💡First, I need to evaluate ( h(3) ) and ( k(3) ) using the given functions.[ h(3) = 3(3) + 4 = 13 ][ k(3) = 4(3) - 3 = 9 ]Next, I'll use these results to find ( h(k(h(3))) ) and ( k(h(k(3))) ).For ( h(k(h(3))) ):[ h(k(13)) = h(4(13) - 3) = h(49) = 3(49) + 4 = 151 ]For ( k(h(k(3))) ):[ k(h(9)) = k(3(9) + 4) = k(31) = 4(31) - 3 = 121 ]Finally, I'll form the fraction and simplify it.[ frac{h(k(h(3)))}{k(h(k(3)))} = frac{151}{121} ]

💡First, I need to determine the possible number of heads each person can get. Keiko tosses two pennies, so she can get 0, 1, or 2 heads. Ephraim tosses three pennies, so he can get 0, 1, 2, or 3 heads.Next, I'll calculate the probabilities for each possible number of heads for both Keiko and Ephraim.For Keiko:- Probability of 0 heads (TT): ( frac{1}{4} )- Probability of 1 head (TH or HT): ( frac{2}{4} = frac{1}{2} )- Probability of 2 heads (HH): ( frac{1}{4} )For Ephraim:- Probability of 0 heads (TTT): ( frac{1}{8} )- Probability of 1 head (TTH, THT, HTT): ( frac{3}{8} )- Probability of 2 heads (THT, THT, HTH): ( frac{3}{8} )- Probability of 3 heads (HHH): ( frac{1}{8} )Now, I'll calculate the probabilities that both Keiko and Ephraim get the same number of heads:- Probability both get 0 heads: ( frac{1}{4} times frac{1}{8} = frac{1}{32} )- Probability both get 1 head: ( frac{1}{2} times frac{3}{8} = frac{3}{16} )- Probability both get 2 heads: ( frac{1}{4} times frac{3}{8} = frac{3}{32} )Finally, I'll sum these probabilities to find the total probability that Ephraim gets the same number of heads as Keiko:- Total probability: ( frac{1}{32} + frac{3}{16} + frac{3}{32} = frac{2}{32} + frac{6}{32} + frac{3}{32} = frac{11}{32} )

💡Okay, so I have this problem here where I need to find the minimum value of the length of the intersection of two sets, A and B. Both A and B are subsets of a universal set U, which is defined as the interval from 1 to 3. First, let me make sure I understand the problem correctly. The set A is defined as all real numbers x such that a ≤ x ≤ a+1, and set B is defined as all real numbers x such that b - 1/2 ≤ x ≤ b + 1. Both A and B are subsets of U, which is the interval [1, 3]. The "length" of a set is defined as the difference between the upper and lower bounds, so for example, the length of A is (a+1) - a = 1, and the length of B is (b + 1) - (b - 1/2) = 3/2. But the question is asking for the minimum value of the length of A ∩ B. So, I need to figure out how small the overlap between A and B can be, given that both are subsets of U.Let me write down the constraints for A and B to be subsets of U. For set A, since it's [a, a+1], to be a subset of [1, 3], the lower bound a must be at least 1, and the upper bound a+1 must be at most 3. So, that gives me two inequalities:1. a ≥ 12. a + 1 ≤ 3 ⇒ a ≤ 2So, a must be between 1 and 2, inclusive.Similarly, for set B, which is [b - 1/2, b + 1], to be a subset of [1, 3], the lower bound b - 1/2 must be at least 1, and the upper bound b + 1 must be at most 3. So, that gives me:1. b - 1/2 ≥ 1 ⇒ b ≥ 1 + 1/2 ⇒ b ≥ 3/22. b + 1 ≤ 3 ⇒ b ≤ 2So, b must be between 3/2 and 2, inclusive.Alright, so now I know that a ∈ [1, 2] and b ∈ [3/2, 2]. Now, I need to find the minimum length of A ∩ B. The intersection of two intervals is another interval whose lower bound is the maximum of the lower bounds of A and B, and whose upper bound is the minimum of the upper bounds of A and B. So, the length of A ∩ B is:Length(A ∩ B) = max(a, b - 1/2) ≤ x ≤ min(a + 1, b + 1)So, the length is min(a + 1, b + 1) - max(a, b - 1/2)I need to find the minimum value of this expression over all possible a in [1, 2] and b in [3/2, 2].Hmm, okay. So, to minimize the length of the intersection, I need to position A and B such that their overlap is as small as possible. Since both A and B are intervals within [1, 3], the minimal overlap would occur when A and B are as far apart as possible within U.But since both A and B have fixed lengths (1 and 3/2 respectively), their positions are constrained by the universal set U. So, I need to see how much I can shift A and B within U to minimize their overlap.Let me visualize this. Let me draw the universal set U as a line from 1 to 3. Set A is an interval of length 1, so it can be placed anywhere from [1, 2] to [2, 3]. Similarly, set B is an interval of length 3/2, so it can be placed from [1, 2.5] to [1.5, 3].Wait, actually, since B is [b - 1/2, b + 1], when b is at its minimum, which is 3/2, B becomes [1, 2.5]. When b is at its maximum, which is 2, B becomes [1.5, 3].Similarly, A when a is 1 is [1, 2], and when a is 2 is [2, 3].So, to minimize the overlap between A and B, I should try to place A as far to the left as possible and B as far to the right as possible, or vice versa.But let's check both possibilities.First, let's try placing A as far left as possible, which is [1, 2], and B as far right as possible, which is [1.5, 3]. Then, the intersection of [1, 2] and [1.5, 3] is [1.5, 2], which has a length of 0.5.Alternatively, if I place A as far right as possible, which is [2, 3], and B as far left as possible, which is [1, 2.5]. The intersection of [2, 3] and [1, 2.5] is [2, 2.5], which also has a length of 0.5.So, in both cases, the minimal overlap is 0.5.Is this the minimal possible? Let me see if I can get a smaller overlap by shifting A and B somewhere else.Suppose I place A somewhere in the middle, say a = 1.5, so A is [1.5, 2.5]. Then, if I place B at [1.5, 3], their intersection is [1.5, 2.5], which is length 1, which is larger than 0.5.Alternatively, if I place A at [1, 2] and B at [1.5, 3], as before, the intersection is [1.5, 2], which is 0.5.Similarly, if I move A slightly to the right, say a = 1.2, so A is [1.2, 2.2], and B is [1.5, 3], then the intersection is [1.5, 2.2], which is 0.7, which is larger than 0.5.Alternatively, if I place B at [1, 2.5] and A at [2, 3], their intersection is [2, 2.5], which is 0.5.So, it seems that 0.5 is indeed the minimal overlap.Wait, but let me think again. Is there a way to make the overlap smaller than 0.5?Suppose I shift A and B such that their intervals just barely touch each other. For example, if A is [1, 2] and B is [2, 3], but wait, B cannot be [2, 3] because B has a length of 1.5, so the lower bound would be 2 - 0.5 = 1.5, so B would be [1.5, 3]. So, if A is [1, 2], and B is [1.5, 3], their intersection is [1.5, 2], which is 0.5.Alternatively, if A is [2, 3], and B is [1, 2.5], their intersection is [2, 2.5], which is 0.5.So, in both cases, the minimal overlap is 0.5.Wait, but what if I place A somewhere else? Let me try a = 1.5, so A is [1.5, 2.5]. Then, if I place B at [1.5, 3], their intersection is [1.5, 2.5], which is 1.0. If I place B at [1, 2.5], their intersection is [1.5, 2.5], which is 1.0 as well. So, no improvement.Alternatively, if I place A at a = 1.25, so A is [1.25, 2.25]. Then, placing B at [1.5, 3], the intersection is [1.5, 2.25], which is 0.75. Placing B at [1, 2.5], the intersection is [1.25, 2.25], which is 1.0. So, still, the minimal overlap is 0.5.Wait, perhaps I can shift A and B such that their overlap is less than 0.5? Let me see.Suppose I place A at [1, 2], and B at [1.5, 3]. The overlap is [1.5, 2], which is 0.5.If I try to shift A slightly to the right, say a = 1.1, so A is [1.1, 2.1]. Then, B is [1.5, 3], so the intersection is [1.5, 2.1], which is 0.6. That's larger than 0.5.Alternatively, if I shift A to the left, but A can't go beyond 1, so that doesn't help.Similarly, if I shift B to the left, but B can't go beyond 1.5 on the lower side, so if I place B at [1.5, 3], that's as far left as it can go without violating the subset condition.Wait, actually, B can be placed anywhere from [1, 2.5] to [1.5, 3]. So, if I place B at [1, 2.5], and A at [2, 3], their intersection is [2, 2.5], which is 0.5.Alternatively, if I place B at [1, 2.5] and A at [1, 2], their intersection is [1, 2], which is 1.0.So, the minimal overlap is when A is as far left as possible and B as far right as possible, or vice versa.Therefore, the minimal length of A ∩ B is 0.5.Wait, but let me confirm this with the formula I had earlier.Length(A ∩ B) = min(a + 1, b + 1) - max(a, b - 1/2)So, to minimize this, I need to maximize the difference between min(a + 1, b + 1) and max(a, b - 1/2). Let me consider different cases.Case 1: a + 1 ≤ b + 1 and a ≥ b - 1/2Then, min(a + 1, b + 1) = a + 1max(a, b - 1/2) = aSo, Length = (a + 1) - a = 1Case 2: a + 1 ≤ b + 1 and a < b - 1/2Then, min(a + 1, b + 1) = a + 1max(a, b - 1/2) = b - 1/2So, Length = (a + 1) - (b - 1/2) = a + 1 - b + 1/2 = a - b + 3/2To minimize this, since a is as small as possible (1) and b as large as possible (2), so Length = 1 - 2 + 3/2 = (1 - 2) + 1.5 = (-1) + 1.5 = 0.5Case 3: a + 1 > b + 1 and a ≥ b - 1/2Then, min(a + 1, b + 1) = b + 1max(a, b - 1/2) = aSo, Length = (b + 1) - aTo minimize this, we need to maximize a and minimize b.a is at most 2, b is at least 3/2.So, Length = (3/2 + 1) - 2 = (5/2) - 2 = 1/2Case 4: a + 1 > b + 1 and a < b - 1/2Then, min(a + 1, b + 1) = b + 1max(a, b - 1/2) = b - 1/2So, Length = (b + 1) - (b - 1/2) = 1 + 1/2 = 3/2So, in this case, the length is 3/2, which is larger than 0.5.Therefore, the minimal length occurs in both Case 2 and Case 3, giving a minimal length of 0.5.So, the minimal length of A ∩ B is 1/2.I think that's consistent with my earlier reasoning. So, the answer should be 1/2.

💡Okay, so I have this problem about an ellipse. Let me try to figure it out step by step. First, part (1) asks for the equation of the ellipse. The ellipse is centered at the origin, with its major axis along the x-axis. They mention the focal distance is 6, and the minor axis is 8. Hmm, okay.I remember that for an ellipse, the standard equation is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), where (a) is the semi-major axis and (b) is the semi-minor axis. Since the major axis is along the x-axis, that means (a > b).They gave the focal distance as 6. I think the focal distance is the distance between the two foci, which is (2c). So if (2c = 6), then (c = 3). Also, the minor axis is 8, which is the length of the minor axis. The minor axis is (2b), so (2b = 8) which means (b = 4).Now, I need to find (a). I remember the relationship between (a), (b), and (c) in an ellipse: (c^2 = a^2 - b^2). So plugging in the values I have:(c = 3), so (c^2 = 9)(b = 4), so (b^2 = 16)So, (9 = a^2 - 16)Adding 16 to both sides: (a^2 = 25), so (a = 5).Therefore, the equation of the ellipse is (frac{x^2}{25} + frac{y^2}{16} = 1). That seems straightforward.Moving on to part (2). It says a line passes through the point ((-5, 0)) with a slope of (frac{pi}{4}). Wait, slope is usually a number, not an angle. Maybe it's a typo? Because (frac{pi}{4}) is an angle, 45 degrees, whose tangent is 1. So perhaps the slope is 1? That would make sense because the tangent of (frac{pi}{4}) is 1.So, assuming the slope is 1, the line passes through ((-5, 0)). Let me write the equation of this line. Using the point-slope form: (y - y_1 = m(x - x_1)). Here, (m = 1), (x_1 = -5), (y_1 = 0). So plugging in:(y - 0 = 1 cdot (x - (-5))), which simplifies to (y = x + 5).Now, this line intersects the ellipse at points (A) and (B). We need to find these points of intersection. So, I need to solve the system of equations:1. (frac{x^2}{25} + frac{y^2}{16} = 1)2. (y = x + 5)I can substitute equation 2 into equation 1 to solve for (x). Let's do that.Substituting (y = x + 5) into the ellipse equation:(frac{x^2}{25} + frac{(x + 5)^2}{16} = 1)Let me expand this:First, expand ((x + 5)^2):((x + 5)^2 = x^2 + 10x + 25)So, plugging back into the equation:(frac{x^2}{25} + frac{x^2 + 10x + 25}{16} = 1)To combine these, I need a common denominator. The denominators are 25 and 16, so the least common multiple is 400. Let me multiply each term by 400 to eliminate the denominators:(400 cdot frac{x^2}{25} + 400 cdot frac{x^2 + 10x + 25}{16} = 400 cdot 1)Simplify each term:(16x^2 + 25(x^2 + 10x + 25) = 400)Now distribute the 25:(16x^2 + 25x^2 + 250x + 625 = 400)Combine like terms:(16x^2 + 25x^2 = 41x^2)So, (41x^2 + 250x + 625 = 400)Subtract 400 from both sides:(41x^2 + 250x + 225 = 0)Now, we have a quadratic equation in terms of (x): (41x^2 + 250x + 225 = 0)Let me try to solve this using the quadratic formula. The quadratic formula is (x = frac{-b pm sqrt{b^2 - 4ac}}{2a}), where (a = 41), (b = 250), and (c = 225).First, compute the discriminant:(D = b^2 - 4ac = 250^2 - 4 cdot 41 cdot 225)Calculate (250^2): 250 * 250 = 62,500Calculate (4 * 41 * 225): 4 * 41 = 164; 164 * 225. Let's compute that:225 * 160 = 36,000225 * 4 = 900So, 36,000 + 900 = 36,900So, D = 62,500 - 36,900 = 25,600Square root of 25,600 is 160, since 160^2 = 25,600.So, the solutions are:(x = frac{-250 pm 160}{2 cdot 41})Compute both possibilities:First solution: (-250 + 160 = -90), so (x = frac{-90}{82} = frac{-45}{41})Second solution: (-250 - 160 = -410), so (x = frac{-410}{82} = frac{-205}{41} = -5)Wait, so one of the solutions is (x = -5). That makes sense because the line passes through ((-5, 0)), which is a point on the ellipse. So, that's one intersection point, which is given. The other intersection point is at (x = -frac{45}{41}).Now, let's find the corresponding (y) values for each (x).For (x = -5): (y = (-5) + 5 = 0). So, that's the point ((-5, 0)), which we already knew.For (x = -frac{45}{41}): (y = (-frac{45}{41}) + 5 = (-frac{45}{41}) + frac{205}{41} = frac{160}{41})So, the two points of intersection are ((-5, 0)) and ((- frac{45}{41}, frac{160}{41})). So, points (A) and (B) are these two points.Wait, the problem says the line intersects the ellipse at points (A) and (B). So, one of them is ((-5, 0)), which is given, and the other is ((- frac{45}{41}, frac{160}{41})). Now, the question is to find the area of triangle (ABO), where (O) is the origin.So, points (A), (B), and (O) form a triangle. Let's note the coordinates:- (O) is at ((0, 0))- (A) is at ((-5, 0))- (B) is at ((- frac{45}{41}, frac{160}{41}))I need to find the area of triangle (ABO). There are a few ways to do this. One way is to use the formula for the area of a triangle given three vertices: [text{Area} = frac{1}{2} |x_A(y_B - y_O) + x_B(y_O - y_A) + x_O(y_A - y_B)|]But since (O) is the origin, (x_O = 0) and (y_O = 0), so the formula simplifies.Alternatively, since points (A) and (O) are on the x-axis, the base of the triangle can be considered as the distance between (A) and (O), which is 5 units. The height would be the y-coordinate of point (B), since the height is the perpendicular distance from (B) to the base (AO).Wait, that might be simpler. Let me visualize the triangle. Points (A) and (O) are on the x-axis, so the base is along the x-axis from ((-5, 0)) to ((0, 0)). The third point (B) is somewhere in the plane, and the height of the triangle would be the vertical distance from (B) to the x-axis, which is just the absolute value of the y-coordinate of (B).So, the area would be (frac{1}{2} times text{base} times text{height}).Here, the base is the distance from (A) to (O), which is 5 units. The height is (|y_B| = frac{160}{41}).So, plugging into the formula:Area = (frac{1}{2} times 5 times frac{160}{41})Compute that:First, multiply 5 and (frac{160}{41}): (5 times frac{160}{41} = frac{800}{41})Then, multiply by (frac{1}{2}): (frac{800}{41} times frac{1}{2} = frac{400}{41})So, the area is (frac{400}{41}).Wait, let me double-check this. Alternatively, using the determinant method:The coordinates are (O(0, 0)), (A(-5, 0)), (B(-frac{45}{41}, frac{160}{41})).The area can be calculated as:[text{Area} = frac{1}{2} |x_A y_B - x_B y_A|]Since (y_A = 0) and (y_O = 0), this simplifies.Plugging in the values:[text{Area} = frac{1}{2} |(-5) times frac{160}{41} - (-frac{45}{41}) times 0| = frac{1}{2} | -frac{800}{41} - 0 | = frac{1}{2} times frac{800}{41} = frac{400}{41}]Same result. So, that seems correct.Alternatively, using vectors or coordinates, but I think this is solid.So, the area is (frac{400}{41}).Wait, just to make sure, let me think if there's another way to compute this. Maybe using the distance between points (A) and (B) and then using Heron's formula? That might be more complicated, but let's see.First, compute the lengths of sides:- (OA): distance from (O) to (A) is 5.- (OB): distance from (O) to (B) is (sqrt{(-frac{45}{41})^2 + (frac{160}{41})^2})Compute that:[sqrt{left(frac{2025}{1681}right) + left(frac{25600}{1681}right)} = sqrt{frac{2025 + 25600}{1681}} = sqrt{frac{27625}{1681}} = frac{sqrt{27625}}{41}]Simplify (sqrt{27625}):27625 divided by 25 is 1105. Hmm, 1105 factors: 5 * 221, which is 13 * 17. So, it doesn't simplify further. So, (sqrt{27625} = 5 sqrt{1105}). So, (OB = frac{5 sqrt{1105}}{41}).Then, distance (AB): between (A(-5, 0)) and (B(-frac{45}{41}, frac{160}{41})).Compute the differences:Δx = (-frac{45}{41} - (-5) = -frac{45}{41} + 5 = -frac{45}{41} + frac{205}{41} = frac{160}{41})Δy = (frac{160}{41} - 0 = frac{160}{41})So, distance (AB = sqrt{(frac{160}{41})^2 + (frac{160}{41})^2} = sqrt{2 times (frac{25600}{1681})} = sqrt{frac{51200}{1681}} = frac{sqrt{51200}}{41} = frac{160 sqrt{2}}{41})So, sides are:- OA = 5- OB = (frac{5 sqrt{1105}}{41})- AB = (frac{160 sqrt{2}}{41})Now, using Heron's formula:First, compute the semi-perimeter (s):(s = frac{OA + OB + AB}{2})But this seems messy. Maybe it's better to stick with the previous method.Alternatively, using coordinates, the area can be calculated with the shoelace formula.Coordinates:O(0, 0), A(-5, 0), B(-45/41, 160/41)Shoelace formula:Area = (frac{1}{2} |x_O(y_A - y_B) + x_A(y_B - y_O) + x_B(y_O - y_A)|)Plugging in:= (frac{1}{2} |0(0 - 160/41) + (-5)(160/41 - 0) + (-45/41)(0 - 0)|)Simplify:= (frac{1}{2} |0 + (-5)(160/41) + 0| = frac{1}{2} | -800/41 | = frac{1}{2} times 800/41 = 400/41)Same result. Okay, so that's consistent.Therefore, I'm confident that the area is (frac{400}{41}).**Final Answer**The area of triangle (ABO) is (boxed{dfrac{400}{41}}).

💡Okay, so I have this equation here: x squared minus k times x plus 1 equals zero. And I'm told that negative one is a root of this equation. I need to find the value of k. Hmm, let me think about how to approach this.First, I remember that if a number is a root of a polynomial equation, then substituting that number into the equation should satisfy it, meaning the equation should equal zero. So, since negative one is a root, I can plug x equals negative one into the equation and solve for k.Alright, let's write that out. If x is negative one, then the equation becomes:(-1)^2 - k*(-1) + 1 = 0.Let me compute each term step by step. The first term is (-1) squared. Squaring a negative number gives a positive result, so (-1)^2 is 1.The second term is -k times (-1). Multiplying two negative numbers gives a positive, so this becomes +k.The third term is just 1.Putting it all together, the equation becomes:1 + k + 1 = 0.Now, let's simplify this. 1 plus 1 is 2, so the equation simplifies to:2 + k = 0.To solve for k, I need to isolate it. I can subtract 2 from both sides of the equation:2 + k - 2 = 0 - 2.Simplifying both sides, 2 minus 2 is 0, so we have:k = -2.Wait, let me double-check that to make sure I didn't make any mistakes. Starting from the beginning, substituting x = -1 into the equation:(-1)^2 is 1, correct. Then, -k*(-1) is +k, right. And then +1. So 1 + k + 1 equals 0. That adds up to 2 + k = 0. Subtracting 2 gives k = -2. Yeah, that seems right.Just to be thorough, maybe I can plug k = -2 back into the original equation and see if x = -1 is indeed a root. Let's do that.Original equation: x^2 - kx + 1 = 0.Substituting k = -2, we get:x^2 - (-2)x + 1 = 0.Simplifying, that becomes:x^2 + 2x + 1 = 0.Now, let's plug x = -1 into this equation:(-1)^2 + 2*(-1) + 1.Calculating each term:(-1)^2 is 1.2*(-1) is -2.And then +1.Adding them up: 1 - 2 + 1 = 0.Yes, that works out to zero, which confirms that x = -1 is indeed a root when k is -2. So, my answer seems correct.I guess another way to think about this is using the quadratic formula or factoring, but since we already know one of the roots, substitution seems straightforward and efficient. Plus, it's always good to verify the solution by plugging it back into the original equation to ensure there were no calculation errors.I don't think I made any mistakes here. The steps seem logical and the verification confirms the solution. So, I'm confident that k equals -2 is the correct answer.

💡Okay, so I have this problem about a barometer that's not working perfectly because some air got inside it. I need to figure out the true barometric pressure based on the readings at two different temperatures. Hmm, let me try to break this down step by step.First, let's understand what's happening. A barometer measures atmospheric pressure by balancing it against the weight of a mercury column. If some air gets into the barometer, it affects the reading because the air exerts some pressure too. So, the barometer isn't just measuring the atmospheric pressure anymore; it's measuring the atmospheric pressure minus the pressure from the trapped air.At 15°C, the barometer shows 704 mm when the actual atmospheric pressure is 762 mm. The height of the air-free space is 143 mm. That means that the air is occupying some space in the barometer, reducing the effective height of the mercury column. So, the pressure from the trapped air must be the difference between the actual pressure and the reading, right? So, the trapped air pressure would be 762 mm - 704 mm = 58 mm Hg.Now, when the temperature increases to 30°C, the barometer shows 692 mm. I need to find out what the true barometric pressure is now. I think I can use the combined gas law here because the amount of air is constant, but the temperature and pressure are changing. The combined gas law is:[ frac{P_1 V_1}{T_1} = frac{P_2 V_2}{T_2} ]Where:- ( P_1 ) is the initial pressure of the trapped air,- ( V_1 ) is the initial volume of the trapped air,- ( T_1 ) is the initial temperature in Kelvin,- ( P_2 ) is the final pressure of the trapped air,- ( V_2 ) is the final volume of the trapped air,- ( T_2 ) is the final temperature in Kelvin.Let me plug in the values I have.First, convert temperatures to Kelvin:- ( T_1 = 15 + 273 = 288 ) K,- ( T_2 = 30 + 273 = 303 ) K.We already found ( P_1 = 58 ) mm Hg. The initial volume ( V_1 ) is the height of the air-free space, which is 143 mm. Wait, is that the volume? Actually, since the cross-sectional area of the barometer is constant, the volume is proportional to the height. So, I can treat the height as a measure of volume here.Now, at the second temperature, the barometer shows 692 mm. The height of the mercury column has changed, so the volume of the trapped air has changed too. The change in height is 704 mm - 692 mm = 12 mm. So, the volume of the trapped air has increased by 12 mm. Therefore, the final volume ( V_2 = 143 mm + 12 mm = 155 mm ).Now, plug these into the combined gas law:[ frac{58 times 143}{288} = frac{P_2 times 155}{303} ]Let me solve for ( P_2 ):First, multiply both sides by 303:[ frac{58 times 143 times 303}{288} = P_2 times 155 ]Then, divide both sides by 155:[ P_2 = frac{58 times 143 times 303}{288 times 155} ]Let me calculate this step by step.First, multiply 58 and 143:58 * 143 = 8294Then, multiply that by 303:8294 * 303 = 2,512, something. Wait, let me do this more carefully.8294 * 300 = 2,488,2008294 * 3 = 24,882So, total is 2,488,200 + 24,882 = 2,513,082Now, divide by (288 * 155):288 * 155 = let's see, 288 * 100 = 28,800; 288 * 50 = 14,400; 288 * 5 = 1,440. So, total is 28,800 + 14,400 + 1,440 = 44,640.So, P2 = 2,513,082 / 44,640 ≈ 56.3 mm HgSo, approximately 56 mm Hg.Now, the barometer reading at 30°C is 692 mm. But this is the effective pressure, which is the actual atmospheric pressure minus the trapped air pressure. So, the true barometric pressure should be 692 mm + 56 mm = 748 mm Hg.Wait, let me double-check my calculations because I might have made a mistake somewhere.First, 58 * 143: 50*143=7,150; 8*143=1,144; total=7,150+1,144=8,294.Then, 8,294 * 303: 8,294*300=2,488,200; 8,294*3=24,882; total=2,488,200+24,882=2,513,082.Divide by 288*155: 288*155=44,640.2,513,082 / 44,640 ≈ 56.3 mm Hg.Yes, that seems correct.So, the trapped air pressure at 30°C is approximately 56 mm Hg. Therefore, the true barometric pressure is 692 + 56 = 748 mm Hg.I think that makes sense. The trapped air pressure increased slightly due to the temperature increase, but the barometer reading decreased because the mercury expanded. So, the true pressure is higher than the reading, which aligns with our calculation.I don't see any mistakes in my reasoning, so I think the true barometric pressure is 748 mm Hg.