💡Okay, so I have this problem here about Tom's age and his four children's ages. Let me try to understand it step by step. First, it says Tom's age is T years, and that's also the sum of the ages of his four children. So, right now, Tom is T years old, and if I add up the ages of all four of his kids, that total is also T. That means Tom is as old as the combined ages of his four children. Interesting.Then, the problem says that N years ago, Tom's age was three times the sum of his children's ages at that time. So, if I go back N years, Tom was T - N years old, and the sum of his children's ages back then was... Hmm, each child would have been N years younger, right? So, if the total sum of their ages now is T, then N years ago, each child was N years younger, so the total sum of their ages would have been T - 4N. Because there are four children, each N years younger, so 4 times N subtracted from the total.So, N years ago, Tom's age was T - N, and the sum of his children's ages was T - 4N. According to the problem, at that time, Tom's age was three times the sum of his children's ages. So, mathematically, that would be:T - N = 3 * (T - 4N)Alright, so I can set up this equation and solve for T/N. Let's do that step by step.First, expand the right side:T - N = 3T - 12NNow, let's get all the T terms on one side and the N terms on the other. Subtract T from both sides:-T - N = -12NWait, that doesn't look right. Let me check my steps again. Starting with:T - N = 3T - 12NIf I subtract T from both sides:-T - N = -12NThen, add 12N to both sides:-T + 11N = 0Which simplifies to:-T = -11NMultiply both sides by -1:T = 11NSo, T is equal to 11N. Therefore, the ratio T/N is 11/1, which is 11. But wait, that's not one of the options. Let me check my calculations again.Wait, maybe I made a mistake when moving terms. Let me go back.Starting equation:T - N = 3(T - 4N)Expand the right side:T - N = 3T - 12NNow, subtract T from both sides:-T - N = -12NThen, add 12N to both sides:-T + 11N = 0So, -T + 11N = 0Which means T = 11NSo, T/N = 11/1 = 11Hmm, but the options given are 7/2, 9/2, 11/2, 13/2, and 6. So, 11 isn't an option. Did I make a mistake somewhere?Wait a second, maybe I misinterpreted the sum of the children's ages. Let me think again.The problem says Tom's age is T, which is also the sum of the ages of his four children. So, currently, Tom is T years old, and the sum of his four children's ages is also T.N years ago, Tom was T - N years old, and each child was N years younger. So, the sum of the children's ages N years ago would be T - 4N, because each of the four children was N years younger.So, the equation should be:T - N = 3*(T - 4N)Let me solve this again carefully.T - N = 3T - 12NSubtract T from both sides:-T - N = -12NAdd 12N to both sides:-T + 11N = 0So, T = 11NTherefore, T/N = 11/1 = 11But 11 isn't an option. Wait, maybe I need to express T/N as a fraction over 2? Let me see.Wait, perhaps I made a mistake in setting up the equation. Let me check again.Tom's current age: TSum of children's current ages: TN years ago, Tom's age: T - NN years ago, each child was N years younger, so sum of children's ages then: T - 4NGiven that T - N = 3*(T - 4N)So, T - N = 3T - 12NSubtract T from both sides:-N = 2T - 12NAdd 12N to both sides:11N = 2TSo, 2T = 11NTherefore, T/N = 11/2Ah! There we go. I see where I went wrong earlier. When I subtracted T from both sides, I should have gotten:T - N - T = 3T - 12N - TWhich simplifies to:-N = 2T - 12NThen, adding 12N to both sides:11N = 2TSo, T/N = 11/2That makes sense now. So, the ratio T/N is 11/2, which is option C.I think I initially missed the step where I had to subtract T from both sides correctly, leading me to an incorrect conclusion. But after rechecking, I realized the correct ratio is 11/2.

💡Alright, so I have this problem where a shooter is firing at three targets until all are hit. The probability of hitting a target with one shot is ( p ). I need to find two things: a) The probability that exactly 5 shots are required.b) The expected number of shots.Okay, let's start with part (a). I need to figure out the probability that exactly 5 shots are needed to hit all three targets. Hmm, so that means that in the first 4 shots, the shooter must have hit exactly 2 targets, and then on the 5th shot, they hit the last remaining target.So, I think this is a problem that involves combinations and the binomial probability formula. The binomial formula gives the probability of having exactly ( k ) successes in ( n ) trials, which is:[P(k) = binom{n}{k} p^k (1 - p)^{n - k}]In this case, for the first 4 shots, we need exactly 2 hits. So, ( n = 4 ), ( k = 2 ). That would give me the probability of hitting 2 targets in 4 shots. Then, I need to multiply that by the probability of hitting the last target on the 5th shot, which is ( p ).So, let's compute that. First, the combination part:[binom{4}{2} = frac{4!}{2! cdot (4 - 2)!} = frac{24}{2 cdot 2} = 6]Okay, so there are 6 ways to choose which 2 out of the 4 shots are hits. Then, the probability of hitting exactly 2 targets in 4 shots is:[6 cdot p^2 cdot (1 - p)^{2}]Now, since the 5th shot must be a hit, we multiply this by ( p ):[6 cdot p^2 cdot (1 - p)^{2} cdot p = 6 cdot p^3 cdot (1 - p)^2]So, that should be the probability that exactly 5 shots are required. Let me just make sure I didn't miss anything. We're assuming that each shot is independent, which I think is a fair assumption unless stated otherwise. Also, the probability of hitting each target is the same, which is given as ( p ). Yeah, that seems right.Now, moving on to part (b), finding the expected number of shots. Hmm, expectation problems can sometimes be tricky, but I remember that for geometric distributions, the expected number of trials to get the first success is ( frac{1}{p} ). But in this case, we have three targets, so it's like we need three successes. I think this is similar to the coupon collector problem, where you want to collect all three coupons, and each trial gives you a coupon with probability ( p ). The expected number of trials to collect all coupons is the sum of the expected times to collect each new coupon.So, for the first target, the expected number of shots to hit it is ( frac{1}{p} ). Once we've hit one target, we still have two left. The probability of hitting a new target is still ( p ), so the expected number of shots to hit the second target is also ( frac{1}{p} ). Similarly, for the third target, it's ( frac{1}{p} ).Therefore, the total expected number of shots is:[frac{1}{p} + frac{1}{p} + frac{1}{p} = frac{3}{p}]Wait, is that correct? I feel like I might be oversimplifying. Let me think again. Each time, the probability of hitting a new target is ( p ), regardless of how many targets are left. So, yes, each time, it's a geometric distribution with expectation ( frac{1}{p} ). So, adding them up for three targets gives ( frac{3}{p} ).But hold on, is there a different way to model this? Maybe using linearity of expectation. Suppose I define indicator variables for each shot, indicating whether that shot was the one that hit a new target. But that might complicate things more. I think the approach of summing the expectations for each target is valid because each target is independent in terms of the probability of being hit.Yeah, I think that's right. So, the expected number of shots is ( frac{3}{p} ).Let me just recap:a) The probability that exactly 5 shots are required is ( 6p^3(1 - p)^2 ).b) The expected number of shots is ( frac{3}{p} ).I think that makes sense. I don't see any mistakes in the reasoning. Each part was approached step by step, considering the necessary probabilities and expectations, and the results seem consistent with what I know about geometric distributions and the coupon collector problem.**Final Answer**a) The probability that exactly 5 shots are required is (boxed{6 p^3 (1 - p)^2}).b) The expected number of shots is (boxed{dfrac{3}{p}}).

💡Okay, so I have this function f(x) where f(1) equals 1, and there's this equation f(x + y) = f(x) + f(y) + 8xy - 2. Also, it's given that f(7) is 163, and I need to find f(6). Hmm, let me think about how to approach this.First, I know that functional equations often require plugging in specific values for x and y to find a pattern or a general form of the function. Since I have f(1) = 1, maybe I can use that to find other values step by step.Let me try plugging in x = 1 and y = 1 into the given equation. That should give me f(2):f(1 + 1) = f(1) + f(1) + 8*1*1 - 2f(2) = 1 + 1 + 8 - 2f(2) = 8Okay, so f(2) is 8. Now, let's find f(3). I can use x = 2 and y = 1:f(2 + 1) = f(2) + f(1) + 8*2*1 - 2f(3) = 8 + 1 + 16 - 2f(3) = 23Wait, that seems a bit high. Let me double-check my calculations:f(3) = f(2) + f(1) + 8*2*1 - 2f(3) = 8 + 1 + 16 - 2f(3) = 23Yeah, that's correct. Maybe the function grows quickly because of the 8xy term.Let's try f(4). Using x = 3 and y = 1:f(3 + 1) = f(3) + f(1) + 8*3*1 - 2f(4) = 23 + 1 + 24 - 2f(4) = 46Hmm, f(4) is 46. Let's see f(5):f(4 + 1) = f(4) + f(1) + 8*4*1 - 2f(5) = 46 + 1 + 32 - 2f(5) = 77Okay, f(5) is 77. Now, f(6):f(5 + 1) = f(5) + f(1) + 8*5*1 - 2f(6) = 77 + 1 + 40 - 2f(6) = 116Wait a minute, but the problem also gives f(7) = 163. Let me check if my f(6) leads to f(7) correctly:f(6 + 1) = f(6) + f(1) + 8*6*1 - 2f(7) = 116 + 1 + 48 - 2f(7) = 163Yes, that matches the given f(7) = 163. So, f(6) must be 116.Alternatively, maybe there's a general formula for f(x). Let me see if I can find a pattern or derive a formula.Looking at the values:f(1) = 1f(2) = 8f(3) = 23f(4) = 46f(5) = 77f(6) = 116f(7) = 163Let me see the differences between consecutive terms:f(2) - f(1) = 8 - 1 = 7f(3) - f(2) = 23 - 8 = 15f(4) - f(3) = 46 - 23 = 23f(5) - f(4) = 77 - 46 = 31f(6) - f(5) = 116 - 77 = 39f(7) - f(6) = 163 - 116 = 47So the differences are 7, 15, 23, 31, 39, 47. These seem to be increasing by 8 each time. That makes sense because the functional equation has an 8xy term, which would contribute to the differences increasing linearly.If I think about the function f(x), it might be a quadratic function because the differences of differences are constant (second differences are constant in quadratic functions). Let's assume f(x) is a quadratic function of the form f(x) = ax² + bx + c.We can use the known values to set up equations and solve for a, b, and c.Using f(1) = 1:a(1)² + b(1) + c = 1a + b + c = 1 ...(1)Using f(2) = 8:a(2)² + b(2) + c = 84a + 2b + c = 8 ...(2)Using f(3) = 23:a(3)² + b(3) + c = 239a + 3b + c = 23 ...(3)Now, let's subtract equation (1) from equation (2):(4a + 2b + c) - (a + b + c) = 8 - 13a + b = 7 ...(4)Subtract equation (2) from equation (3):(9a + 3b + c) - (4a + 2b + c) = 23 - 85a + b = 15 ...(5)Now, subtract equation (4) from equation (5):(5a + b) - (3a + b) = 15 - 72a = 8a = 4Now plug a = 4 into equation (4):3*4 + b = 712 + b = 7b = 7 - 12b = -5Now plug a = 4 and b = -5 into equation (1):4 - 5 + c = 1-1 + c = 1c = 2So, the function is f(x) = 4x² - 5x + 2.Let's verify this with f(4):f(4) = 4*(4)² - 5*4 + 2 = 4*16 - 20 + 2 = 64 - 20 + 2 = 46, which matches.Similarly, f(5) = 4*25 - 25 + 2 = 100 - 25 + 2 = 77, which matches.f(6) = 4*36 - 30 + 2 = 144 - 30 + 2 = 116, which is what I found earlier.And f(7) = 4*49 - 35 + 2 = 196 - 35 + 2 = 163, which matches the given value.So, the general formula is f(x) = 4x² - 5x + 2, and thus f(6) is 116.

💡Let ( x ) be the initial number of candies Tracy had.1. **Tracy ate ( frac{1}{4} ) of the candies:** [ text{Remaining candies} = x - frac{1}{4}x = frac{3}{4}x ] Since ( frac{3}{4}x ) must be an integer, ( x ) is divisible by 4.2. **Tracy gave ( frac{1}{3} ) of the remaining candies to Rachel:** [ text{Remaining candies} = frac{3}{4}x - frac{1}{3} cdot frac{3}{4}x = frac{2}{3} cdot frac{3}{4}x = frac{1}{2}x ] Since ( frac{1}{2}x ) is also an integer, ( x ) must be divisible by 2.3. **Tracy and her mom each ate 12 candies:** [ text{Remaining candies} = frac{1}{2}x - 12 - 12 = frac{1}{2}x - 24 ]4. **Her brother took between 2 to 6 candies, leaving Tracy with 5 candies:** [ 7 leq frac{1}{2}x - 24 leq 11 ] [ 31 leq frac{1}{2}x leq 35 ] [ 62 leq x leq 70 ] Since ( x ) must be divisible by 4, the possible values are 64 and 68.5. **Checking both values:** - If ( x = 64 ): [ frac{1}{2}x - 24 = 8 ] Subtracting between 2 to 6 candies leaves 2 to 6 candies, which is inconsistent with having exactly 5. - If ( x = 68 ): [ frac{1}{2}x - 24 = 10 ] Subtracting between 2 to 6 candies leaves 4 to 8 candies, which includes the possibility of having exactly 5.Thus, Tracy initially had (boxed{68}) candies.

💡First, I recognize that Dorothy is currently 15 years old and three times as old as her sister. This means her sister is 5 years old.Next, I need to determine how many years, ( Y ), it will take for Dorothy to be twice as old as her sister. In ( Y ) years, Dorothy will be ( 15 + Y ) years old, and her sister will be ( 5 + Y ) years old.Setting up the equation for when Dorothy is twice as old as her sister:[15 + Y = 2(5 + Y)]Solving for ( Y ):[15 + Y = 10 + 2Y][15 - 10 = 2Y - Y][5 = Y]Therefore, in 5 years, Dorothy will be twice as old as her sister.

💡Alright, so I have this function here: ( f(x) = log_{2}(2sin x - 1) + sqrt{sqrt{2} + 2cos x} ). I need to find the domain of this function. Hmm, okay. The domain of a function is basically all the real numbers x for which the function is defined. Since this function has two parts, a logarithm and a square root, I need to make sure both parts are defined. Let me start by recalling that the logarithm function, ( log_{a}(b) ), is defined only when the argument ( b ) is positive. So, for ( log_{2}(2sin x - 1) ) to be defined, the expression inside the log, which is ( 2sin x - 1 ), must be greater than zero. That gives me the first condition: ( 2sin x - 1 > 0 ).Next, the square root function, ( sqrt{c} ), is defined when the expression inside the square root, ( c ), is greater than or equal to zero. So, for ( sqrt{sqrt{2} + 2cos x} ) to be defined, the expression inside the square root, ( sqrt{2} + 2cos x ), must be greater than or equal to zero. That gives me the second condition: ( sqrt{2} + 2cos x geq 0 ).So, to find the domain of ( f(x) ), I need to solve these two inequalities simultaneously:1. ( 2sin x - 1 > 0 )2. ( sqrt{2} + 2cos x geq 0 )Let me tackle the first inequality: ( 2sin x - 1 > 0 ). Simplifying this, I get ( 2sin x > 1 ), which means ( sin x > frac{1}{2} ). I know that the sine function is greater than ( frac{1}{2} ) in specific intervals. The sine function is positive in the first and second quadrants, so between ( 0 ) and ( pi ). Specifically, ( sin x = frac{1}{2} ) at ( x = frac{pi}{6} ) and ( x = frac{5pi}{6} ) in the interval ( [0, 2pi] ). Therefore, ( sin x > frac{1}{2} ) when ( x ) is between ( frac{pi}{6} ) and ( frac{5pi}{6} ). Since the sine function is periodic with period ( 2pi ), this interval repeats every ( 2pi ). So, in general, the solution for the first inequality is ( frac{pi}{6} + 2kpi < x < frac{5pi}{6} + 2kpi ) for any integer ( k ).Now, moving on to the second inequality: ( sqrt{2} + 2cos x geq 0 ). Let me simplify this. Subtracting ( sqrt{2} ) from both sides gives ( 2cos x geq -sqrt{2} ). Dividing both sides by 2, I get ( cos x geq -frac{sqrt{2}}{2} ). I know that ( cos x = -frac{sqrt{2}}{2} ) at ( x = frac{3pi}{4} ) and ( x = frac{5pi}{4} ) in the interval ( [0, 2pi] ). The cosine function is greater than or equal to ( -frac{sqrt{2}}{2} ) in the intervals ( [0, frac{3pi}{4}] ) and ( [frac{5pi}{4}, 2pi] ). Again, since cosine is periodic with period ( 2pi ), this pattern repeats every ( 2pi ). So, the solution for the second inequality is ( -frac{3pi}{4} + 2kpi leq x leq frac{3pi}{4} + 2kpi ) for any integer ( k ).Now, I need to find the intersection of the solutions to both inequalities because both conditions must be satisfied simultaneously for the function ( f(x) ) to be defined. So, let me visualize this. The first inequality gives intervals where ( x ) is between ( frac{pi}{6} + 2kpi ) and ( frac{5pi}{6} + 2kpi ). The second inequality gives intervals where ( x ) is between ( -frac{3pi}{4} + 2kpi ) and ( frac{3pi}{4} + 2kpi ). To find the overlap, I can consider one period, say ( [0, 2pi] ), and then generalize it. In the interval ( [0, 2pi] ), the first inequality gives ( (frac{pi}{6}, frac{5pi}{6}) ) and the second inequality gives ( [0, frac{3pi}{4}] ) and ( [frac{5pi}{4}, 2pi] ). So, the overlap between ( (frac{pi}{6}, frac{5pi}{6}) ) and ( [0, frac{3pi}{4}] ) is ( (frac{pi}{6}, frac{3pi}{4}] ). There is no overlap between ( (frac{pi}{6}, frac{5pi}{6}) ) and ( [frac{5pi}{4}, 2pi] ) because ( frac{5pi}{6} ) is approximately 2.618, and ( frac{5pi}{4} ) is approximately 3.927, so they don't overlap in this period.Therefore, in the interval ( [0, 2pi] ), the domain is ( (frac{pi}{6}, frac{3pi}{4}] ). Since both inequalities are periodic with period ( 2pi ), this interval repeats every ( 2pi ). So, in general, the domain is all real numbers ( x ) such that ( frac{pi}{6} + 2kpi < x leq frac{3pi}{4} + 2kpi ) for some integer ( k ).Let me just double-check my work. For the first inequality, ( sin x > frac{1}{2} ) is indeed between ( frac{pi}{6} ) and ( frac{5pi}{6} ) in each period. For the second inequality, ( cos x geq -frac{sqrt{2}}{2} ) is true in ( [0, frac{3pi}{4}] ) and ( [frac{5pi}{4}, 2pi] ). The overlap between these intervals is ( (frac{pi}{6}, frac{3pi}{4}] ), which seems correct.I also need to make sure that I didn't make any mistakes in translating the inequalities into interval notation. The first inequality is strict inequality, so open intervals, while the second inequality includes equality, so closed intervals. Therefore, the intersection should be ( (frac{pi}{6}, frac{3pi}{4}] ), which is what I have.So, putting it all together, the domain of ( f(x) ) is all real numbers ( x ) such that ( frac{pi}{6} + 2kpi < x leq frac{3pi}{4} + 2kpi ) for any integer ( k ). I think that's it. I don't see any mistakes in my reasoning, so I feel confident about this answer.**Final Answer**The domain of the function ( f(x) ) is boxed{left{xmiddle| frac{pi}{6}+2kpi<xleq frac{3pi}{4}+2kpi, kin mathbb{Z}right}}.