'm pretty sure the probability is 5/12.

💡Alright, so I've got this problem here: "Given that the converse of the statement 'If ( m - 1 < x < m + 1 ), then ( 1 < x < 2 )' is true, find the range of values for ( m )." Hmm, okay. Let me try to unpack this step by step.First off, I need to recall what a converse of a statement is. From what I remember, the converse of an implication "If P, then Q" is "If Q, then P." So, in this case, the original statement is "If ( m - 1 < x < m + 1 ), then ( 1 < x < 2 )." Therefore, the converse should be "If ( 1 < x < 2 ), then ( m - 1 < x < m + 1 )." Got that down.Now, the problem says that this converse is true. So, I need to find all values of ( m ) such that whenever ( x ) is between 1 and 2, it's also between ( m - 1 ) and ( m + 1 ). Essentially, the interval ( (1, 2) ) must be entirely contained within the interval ( (m - 1, m + 1) ).Let me visualize this. If I have two intervals on the number line, one from ( m - 1 ) to ( m + 1 ), and another from 1 to 2, I want the second interval to be completely inside the first one. That means the left endpoint of the first interval, ( m - 1 ), has to be less than or equal to 1, and the right endpoint of the first interval, ( m + 1 ), has to be greater than or equal to 2. Because if ( m - 1 ) is less than 1, then the interval ( (m - 1, m + 1) ) starts before 1, and if ( m + 1 ) is greater than 2, it ends after 2. So, the entire ( (1, 2) ) interval is inside ( (m - 1, m + 1) ).Let me write that down as inequalities:1. ( m - 1 leq 1 )2. ( m + 1 geq 2 )Now, solving these inequalities for ( m ):Starting with the first inequality:( m - 1 leq 1 )Add 1 to both sides:( m leq 2 )Okay, that's straightforward. Now, the second inequality:( m + 1 geq 2 )Subtract 1 from both sides:( m geq 1 )So, combining these two results, we have:( 1 leq m leq 2 )Hmm, so ( m ) must be between 1 and 2, inclusive. That makes sense because if ( m ) is exactly 1, then the interval ( (m - 1, m + 1) ) becomes ( (0, 2) ), which still contains ( (1, 2) ). Similarly, if ( m ) is exactly 2, the interval becomes ( (1, 3) ), which also contains ( (1, 2) ).Wait a second, let me double-check that. If ( m = 1 ), then ( m - 1 = 0 ) and ( m + 1 = 2 ), so the interval is ( (0, 2) ). Since ( 1 < x < 2 ) is entirely within ( (0, 2) ), that works. If ( m = 2 ), then ( m - 1 = 1 ) and ( m + 1 = 3 ), so the interval is ( (1, 3) ). Again, ( 1 < x < 2 ) is entirely within ( (1, 3) ), so that also works.What if ( m ) is less than 1? Let's say ( m = 0.5 ). Then, ( m - 1 = -0.5 ) and ( m + 1 = 1.5 ). So, the interval is ( (-0.5, 1.5) ). But ( 1 < x < 2 ) is not entirely within ( (-0.5, 1.5) ) because part of it (from 1.5 to 2) is outside. So, that doesn't satisfy the converse statement.Similarly, if ( m ) is greater than 2, say ( m = 3 ), then ( m - 1 = 2 ) and ( m + 1 = 4 ). The interval is ( (2, 4) ). But ( 1 < x < 2 ) is not entirely within ( (2, 4) ) because the entire interval ( (1, 2) ) is to the left of 2. So, that also doesn't satisfy the converse.Therefore, it seems that ( m ) must be between 1 and 2, inclusive, for the converse statement to hold true.Just to be thorough, let me consider the endpoints. If ( m = 1 ), as I did before, the interval is ( (0, 2) ), which contains ( (1, 2) ). If ( m = 2 ), the interval is ( (1, 3) ), which also contains ( (1, 2) ). So, including the endpoints is correct.Is there any other value of ( m ) that could possibly work? Let's see. If ( m ) is exactly 1.5, then ( m - 1 = 0.5 ) and ( m + 1 = 2.5 ). The interval ( (0.5, 2.5) ) certainly contains ( (1, 2) ). So, that works too.What if ( m ) is a fraction between 1 and 2, like ( m = 1.2 )? Then, ( m - 1 = 0.2 ) and ( m + 1 = 2.2 ). The interval ( (0.2, 2.2) ) contains ( (1, 2) ). Yep, that works.Similarly, if ( m = 1.9 ), ( m - 1 = 0.9 ) and ( m + 1 = 2.9 ). The interval ( (0.9, 2.9) ) contains ( (1, 2) ). So, that's good.I think I've covered all the cases. Whether ( m ) is exactly 1, exactly 2, or somewhere in between, as long as it's within that range, the converse statement holds true. If ( m ) is outside of that range, either less than 1 or greater than 2, the converse statement doesn't hold because the interval ( (1, 2) ) wouldn't be entirely contained within ( (m - 1, m + 1) ).So, putting it all together, the range of values for ( m ) is from 1 to 2, inclusive. Therefore, ( m ) must satisfy ( 1 leq m leq 2 ).**Final Answer**The range of values for ( m ) is boxed{[1, 2]}.

💡Okay, so I have this problem here where I'm given positive integers (1 < k leq n) and positive real numbers (x_1, x_2, ldots, x_k) such that their product equals their sum. That is,[x_1 cdot x_2 cdot ldots cdot x_k = x_1 + x_2 + ldots + x_k.]I need to show two things: first, that the sum of each (x_i) raised to the power (n-1) is at least (kn), and second, find all cases where equality holds.Starting with part (a), I think I should use some inequality principles here. The condition given relates the product and the sum of the variables, which makes me think of the AM-GM inequality. AM-GM often comes into play when dealing with sums and products.So, recalling the AM-GM inequality, for positive real numbers, the arithmetic mean is always greater than or equal to the geometric mean. Specifically,[frac{x_1 + x_2 + ldots + x_k}{k} geq sqrt[k]{x_1 x_2 ldots x_k}.]But in our case, we know that the product of the (x_i)s equals their sum. Let me denote the sum as (S = x_1 + x_2 + ldots + x_k). Then, the product is also (S), so we have:[sqrt[k]{S} leq frac{S}{k}.]Wait, that seems a bit off. Let me double-check. If I apply AM-GM to the (x_i)s, it's:[frac{S}{k} geq sqrt[k]{S}.]So, that simplifies to:[frac{S}{k} geq S^{1/k}.]Hmm, that's interesting. Let me raise both sides to the power of (k) to eliminate the root:[left(frac{S}{k}right)^k geq S.]Simplifying the left side:[frac{S^k}{k^k} geq S.]Multiply both sides by (k^k) (which is positive, so inequality remains the same):[S^k geq S cdot k^k.]Assuming (S neq 0) (which it isn't because all (x_i) are positive), we can divide both sides by (S):[S^{k-1} geq k^k.]Taking the ((k-1))-th root of both sides:[S geq k^{frac{k}{k-1}}.]So, the sum (S) is at least (k^{frac{k}{k-1}}). That's a useful inequality.Now, moving on to the expression we need to bound: (x_1^{n-1} + x_2^{n-1} + ldots + x_k^{n-1}). I need to show this is at least (kn).I wonder if I can apply another inequality here, maybe Hölder's inequality or Power Mean. Let me think about Power Mean. The Power Mean inequality states that for positive real numbers and exponents (r > s),[left(frac{x_1^r + x_2^r + ldots + x_k^r}{k}right)^{1/r} geq left(frac{x_1^s + x_2^s + ldots + x_k^s}{k}right)^{1/s}.]If I set (r = n-1) and (s = 1), then since (n-1 > 1) (because (n > k > 1)), we have:[left(frac{x_1^{n-1} + x_2^{n-1} + ldots + x_k^{n-1}}{k}right)^{1/(n-1)} geq left(frac{x_1 + x_2 + ldots + x_k}{k}right)^{1/1}.]Simplifying, this becomes:[left(frac{sum x_i^{n-1}}{k}right)^{1/(n-1)} geq frac{S}{k}.]Raising both sides to the power of (n-1):[frac{sum x_i^{n-1}}{k} geq left(frac{S}{k}right)^{n-1}.]Multiplying both sides by (k):[sum x_i^{n-1} geq k left(frac{S}{k}right)^{n-1}.]So, now I have:[sum x_i^{n-1} geq k left(frac{S}{k}right)^{n-1}.]But earlier, I found that (S geq k^{frac{k}{k-1}}). Let me substitute that into the inequality:[sum x_i^{n-1} geq k left(frac{k^{frac{k}{k-1}}}{k}right)^{n-1} = k left(k^{frac{k}{k-1} - 1}right)^{n-1}.]Simplifying the exponent:[frac{k}{k-1} - 1 = frac{k - (k - 1)}{k - 1} = frac{1}{k - 1}.]So, we have:[sum x_i^{n-1} geq k left(k^{frac{1}{k - 1}}right)^{n - 1} = k cdot k^{frac{n - 1}{k - 1}}.]Hmm, this is getting a bit complicated. I need to show that this is at least (kn). So, let's see:[k cdot k^{frac{n - 1}{k - 1}} geq kn.]Dividing both sides by (k):[k^{frac{n - 1}{k - 1}} geq n.]Taking natural logarithm on both sides:[frac{n - 1}{k - 1} ln k geq ln n.]Simplify:[(n - 1) ln k geq (k - 1) ln n.]Hmm, not sure if this is necessarily true for all (1 < k leq n). Maybe I took a wrong turn somewhere.Let me backtrack. Maybe instead of using Power Mean, I should consider another approach. Since the product of the (x_i)s equals their sum, perhaps I can use substitution or Lagrange multipliers to find the minimum of the sum of (x_i^{n-1}).Alternatively, maybe I can use the AM-GM inequality directly on the terms (x_i^{n-1}). Let's see:By AM-GM,[frac{x_1^{n-1} + x_2^{n-1} + ldots + x_k^{n-1}}{k} geq sqrt[k]{x_1^{n-1} x_2^{n-1} ldots x_k^{n-1}} = left(prod x_iright)^{frac{n - 1}{k}}.]But we know that (prod x_i = S), so:[frac{sum x_i^{n-1}}{k} geq S^{frac{n - 1}{k}}.]Thus,[sum x_i^{n-1} geq k S^{frac{n - 1}{k}}.]Now, from earlier, we have (S geq k^{frac{k}{k - 1}}). So,[S^{frac{n - 1}{k}} geq left(k^{frac{k}{k - 1}}right)^{frac{n - 1}{k}} = k^{frac{n - 1}{k - 1}}.]Therefore,[sum x_i^{n-1} geq k cdot k^{frac{n - 1}{k - 1}} = k^{frac{n - 1}{k - 1} + 1}.]Simplify the exponent:[frac{n - 1}{k - 1} + 1 = frac{n - 1 + k - 1}{k - 1} = frac{n + k - 2}{k - 1}.]So,[sum x_i^{n-1} geq k^{frac{n + k - 2}{k - 1}}.]Hmm, I need this to be at least (kn). So,[k^{frac{n + k - 2}{k - 1}} geq kn.]Taking logarithms again:[frac{n + k - 2}{k - 1} ln k geq ln k + ln n.]Simplify:[frac{n + k - 2}{k - 1} ln k - ln k geq ln n.]Factor out (ln k):[left(frac{n + k - 2}{k - 1} - 1right) ln k geq ln n.]Simplify the fraction:[frac{n + k - 2 - (k - 1)}{k - 1} = frac{n - 1}{k - 1}.]So,[frac{n - 1}{k - 1} ln k geq ln n.]Which is the same inequality as before. So, I end up at the same point. Maybe I need a different approach.Let me consider specific cases to get some intuition. Suppose (k = 2). Then, the condition becomes (x_1 x_2 = x_1 + x_2). Let me solve for (x_2) in terms of (x_1):[x_1 x_2 = x_1 + x_2 implies x_2(x_1 - 1) = x_1 implies x_2 = frac{x_1}{x_1 - 1}.]So, (x_2) is expressed in terms of (x_1). Now, the expression we need to bound is (x_1^{n-1} + x_2^{n-1}). Let's substitute (x_2):[x_1^{n-1} + left(frac{x_1}{x_1 - 1}right)^{n-1}.]I need to show that this is at least (2n). Maybe I can set (x_1 = t) and analyze the function (f(t) = t^{n-1} + left(frac{t}{t - 1}right)^{n-1}) for (t > 1).Taking derivative to find minimum might be complicated, but perhaps I can test (t = 2):[f(2) = 2^{n-1} + left(frac{2}{1}right)^{n-1} = 2^{n-1} + 2^{n-1} = 2^n.]But (2^n) is greater than (2n) for (n geq 2), which is true since (k = 2 leq n). So, in this case, the inequality holds.What if (t) approaches 1 from above? Then (x_2) approaches infinity, so (x_2^{n-1}) also approaches infinity, making the sum go to infinity, which is certainly greater than (2n).If (t) approaches infinity, then (x_2 = frac{t}{t - 1} approx 1), so (x_1^{n-1}) approaches infinity, again making the sum go to infinity.So, for (k = 2), the inequality seems to hold. Maybe the minimum occurs somewhere in between. Let me try (t = phi) where (phi) is the golden ratio, approximately 1.618. But this might not lead me anywhere.Alternatively, perhaps the minimum occurs when all (x_i) are equal. Let me assume (x_1 = x_2 = ldots = x_k = x). Then, the condition becomes:[x^k = kx implies x^{k-1} = k implies x = k^{1/(k-1)}.]So, each (x_i = k^{1/(k-1)}). Then, the sum we need is:[k cdot left(k^{1/(k-1)}right)^{n-1} = k cdot k^{(n - 1)/(k - 1)} = k^{1 + (n - 1)/(k - 1)}.]Simplify the exponent:[1 + frac{n - 1}{k - 1} = frac{(k - 1) + (n - 1)}{k - 1} = frac{n + k - 2}{k - 1}.]So, the sum is (k^{(n + k - 2)/(k - 1)}). Earlier, I had that (sum x_i^{n-1} geq k^{(n + k - 2)/(k - 1)}), so equality holds when all (x_i) are equal. Therefore, the minimum is achieved when all (x_i) are equal to (k^{1/(k-1)}).But wait, I need to show that this minimum is at least (kn). So,[k^{(n + k - 2)/(k - 1)} geq kn.]Taking logarithms:[frac{n + k - 2}{k - 1} ln k geq ln k + ln n.]Simplify:[frac{n + k - 2}{k - 1} ln k - ln k geq ln n.]Factor out (ln k):[left(frac{n + k - 2}{k - 1} - 1right) ln k geq ln n.]Simplify the fraction:[frac{n + k - 2 - (k - 1)}{k - 1} = frac{n - 1}{k - 1}.]So,[frac{n - 1}{k - 1} ln k geq ln n.]This is the same inequality as before. So, for the equality case, we need:[frac{n - 1}{k - 1} ln k = ln n.]Which can be rewritten as:[left(frac{n - 1}{k - 1}right) ln k = ln n.]Exponentiating both sides:[k^{frac{n - 1}{k - 1}} = n.]So, equality holds when (k^{frac{n - 1}{k - 1}} = n). Let me see for which (k) and (n) this is true.Suppose (k = 2). Then,[2^{frac{n - 1}{1}} = n implies 2^{n - 1} = n.]Solving for (n), we can test small integers:- (n = 1): (2^{0} = 1), which is true, but (k = 2 > n = 1), which violates (k leq n).- (n = 2): (2^{1} = 2), which is true.- (n = 3): (2^{2} = 4 neq 3).- (n = 4): (2^{3} = 8 neq 4).So, only (n = 2) works for (k = 2).Similarly, for (k = 3):[3^{frac{n - 1}{2}} = n.]Testing small (n):- (n = 3): (3^{1} = 3), which is true.- (n = 4): (3^{1.5} approx 5.196 neq 4).- (n = 5): (3^{2} = 9 neq 5).So, only (n = 3) works for (k = 3).Continuing this pattern, it seems that for each (k), the equality holds when (n = k). Let me check:If (n = k), then:[k^{frac{k - 1}{k - 1}} = k^1 = k = n.]Yes, that works. So, equality holds when (n = k).Wait, but earlier for (k = 2), (n = 2) worked, and for (k = 3), (n = 3) worked. So, in general, equality holds when (n = k), and all (x_i = k^{1/(k-1)}).But let me verify this. If (n = k), then the inequality we need to show is:[sum x_i^{k-1} geq k cdot k = k^2.]But from the equality case, when all (x_i = k^{1/(k-1)}), the sum is:[k cdot left(k^{1/(k-1)}right)^{k-1} = k cdot k = k^2.]So, equality holds when (n = k) and all (x_i = k^{1/(k-1)}).But what about when (n > k)? For example, if (k = 2) and (n = 3), does equality hold? Earlier, when (k = 2) and (n = 3), we saw that (2^{2} = 4 neq 3), so equality doesn't hold. Therefore, equality only holds when (n = k).So, putting it all together:For part (a), using AM-GM and Power Mean inequalities, we can show that:[sum_{i=1}^k x_i^{n-1} geq kn,]with equality if and only if (n = k) and all (x_i = k^{1/(k-1)}).I think this makes sense. I went through specific cases, applied inequalities, and checked the equality conditions. It seems consistent.

💡Okay, so I have this problem about balls of different colors in a box, and I need to figure out the minimum number of balls I must draw without replacement to guarantee that at least 18 balls of one color are drawn. Let me try to break this down step by step.First, let me list out the number of balls of each color in the box:- Red: 30- Green: 25- Yellow: 20- Blue: 15- White: 10- Black: 5So, there are six different colors with varying quantities. The question is asking for the minimum number of balls I need to draw to ensure that I have at least 18 balls of a single color. Hmm, this sounds like a problem that involves the pigeonhole principle, where I have to consider the worst-case scenario.The pigeonhole principle basically says that if you have more pigeons than pigeonholes, at least one pigeonhole must contain more than one pigeon. Translating that into this problem, if I draw a certain number of balls, I want to make sure that at least one color has 18 balls. So, I need to consider the worst-case scenario where I draw as many balls as possible without getting 18 of any single color, and then add one more ball to ensure that I cross that threshold.Let me think about how that would work. For each color, I want to draw as many balls as possible without reaching 18. So, for red, green, and yellow, which have more than 18 balls, I can draw up to 17 balls each without hitting 18. For blue, white, and black, there are fewer than 18 balls, so I can draw all of them without reaching 18.Let me write that down:- Red: 17 balls- Green: 17 balls- Yellow: 17 balls- Blue: 15 balls (since there are only 15)- White: 10 balls (since there are only 10)- Black: 5 balls (since there are only 5)Now, if I add all these up, that should give me the maximum number of balls I can draw without having 18 of any single color. Let me calculate that:17 (red) + 17 (green) + 17 (yellow) + 15 (blue) + 10 (white) + 5 (black) = ?Let me add them step by step:17 + 17 = 3434 + 17 = 5151 + 15 = 6666 + 10 = 7676 + 5 = 81So, 81 balls can be drawn without necessarily having 18 of any single color. That means if I draw 81 balls, it's possible that I have 17 red, 17 green, 17 yellow, 15 blue, 10 white, and 5 black balls. In this case, I still don't have 18 of any single color.But the question is asking for the minimum number of balls needed to guarantee that I have at least 18 of a single color. So, if I draw one more ball after 81, that would be the 82nd ball. At this point, no matter which color that 82nd ball is, it has to push one of the colors over the 18 threshold.Wait, let me make sure about that. If I already have 17 red, 17 green, and 17 yellow, and I draw one more, it could be red, green, or yellow, making it 18 of that color. If I have already drawn all the blue, white, and black balls, then the next ball has to be either red, green, or yellow. So, yes, the 82nd ball would definitely give me 18 of one color.But hold on, let me double-check my calculations to make sure I didn't make a mistake. So, 17 red + 17 green + 17 yellow + 15 blue + 10 white + 5 black equals 81. Yes, that seems right. So, 81 is the maximum number of balls you can draw without having 18 of any single color. Therefore, drawing one more ball, which is 82, would ensure that you have at least 18 of one color.Is there any other color that could interfere with this? For example, what if the 82nd ball is blue, white, or black? But wait, there are only 15 blue, 10 white, and 5 black balls. So, in the worst-case scenario, you've already drawn all of them, so the next ball has to be red, green, or yellow. Therefore, it's impossible for the 82nd ball to be blue, white, or black because you've already drawn all of them. So, the 82nd ball has to be red, green, or yellow, each of which already has 17 balls drawn. Therefore, adding one more would make it 18.So, I think my reasoning is correct. The minimum number of balls that must be drawn to guarantee at least 18 balls of a single color is 82.But just to be thorough, let me consider if there's any other way this could go. Suppose I have a different distribution of balls. For example, what if I drew more of one color and less of another? But in the worst-case scenario, we have to assume that we're trying to avoid getting 18 of any color as much as possible. So, the way to do that is to draw as many as possible without reaching 18 in any color, which is exactly what I did.Therefore, I'm confident that 82 is the correct answer.**Final Answer**boxed{C}

💡Okay, so I've got this math problem here with four options: A, B, C, and D. I need to figure out which one is correct. Let me take it step by step.Starting with option A: (a^{2} cdot a^{3} = a^{5}). Hmm, I remember that when you multiply exponents with the same base, you add the exponents. So, (a^{2} cdot a^{3}) should be (a^{2+3}), which is (a^{5}). That seems right. So, A looks correct.Moving on to option B: ((a^{3})^{2} = a^{5}). I think when you raise a power to another power, you multiply the exponents. So, ((a^{3})^{2}) should be (a^{3 cdot 2}), which is (a^{6}). But the option says it's (a^{5}), so that doesn't match. Therefore, B is incorrect.Next is option C: ((2a)^{5} = 10a^{5}). Okay, when you have a product inside parentheses raised to a power, both the coefficient and the variable get raised to that power. So, ((2a)^{5}) should be (2^{5} cdot a^{5}). Calculating (2^{5}), that's 32, so it should be (32a^{5}). But the option says it's (10a^{5}), which is wrong. So, C is incorrect.Finally, option D: (a^{4} + a^{4} = a^{8}). Adding like terms, right? (a^{4} + a^{4}) should be (2a^{4}), not (a^{8}). So, D is also incorrect.Putting it all together, only option A is correct.

💡First, I need to determine how many times each team faces the other teams in the league. There are 15 teams in total.If each team plays every other team once, the total number of unique games can be calculated using the combination formula ( frac{n(n-1)}{2} ), where ( n ) is the number of teams. Plugging in the numbers:[frac{15 times 14}{2} = 105 text{ games}]However, the total number of games played in the season is 1050. To find out how many times each pair of teams plays against each other, I divide the total number of games by the number of unique pairings:[frac{1050}{105} = 10]Therefore, each team faces every other team 10 times during the season.