🧠:<think>Alright, so I have this problem about an offshore drilling engineer analyzing an underwater oil field. It's modeled as a cylindrical reservoir, and I need to calculate two things: the total volume of oil that can be extracted and the volumetric flow rate using Darcy's Law. Hmm, okay, let me break this down step by step.First, for part 1, I need to find the total volume of oil. The reservoir is cylindrical, so I remember the formula for the volume of a cylinder is πr²h. The radius is given as 1.5 kilometers, which I should convert to meters because the height is in meters. So, 1.5 km is 1500 meters. The height is 200 meters. Let me write that down:Radius, r = 1500 mHeight, h = 200 mSo, volume of the cylinder, V = π * (1500)^2 * 200. Let me calculate that. First, 1500 squared is 2,250,000. Multiply that by 200, which gives 450,000,000. Then multiply by π, approximately 3.1416. So, 450,000,000 * 3.1416 ≈ 1,413,716,694 cubic meters. That's the total volume of the reservoir.But wait, not all of that is oil. The porosity is 25%, which means only 25% of the reservoir's volume is pore space. So, the pore volume is 0.25 * 1,413,716,694 ≈ 353,429,173.5 cubic meters.And then, only 60% of that pore space is filled with oil. So, the oil volume is 0.6 * 353,429,173.5. Let me compute that: 353,429,173.5 * 0.6 ≈ 212,057,504.1 cubic meters. So, that's the total oil volume.Wait, let me double-check my calculations. 1500 squared is indeed 2,250,000. Multiply by 200 is 450,000,000. Multiply by π is roughly 1.4137e9. Then, 25% of that is 3.534e8, and 60% of that is 2.1205e8. Yeah, that seems right.Okay, moving on to part 2. Now, the engineer wants to model the oil flow using Darcy's Law. I remember Darcy's Law formula is Q = (k * A * ΔP) / (μ * L), where Q is the volumetric flow rate, k is permeability, A is the cross-sectional area, ΔP is the pressure gradient, and L is the length or distance.Wait, actually, hold on. Darcy's Law is usually written as Q = (k * A * ΔP) / (μ * L). But sometimes, ΔP is given per unit length, so maybe I need to adjust that. Let me confirm.Given:Permeability, k = 150 millidarcies. I need to convert that to darcies because usually, the units for k in Darcy's Law are in darcies. 150 millidarcies is 0.15 darcies.Viscosity, μ = 0.8 centipoise. I should convert that to poise because the standard unit for viscosity in Darcy's Law is poise. 0.8 centipoise is 0.008 poise. Wait, no, 1 centipoise is 0.01 poise, so 0.8 centipoise is 0.008 poise? Wait, no, 1 centipoise = 0.01 poise, so 0.8 centipoise = 0.008 poise. Hmm, that seems correct.Pressure gradient is 500 kilopascals per kilometer. So, that's ΔP per L, where L is in kilometers. Since the wellbore is at the center, I think the distance from the center to the edge is the radius, which is 1500 meters, or 1.5 kilometers. So, the pressure gradient is 500 kPa per km, so over 1.5 km, the total pressure drop would be 500 * 1.5 = 750 kPa. Wait, is that right?Wait, actually, Darcy's Law uses the pressure gradient, which is ΔP per unit length. So, if the pressure gradient is 500 kPa per km, that's 500 kPa/km. Since the well is at the center, the distance from the well to the edge is 1.5 km, so the total pressure drop across the reservoir would be 500 kPa/km * 1.5 km = 750 kPa. But in Darcy's Law, do we use the total pressure drop or the gradient?Wait, let me recall. Darcy's Law is Q = (k * A * ΔP) / (μ * L). Here, ΔP is the total pressure drop over the length L. So, if the pressure gradient is 500 kPa/km, and the length is 1.5 km, then ΔP = 500 * 1.5 = 750 kPa. So, ΔP is 750,000 Pa (since 1 kPa = 1000 Pa). So, 750,000 Pa.Cross-sectional area, A = 0.1 m².So, plugging into Darcy's Law:Q = (k * A * ΔP) / (μ * L)Wait, but hold on, I need to make sure all units are consistent. Let me list out the units:k is in darcies. 1 darcy is 1e-12 m². So, 0.15 darcies is 0.15e-12 m².μ is in poise. 1 poise = 0.1 Pa·s. So, 0.008 poise is 0.008 * 0.1 = 0.0008 Pa·s.ΔP is in Pascals: 750,000 Pa.L is in meters: 1.5 km is 1500 m.So, let me plug in the numbers:Q = (0.15e-12 m² * 0.1 m² * 750,000 Pa) / (0.0008 Pa·s * 1500 m)Wait, hold on, A is 0.1 m², right? So, A is 0.1 m².So, let me write it step by step:k = 0.15e-12 m²A = 0.1 m²ΔP = 750,000 Paμ = 0.0008 Pa·sL = 1500 mSo,Q = (0.15e-12 * 0.1 * 750,000) / (0.0008 * 1500)Let me compute numerator and denominator separately.Numerator:0.15e-12 * 0.1 = 0.015e-120.015e-12 * 750,000 = 0.015e-12 * 7.5e5 = (0.015 * 7.5) * 1e-7 = 0.1125 * 1e-7 = 1.125e-8Denominator:0.0008 * 1500 = 1.2So, Q = 1.125e-8 / 1.2 ≈ 0.9375e-8 m³/sWait, that seems really low. Let me check my calculations again.Wait, 0.15e-12 * 0.1 is 0.015e-12, correct. Then, 0.015e-12 * 750,000.Wait, 750,000 is 7.5e5, so 0.015e-12 * 7.5e5 = (0.015 * 7.5) * 1e-7 = 0.1125 * 1e-7 = 1.125e-8. That seems correct.Denominator: 0.0008 * 1500 = 1.2. Correct.So, Q = 1.125e-8 / 1.2 ≈ 0.9375e-8 m³/s.But wait, that's in cubic meters per second. The question asks for cubic meters per day. So, I need to convert seconds to days.There are 60 seconds in a minute, 60 minutes in an hour, 24 hours in a day. So, 60*60*24 = 86,400 seconds in a day.So, Q = 0.9375e-8 m³/s * 86,400 s/day ≈ 0.9375e-8 * 8.64e4 = (0.9375 * 8.64) * 1e-4 ≈ 8.1 * 1e-4 ≈ 0.00081 m³/day.Wait, that seems extremely low. Is that correct? 0.00081 cubic meters per day? That can't be right. Maybe I messed up the units somewhere.Let me double-check the units:k is in m², correct. A is in m², correct. ΔP is in Pa, correct. μ is in Pa·s, correct. L is in meters, correct.Wait, 1 darcy is 1e-12 m², so 0.15 darcies is 0.15e-12 m², correct.Viscosity: 0.8 centipoise is 0.008 poise, and 1 poise is 0.1 Pa·s, so 0.008 * 0.1 = 0.0008 Pa·s, correct.Pressure gradient: 500 kPa/km, converted to total pressure drop over 1.5 km is 750 kPa, which is 750,000 Pa, correct.Cross-sectional area: 0.1 m², correct.So, plugging into Darcy's Law:Q = (k * A * ΔP) / (μ * L) = (0.15e-12 * 0.1 * 750,000) / (0.0008 * 1500)Compute numerator:0.15e-12 * 0.1 = 0.015e-120.015e-12 * 750,000 = 0.015 * 750,000 * 1e-12 = 11,250 * 1e-12 = 1.125e-8Denominator:0.0008 * 1500 = 1.2So, Q = 1.125e-8 / 1.2 ≈ 0.9375e-8 m³/sConvert to m³/day:0.9375e-8 * 86,400 ≈ 0.9375 * 86.4e-4 ≈ 0.9375 * 0.00864 ≈ 0.0081 m³/dayWait, that's 0.0081 m³/day, not 0.00081. Because 0.9375e-8 * 86,400 = 0.9375 * 86,400 * 1e-8 = (0.9375 * 86.4) * 1e-4 ≈ 81 * 1e-4 = 0.0081.Ah, okay, so 0.0081 m³/day. Still seems low, but maybe that's correct? Let me think about the scale.Given that the permeability is 150 millidarcies, which is quite low. 1 darcy is already a low permeability, so 0.15 darcies is very low. The viscosity is 0.8 centipoise, which is relatively low for oil, but still, with such low permeability, the flow rate would be very low.Alternatively, maybe I made a mistake in converting units somewhere. Let me check the Darcy's Law formula again.Wait, another version of Darcy's Law is Q = (k * A * (ΔP / L)) / μ. So, sometimes it's written as Q = (k * A * ΔP) / (μ * L), which is the same as I used. So, that seems correct.Alternatively, sometimes people use different units, like using pressure gradient in kPa/km, but in that case, the formula would be Q = (k * A * (ΔP / L)) / μ, where ΔP is in Pascals and L is in meters. Wait, let me see.If pressure gradient is given as 500 kPa/km, that's equivalent to 500,000 Pa per 1000 meters, which is 500 Pa/m. So, ΔP per unit length is 500 Pa/m.So, in that case, the formula would be Q = (k * A * (ΔP / L)) / μ, but wait, no, because ΔP is already per unit length. Wait, no, if ΔP is given per unit length, then Q = (k * A * (ΔP / L)) / μ is not correct. Wait, no, if ΔP is per unit length, then Q = (k * A * (ΔP)) / μ, because ΔP is already per unit length.Wait, I think I confused myself. Let me clarify.Darcy's Law is Q = (k * A * (ΔP / L)) / μ, where ΔP is the total pressure drop over length L. Alternatively, if the pressure gradient is given as ΔP per unit length (like 500 kPa/km), then Q = (k * A * (ΔP / L)) / μ becomes Q = (k * A * (ΔP_gradient * L)) / μ, because ΔP = ΔP_gradient * L.Wait, no, if ΔP_gradient is given as 500 kPa/km, then ΔP = 500 kPa/km * L, where L is in km. So, if L is 1.5 km, then ΔP = 500 * 1.5 = 750 kPa, as I did before.So, then Q = (k * A * ΔP) / (μ * L), with ΔP in Pascals and L in meters.So, that's what I did earlier, and the result was 0.0081 m³/day. Hmm.Alternatively, maybe I should use the pressure gradient directly without multiplying by L. Let me see.If I use Q = (k * A * (ΔP_gradient)) / μ, where ΔP_gradient is in Pascals per meter.Given that ΔP_gradient is 500 kPa/km, which is 500,000 Pa / 1000 m = 500 Pa/m.So, ΔP_gradient = 500 Pa/m.Then, Q = (k * A * ΔP_gradient) / μSo, plugging in:k = 0.15e-12 m²A = 0.1 m²ΔP_gradient = 500 Pa/mμ = 0.0008 Pa·sSo,Q = (0.15e-12 * 0.1 * 500) / 0.0008Compute numerator:0.15e-12 * 0.1 = 0.015e-120.015e-12 * 500 = 0.0075e-12Denominator: 0.0008So, Q = 0.0075e-12 / 0.0008 ≈ 0.009375e-12 m³/sConvert to m³/day:0.009375e-12 * 86,400 ≈ 0.009375 * 86.4e-9 ≈ 0.81e-9 m³/dayWait, that's even smaller, 0.00000000081 m³/day. That can't be right either.Hmm, so which approach is correct? Using ΔP as total pressure drop over L, or using ΔP_gradient directly?I think the confusion comes from the form of Darcy's Law. Let me recall.Darcy's Law is:Q = (k * A * (ΔP / L)) / μWhere ΔP is the total pressure drop over length L.Alternatively, if the pressure gradient is given as ΔP per unit length (like 500 kPa/km), then ΔP = gradient * L, so:Q = (k * A * (gradient * L) / L) / μ = (k * A * gradient) / μSo, in that case, Q = (k * A * gradient) / μSo, which one is correct? It depends on how the pressure gradient is defined.In the problem, it says "pressure gradient of 500 kilopascals per kilometer". So, that's 500 kPa/km, which is the same as 500,000 Pa/km or 500 Pa/m.So, if I use the second form, Q = (k * A * gradient) / μ, then:Q = (0.15e-12 m² * 0.1 m² * 500 Pa/m) / 0.0008 Pa·sWait, but units: k is m², A is m², gradient is Pa/m, μ is Pa·s.So, units:(m² * m² * Pa/m) / (Pa·s) = (m^4 * Pa) / (Pa·s·m) ) = m^3 / sWhich is correct for Q.So, let's compute:Numerator: 0.15e-12 * 0.1 * 500 = 0.15e-12 * 50 = 7.5e-12Denominator: 0.0008So, Q = 7.5e-12 / 0.0008 ≈ 9.375e-9 m³/sConvert to m³/day:9.375e-9 * 86,400 ≈ 9.375 * 86.4e-4 ≈ 9.375 * 0.00864 ≈ 0.081 m³/dayWait, that's 0.081 m³/day. That's still low, but higher than the previous 0.0081.Wait, so which one is correct? The first approach gave me 0.0081 m³/day, the second approach gave me 0.081 m³/day. The difference is a factor of 10.Looking back, in the first approach, I used ΔP = 750,000 Pa and L = 1500 m, so Q = (k * A * ΔP) / (μ * L) = (0.15e-12 * 0.1 * 750,000) / (0.0008 * 1500) ≈ 0.0081 m³/day.In the second approach, using gradient directly, Q = (k * A * gradient) / μ = (0.15e-12 * 0.1 * 500) / 0.0008 ≈ 0.081 m³/day.So, which one is correct? I think the second approach is correct because the pressure gradient is given per kilometer, so we don't need to multiply by L again. Because in the formula Q = (k * A * (ΔP / L)) / μ, ΔP is the total pressure drop, which is gradient * L. So, if we already have the gradient, we can use it directly without multiplying by L again.Therefore, the correct calculation is the second one, giving Q ≈ 0.081 m³/day.But wait, let me check the units again. If I use gradient in Pa/m, then:k (m²) * A (m²) * gradient (Pa/m) = m² * m² * Pa/m = m^4 * Pa/m = m^3 * PaDivide by μ (Pa·s): m^3 * Pa / (Pa·s) = m³/sSo, correct.Therefore, the correct flow rate is approximately 0.081 m³/day.But that still seems low. Let me think about typical flow rates. For a well with such low permeability, maybe it's correct. 150 millidarcies is 0.15 darcies, which is very low. For comparison, a good oil reservoir might have permeability in the range of 100-1000 millidarcies, so 150 is on the lower side. So, a low flow rate makes sense.Alternatively, maybe I made a mistake in the exponent for k. 1 darcy is 1e-12 m², so 0.15 darcies is 0.15e-12 m². Correct.Alternatively, maybe the cross-sectional area is not 0.1 m²? Wait, the wellbore has a cross-sectional area of 0.1 m². That seems reasonable for a wellbore.Alternatively, maybe the pressure gradient is 500 kPa per kilometer, which is quite steep. 500 kPa over 1.5 km is 750 kPa, which is a significant pressure drop. But with such low permeability, the flow rate remains low.Alternatively, maybe I should use a different formula, like the one considering the radius of the reservoir. Wait, no, Darcy's Law is for flow through a porous medium, and in this case, the wellbore is at the center, so the flow is radial. But the problem says to model the oil flow using Darcy's Law, so maybe it's assuming linear flow, not radial.Wait, actually, in reality, flow in a reservoir is radial, but Darcy's Law in its basic form is for linear flow. So, maybe the problem is simplifying it to linear flow, treating the wellbore as a pipe with cross-sectional area A, and the pressure gradient is given as 500 kPa/km, so over the radius of 1.5 km, the total pressure drop is 750 kPa.But in that case, using Q = (k * A * ΔP) / (μ * L), with L = 1.5 km = 1500 m, which is what I did first, giving 0.0081 m³/day.But now I'm confused because two different approaches give different results.Wait, let me look up Darcy's Law formula to confirm.Darcy's Law is typically written as:Q = (k * A * (ΔP / L)) / μWhere:- Q is the volumetric flow rate (m³/s)- k is the permeability (m²)- A is the cross-sectional area (m²)- ΔP is the pressure drop (Pa)- L is the length over which the pressure drop occurs (m)- μ is the viscosity (Pa·s)So, in this case, if the pressure gradient is given as 500 kPa/km, which is 500,000 Pa / 1000 m = 500 Pa/m.So, ΔP / L = 500 Pa/m.Therefore, Q = (k * A * (ΔP / L)) / μ = (k * A * gradient) / μSo, plugging in:k = 0.15e-12 m²A = 0.1 m²gradient = 500 Pa/mμ = 0.0008 Pa·sQ = (0.15e-12 * 0.1 * 500) / 0.0008Compute numerator:0.15e-12 * 0.1 = 0.015e-120.015e-12 * 500 = 0.0075e-12Denominator: 0.0008So, Q = 0.0075e-12 / 0.0008 ≈ 0.009375e-12 m³/sConvert to m³/day:0.009375e-12 * 86,400 ≈ 0.009375 * 86.4e-9 ≈ 0.81e-9 m³/dayWait, that's 0.00000000081 m³/day, which is 8.1e-10 m³/day. That seems way too low.Wait, but earlier when I used ΔP = 750,000 Pa and L = 1500 m, I got Q ≈ 0.0081 m³/day.But according to the formula, since ΔP / L is the gradient, and the formula is Q = (k * A * (ΔP / L)) / μ, then using the gradient directly is correct, and the result should be 0.009375e-12 m³/s, which is 8.1e-10 m³/day.But that's conflicting with the other approach.Wait, maybe I made a mistake in the first approach. Let me recast it.If I use Q = (k * A * ΔP) / (μ * L), with ΔP = 750,000 Pa and L = 1500 m.So,Q = (0.15e-12 * 0.1 * 750,000) / (0.0008 * 1500)Compute numerator:0.15e-12 * 0.1 = 0.015e-120.015e-12 * 750,000 = 0.015 * 750,000 * 1e-12 = 11,250 * 1e-12 = 1.125e-8Denominator:0.0008 * 1500 = 1.2So, Q = 1.125e-8 / 1.2 ≈ 0.9375e-8 m³/sConvert to m³/day:0.9375e-8 * 86,400 ≈ 0.9375 * 86.4e-4 ≈ 0.9375 * 0.00864 ≈ 0.0081 m³/daySo, 0.0081 m³/day.But according to the formula, Q = (k * A * (ΔP / L)) / μ, which is the same as Q = (k * A * gradient) / μ.Wait, but in the first approach, I used ΔP = gradient * L, so Q = (k * A * gradient * L) / (μ * L) = (k * A * gradient) / μ, which is the same as the second approach. So, both approaches should give the same result.But in the first approach, I got 0.0081 m³/day, and in the second approach, I got 0.00000000081 m³/day. That can't be.Wait, no, in the second approach, I think I messed up the exponent.Wait, in the second approach:Q = (0.15e-12 * 0.1 * 500) / 0.0008= (0.15 * 0.1 * 500) * 1e-12 / 0.0008= (7.5) * 1e-12 / 0.0008= 7.5 / 0.0008 * 1e-12= 9375 * 1e-12= 9.375e-9 m³/sConvert to m³/day:9.375e-9 * 86,400 ≈ 9.375 * 86.4e-4 ≈ 9.375 * 0.000864 ≈ 0.0081 m³/dayAh! There we go. So, I must have messed up the exponent earlier.So, correct calculation:Q = (0.15e-12 * 0.1 * 500) / 0.0008= (0.15 * 0.1 * 500) * 1e-12 / 0.0008= 7.5 * 1e-12 / 0.0008= 7.5 / 0.0008 * 1e-12= 9375 * 1e-12= 9.375e-9 m³/sConvert to m³/day:9.375e-9 * 86,400 ≈ 9.375 * 86.4e-4 ≈ 9.375 * 0.000864 ≈ 0.0081 m³/daySo, both approaches give the same result: approximately 0.0081 m³/day.Therefore, the correct volumetric flow rate is approximately 0.0081 cubic meters per day.But wait, 0.0081 m³/day is 8.1 liters per day. That seems extremely low for an oil well. Even with low permeability, 8 liters a day seems too low. Maybe I made a mistake in the exponent somewhere.Wait, let's recast the calculation:k = 150 millidarcies = 0.15 darcies = 0.15e-12 m²A = 0.1 m²ΔP_gradient = 500 kPa/km = 500,000 Pa / 1000 m = 500 Pa/mμ = 0.8 centipoise = 0.0008 Pa·sSo, Q = (k * A * gradient) / μ= (0.15e-12 * 0.1 * 500) / 0.0008Compute numerator:0.15e-12 * 0.1 = 0.015e-120.015e-12 * 500 = 0.0075e-12Denominator: 0.0008So, Q = 0.0075e-12 / 0.0008 = (0.0075 / 0.0008) * 1e-12 = 9.375 * 1e-12 = 9.375e-12 m³/sWait, no, that's not right. Wait, 0.0075e-12 divided by 0.0008 is (0.0075 / 0.0008) * 1e-12 = 9.375 * 1e-12 = 9.375e-12 m³/sConvert to m³/day:9.375e-12 * 86,400 ≈ 9.375 * 86.4e-6 ≈ 9.375 * 0.0000864 ≈ 0.00081 m³/dayWait, so now I'm getting 0.00081 m³/day, which is 0.81 liters per day. That's even lower.Wait, I think I'm making a mistake in the exponent handling.Let me compute it step by step without exponents:k = 0.15e-12 m² = 0.15 / 1e12 m² = 1.5e-13 m²A = 0.1 m²gradient = 500 Pa/mμ = 0.0008 Pa·sSo,Q = (1.5e-13 * 0.1 * 500) / 0.0008Compute numerator:1.5e-13 * 0.1 = 1.5e-141.5e-14 * 500 = 7.5e-12Denominator: 0.0008So, Q = 7.5e-12 / 0.0008 = 9.375e-9 m³/sConvert to m³/day:9.375e-9 * 86,400 ≈ 9.375 * 86.4e-4 ≈ 9.375 * 0.000864 ≈ 0.0081 m³/dayAh, there we go. So, 0.0081 m³/day is correct.So, the confusion was due to handling the exponents correctly. So, the correct flow rate is approximately 0.0081 cubic meters per day.But just to get a sense, 0.0081 m³/day is 8.1 liters per day. That seems incredibly low for an oil well, even with low permeability. Maybe the units were misinterpreted.Wait, let me check the units again:- Permeability: 150 millidarcies = 0.15 darcies = 0.15e-12 m². Correct.- Viscosity: 0.8 centipoise = 0.008 poise = 0.0008 Pa·s. Correct.- Pressure gradient: 500 kPa/km = 500,000 Pa / 1000 m = 500 Pa/m. Correct.- Cross-sectional area: 0.1 m². Correct.So, the calculation seems correct, but the result is unexpectedly low. Maybe in reality, the flow rate would be higher because of other factors like multiple wells, but in this case, it's just one well at the center.Alternatively, maybe the pressure gradient is given as 500 kPa per kilometer, but the well is at the center, so the effective length is half the radius? Wait, no, the pressure gradient is given per kilometer, so over the entire radius of 1.5 km, the pressure drop is 750 kPa.Alternatively, maybe the formula should consider the radius in a different way, but since the problem specifies to use Darcy's Law, which is for linear flow, not radial, I think the calculation is correct as is.So, to sum up:1. Total oil volume: approximately 212,057,504 cubic meters.2. Volumetric flow rate: approximately 0.0081 cubic meters per day.But wait, 0.0081 m³/day is 8.1 liters per day. That seems too low. Maybe I made a mistake in the exponent somewhere.Wait, let me compute Q again:k = 0.15e-12 m²A = 0.1 m²gradient = 500 Pa/mμ = 0.0008 Pa·sQ = (0.15e-12 * 0.1 * 500) / 0.0008= (0.15 * 0.1 * 500) * 1e-12 / 0.0008= (7.5) * 1e-12 / 0.0008= 7.5 / 0.0008 * 1e-12= 9375 * 1e-12= 9.375e-9 m³/sConvert to m³/day:9.375e-9 * 86,400 = 9.375 * 86.4e-4 = 9.375 * 0.000864 ≈ 0.0081 m³/dayYes, that's correct. So, despite seeming low, the calculation is accurate given the parameters.Therefore, the answers are:1. Total oil volume: approximately 212,057,504 m³2. Volumetric flow rate: approximately 0.0081 m³/dayBut wait, 0.0081 m³/day is 8.1 liters per day. That's extremely low. Maybe the units for permeability were misinterpreted? Let me check.Permeability is given as 150 millidarcies. 1 millidarcy is 1e-3 darcies, so 150 millidarcies is 0.15 darcies, which is 0.15e-12 m². Correct.Alternatively, sometimes permeability is expressed in units of mD (millidarcies), and 1 mD = 1e-15 m². Wait, no, 1 darcy is 1e-12 m², so 1 millidarcy is 1e-15 m². Wait, is that correct?Wait, no, 1 darcy = 1e-12 m², so 1 millidarcy = 1e-3 darcies = 1e-15 m². So, 150 millidarcies = 150e-15 m² = 1.5e-13 m².Wait, earlier I used 0.15e-12 m², which is 1.5e-13 m². So, that's correct.So, no mistake there.Alternatively, maybe the pressure gradient is 500 kPa per kilometer, which is 500,000 Pa per 1000 meters, so 500 Pa/m. Correct.So, I think the calculation is correct, even though the result seems low.Therefore, the final answers are:1. Total oil volume: approximately 212,057,504 cubic meters.2. Volumetric flow rate: approximately 0.0081 cubic meters per day.But to express them more neatly:1. 212,057,504 m³ ≈ 2.1206e8 m³2. 0.0081 m³/day ≈ 8.1e-3 m³/dayAlternatively, maybe the flow rate should be expressed in a different unit, like barrels per day, but the question asks for cubic meters per day.So, I think that's the answer.</think>

🧠:<think>Okay, so I have this problem about a state representative analyzing reentry programs. She's using a probabilistic model to measure success rates in reducing recidivism. The model is given by the function:[ P(S) = frac{e^{kN}}{1 + e^{kN}} ]where ( N ) is the number of participants and ( k ) is a constant representing the program's effectiveness.There are two parts to the problem. Let me tackle them one by one.1. Determining ( N ) in terms of ( k ) for an 80% success rate.Alright, so the first part is asking: If the representative wants an 80% success rate, what should ( N ) be in terms of ( k )?Let me write down what I know. The success probability ( P(S) ) is 0.8, and the formula is:[ 0.8 = frac{e^{kN}}{1 + e^{kN}} ]I need to solve for ( N ). Hmm, okay, let's see. This looks like a logistic function, which is commonly used in probability models. The form is similar to the sigmoid function.First, I can rewrite the equation:[ 0.8 = frac{e^{kN}}{1 + e^{kN}} ]Let me denote ( e^{kN} ) as ( x ) for simplicity. Then the equation becomes:[ 0.8 = frac{x}{1 + x} ]Now, solving for ( x ):Multiply both sides by ( 1 + x ):[ 0.8(1 + x) = x ]Expanding the left side:[ 0.8 + 0.8x = x ]Subtract ( 0.8x ) from both sides:[ 0.8 = x - 0.8x ][ 0.8 = 0.2x ]Divide both sides by 0.2:[ x = frac{0.8}{0.2} ][ x = 4 ]But remember, ( x = e^{kN} ), so:[ e^{kN} = 4 ]To solve for ( N ), take the natural logarithm of both sides:[ ln(e^{kN}) = ln(4) ][ kN = ln(4) ]Therefore, solving for ( N ):[ N = frac{ln(4)}{k} ]So, that's the value of ( N ) in terms of ( k ) needed to achieve an 80% success rate.Wait, let me double-check my steps. Starting from the equation:[ 0.8 = frac{e^{kN}}{1 + e^{kN}} ]Cross-multiplied:[ 0.8(1 + e^{kN}) = e^{kN} ][ 0.8 + 0.8e^{kN} = e^{kN} ][ 0.8 = e^{kN} - 0.8e^{kN} ][ 0.8 = 0.2e^{kN} ][ e^{kN} = 4 ]Yes, that's correct. So, ( N = frac{ln(4)}{k} ). That seems right.2. Deriving an expression for ( k ) in terms of ( B ), ( C ), and ( P(S) ).Alright, moving on to the second part. The state has a fixed budget ( B ) allocated to the reentry programs, and the cost per participant is ( C ). So, the number of participants ( N ) is given by:[ N = frac{B}{C} ]We need to derive an expression for ( k ) in terms of ( B ), ( C ), and the desired success rate ( P(S) ).From the first part, we have the equation:[ P(S) = frac{e^{kN}}{1 + e^{kN}} ]We can use this equation and substitute ( N ) with ( frac{B}{C} ), then solve for ( k ).Let me write down the equation again:[ P(S) = frac{e^{k cdot frac{B}{C}}}{1 + e^{k cdot frac{B}{C}}} ]Let me denote ( e^{k cdot frac{B}{C}} ) as ( y ) for simplicity. Then:[ P(S) = frac{y}{1 + y} ]Solving for ( y ):Multiply both sides by ( 1 + y ):[ P(S)(1 + y) = y ][ P(S) + P(S)y = y ]Subtract ( P(S)y ) from both sides:[ P(S) = y - P(S)y ][ P(S) = y(1 - P(S)) ]Therefore:[ y = frac{P(S)}{1 - P(S)} ]But ( y = e^{k cdot frac{B}{C}} ), so:[ e^{k cdot frac{B}{C}} = frac{P(S)}{1 - P(S)} ]Now, take the natural logarithm of both sides:[ lnleft(e^{k cdot frac{B}{C}}right) = lnleft(frac{P(S)}{1 - P(S)}right) ][ k cdot frac{B}{C} = lnleft(frac{P(S)}{1 - P(S)}right) ]Solving for ( k ):[ k = frac{C}{B} cdot lnleft(frac{P(S)}{1 - P(S)}right) ]So, that's the expression for ( k ) in terms of ( B ), ( C ), and ( P(S) ).Let me verify my steps again. Starting from:[ P(S) = frac{e^{kN}}{1 + e^{kN}} ]Substitute ( N = frac{B}{C} ):[ P(S) = frac{e^{k cdot frac{B}{C}}}{1 + e^{k cdot frac{B}{C}}} ]Let ( y = e^{k cdot frac{B}{C}} ):[ P(S) = frac{y}{1 + y} ]Solving for ( y ):[ y = frac{P(S)}{1 - P(S)} ]Then, ( y = e^{k cdot frac{B}{C}} ) implies:[ k cdot frac{B}{C} = lnleft(frac{P(S)}{1 - P(S)}right) ]Thus:[ k = frac{C}{B} lnleft(frac{P(S)}{1 - P(S)}right) ]Yes, that seems correct.So, summarizing:1. For an 80% success rate, ( N = frac{ln(4)}{k} ).2. Expressing ( k ) in terms of ( B ), ( C ), and ( P(S) ), we get ( k = frac{C}{B} lnleft(frac{P(S)}{1 - P(S)}right) ).I think that's all. Let me just make sure I didn't make any algebraic errors, especially with the logarithms and exponents.In the first part, starting with 0.8 equals that fraction, solving for ( e^{kN} ) gives 4, so ( kN = ln(4) ). That seems right because ( e^{ln(4)} = 4 ).In the second part, expressing ( k ) in terms of ( B ), ( C ), and ( P(S) ), we manipulate the same equation but substitute ( N ) with ( B/C ). Then, solving for ( k ) gives us the logarithm expression. That also seems consistent.I don't see any mistakes in the reasoning or calculations.Final Answer1. The value of ( N ) is boxed{dfrac{ln(4)}{k}}.2. The expression for ( k ) is boxed{dfrac{C}{B} lnleft(dfrac{P(S)}{1 - P(S)}right)}.</think>

🧠:<think>Okay, so I have this problem about build times for a project, and they follow a normal distribution. I need to find the mean and standard deviation given that 95% of the build times are between 10 and 30 minutes. Then, I have to figure out the new 95% confidence interval after optimizing the process, which reduces the mean by 20% and halves the standard deviation. Hmm, let's break this down step by step.First, part 1: finding μ and σ. I remember that for a normal distribution, about 95% of the data lies within two standard deviations from the mean. So, if 95% of the build times are between 10 and 30 minutes, that interval should correspond to μ ± 2σ. Let me write that down:10 = μ - 2σ 30 = μ + 2σSo, I have two equations here. I can solve these simultaneously to find μ and σ. Let me add the two equations together:10 + 30 = (μ - 2σ) + (μ + 2σ) 40 = 2μ So, μ = 20.Okay, that makes sense. The mean is 20 minutes. Now, to find σ, I can plug μ back into one of the equations. Let's use the first one:10 = 20 - 2σ Subtract 20 from both sides: -10 = -2σ Divide both sides by -2: σ = 5.So, the standard deviation is 5 minutes. Let me just double-check that with the second equation:30 = 20 + 2σ 30 - 20 = 2σ 10 = 2σ σ = 5.Yep, that's consistent. So, part 1 is done: μ = 20, σ = 5.Moving on to part 2: optimizing the process. The mean is reduced by 20%, and the standard deviation is halved. Let me compute the new mean and standard deviation.First, reducing the mean by 20%. The original mean is 20 minutes. So, 20% of 20 is 0.2 * 20 = 4. So, the new mean μ' is 20 - 4 = 16 minutes.Next, halving the standard deviation. The original σ is 5, so half of that is 2.5. So, the new standard deviation σ' is 2.5 minutes.Now, we need the new 95% confidence interval. Again, for a normal distribution, 95% of the data lies within μ ± 2σ. So, let's compute that.Lower bound: μ' - 2σ' = 16 - 2*2.5 = 16 - 5 = 11 minutes. Upper bound: μ' + 2σ' = 16 + 5 = 21 minutes.So, the new 95% confidence interval is between 11 and 21 minutes. Let me just make sure that makes sense. The mean decreased by 4, so from 20 to 16, and the standard deviation decreased from 5 to 2.5. So, the interval should be narrower and shifted to the left. 11 to 21 seems right because it's centered at 16, which is 4 less than 20, and the width is now 10 minutes instead of 20 minutes. That seems correct.Wait, actually, hold on. The original interval was 10 to 30, which is 20 minutes wide. The new interval is 11 to 21, which is 10 minutes wide. Since the standard deviation was halved, the interval should be half as wide, which it is. So, that makes sense.I think that's all. So, summarizing:1. μ = 20, σ = 5 2. New interval: 11 to 21 minutesFinal Answer1. The mean is boxed{20} minutes and the standard deviation is boxed{5} minutes. 2. The new 95% confidence interval is boxed{[11, 21]} minutes.</think>

🧠:<think>Okay, so I have this problem about organizing braille books in a library. It's divided into two parts. Let me tackle them one by one.Starting with the first part: Given that the total number of books is 256, I need to find the minimum length ( n ) of the binary sequence required to uniquely represent each book. Hmm, binary sequences of length ( n ) can represent ( 2^n ) different combinations since each position can be either a 0 or a 1. So, if I have 256 books, each needs a unique sequence. That means ( 2^n ) should be at least 256.Let me write that down:( 2^n geq 256 )I know that 256 is a power of 2. Let me recall, 2^8 is 256 because 2^10 is 1024, so 2^8 is 256. Therefore, ( n = 8 ) should be the minimum length because 2^8 = 256, which exactly matches the number of books. If I used a smaller ( n ), like 7, then 2^7 is 128, which is less than 256, so that wouldn't be enough. So, n has to be 8.Alright, that seems straightforward. Now, moving on to the second part. The volunteer wants to minimize the effort required to find a specific book, and the effort is proportional to the Hamming distance between the book's sequence and a reference sequence ( R ). I need to derive a strategy to choose ( R ) such that the average effort over all possible books is minimized, assuming the binary sequences are uniformly distributed.First, let me recall what Hamming distance is. It's the number of positions at which the corresponding symbols are different. So, for two binary sequences of length ( n ), the Hamming distance is the count of positions where one has a 0 and the other has a 1.The goal is to choose ( R ) such that the average Hamming distance from ( R ) to all other sequences is minimized. Since the sequences are uniformly distributed, each sequence is equally likely. So, I need to find an ( R ) that minimizes the expected Hamming distance.Let me think about how to approach this. The Hamming distance between two sequences is the sum of the differences at each bit position. So, for each bit position, if the reference ( R ) has a 0, then the number of sequences that have a 1 at that position is half of the total, right? Because the sequences are uniformly distributed. Similarly, if ( R ) has a 1, the number of sequences with a 0 at that position is also half.Wait, so for each bit position, the contribution to the Hamming distance is the number of sequences that differ at that bit. Since the sequences are uniformly distributed, for each bit, half of them will differ from ( R ) at that position. So, for each bit, the average contribution to the Hamming distance is ( frac{1}{2} times 1 ) because half the time, it's different, and half the time, it's the same.But hold on, the average Hamming distance is the sum over all bit positions of the average contribution per bit. So, if each bit contributes an average of ( frac{1}{2} ) to the Hamming distance, then the total average Hamming distance would be ( n times frac{1}{2} = frac{n}{2} ).But is this the minimum? Or can we choose ( R ) such that the average is lower?Wait, maybe I need to think about it differently. The average Hamming distance is actually dependent on the choice of ( R ). But since the distribution is uniform, the choice of ( R ) doesn't affect the average Hamming distance. Because regardless of what ( R ) is, each bit is equally likely to be 0 or 1 in the other sequences, so the average Hamming distance remains the same.Let me verify that. Suppose ( R ) is all 0s. Then, for each bit, half of the sequences have a 1, so the average number of differing bits is ( frac{n}{2} ). Similarly, if ( R ) is all 1s, the same applies. If ( R ) is a mix of 0s and 1s, say, half 0s and half 1s, then for each bit, the number of differing sequences is still half, because for each bit, half the sequences will differ from ( R ) at that position.Wait, is that true? Let me think about a specific bit. Suppose ( R ) has a 0 at position 1. Then, half of the sequences have a 1 at position 1, contributing 1 to the Hamming distance for those sequences. Similarly, if ( R ) has a 1 at position 1, half the sequences have a 0, again contributing 1. So, regardless of what ( R ) is, each bit contributes an average of ( frac{1}{2} ) to the Hamming distance.Therefore, the average Hamming distance is always ( frac{n}{2} ), regardless of the choice of ( R ). So, does that mean that all choices of ( R ) are equally good in terms of minimizing the average effort? That seems counterintuitive because I thought maybe choosing a central point or something would help, but in this case, since the distribution is uniform, it doesn't matter.But wait, maybe I'm missing something. The problem says the effort is proportional to the Hamming distance. So, if the average Hamming distance is fixed, then the average effort is fixed too, regardless of ( R ). So, in that case, any ( R ) would be equally good. But that seems odd because in other distributions, the choice of ( R ) can affect the average distance.Wait, but in this case, since the distribution is uniform, every sequence is equally likely, so the average distance is the same from any reference point. So, in that case, the average effort is fixed, and there's no way to choose ( R ) to make it better.But the problem says "derive a strategy to choose ( R ) such that the average effort over all possible books is minimized." So, maybe I'm wrong in assuming that the average is fixed. Let me think again.Alternatively, perhaps the average Hamming distance is minimized when ( R ) is chosen such that it's the most "central" point, but in the hypercube graph, all points are symmetric. So, in the hypercube, the average distance from any node is the same. So, perhaps, regardless of ( R ), the average Hamming distance is the same.Wait, let me calculate it. For a hypercube of dimension ( n ), the average Hamming distance from a given vertex is indeed ( frac{n}{2} ). Because each bit has a 50% chance of being different, so the expected number of differing bits is ( n times frac{1}{2} ).Therefore, the average Hamming distance is ( frac{n}{2} ), and it's the same for any ( R ). So, in that case, the choice of ( R ) doesn't affect the average effort. Hence, any ( R ) would do, and the average effort is minimized at ( frac{n}{2} ).But the problem says "derive a strategy to choose ( R )", which implies that there is a specific way to choose ( R ) to minimize the average effort. Maybe I'm misunderstanding something.Wait, perhaps I need to think about the median or something. In some cases, choosing the median minimizes the average distance. But in this case, since the distribution is uniform, all points are symmetric, so the median is not unique or doesn't help.Alternatively, maybe the problem is referring to minimizing the maximum effort, but the question says average effort. Hmm.Wait, let me think about the variance. If we choose ( R ) such that it's in the middle, maybe the variance is minimized, but the average is still the same.Alternatively, perhaps I need to consider the concept of a centroid or something else. But in Hamming space, the centroid isn't well-defined in the same way as Euclidean space.Wait, another approach: For each bit position, if we set ( R )'s bit to 0 or 1, it affects the Hamming distance. Since the distribution is uniform, for each bit, half the sequences have 0 and half have 1. So, if we set ( R )'s bit to 0, then half the sequences will differ on that bit, contributing 1 to the Hamming distance, and the other half will not. Similarly, if we set it to 1, the same thing happens. So, for each bit, regardless of what we set in ( R ), half the sequences will differ on that bit.Therefore, the total average Hamming distance is ( n times frac{1}{2} ), which is ( frac{n}{2} ). So, no matter what ( R ) we choose, the average Hamming distance remains the same.Therefore, the average effort is fixed, and there's no way to choose ( R ) to make it lower. So, the strategy is that any ( R ) will result in the same average effort, so we can choose any ( R ). But the problem says "derive a strategy to choose ( R )", so maybe I'm missing something.Wait, perhaps the problem is considering the average Hamming distance over all pairs, but no, it's the average effort over all possible books, which is the average Hamming distance from ( R ) to all books.Wait, another thought: Maybe the problem is considering the sum of Hamming distances rather than the average. If it's the sum, then the average is just the sum divided by the number of books. But in that case, the sum is ( 256 times frac{n}{2} ), which is fixed, so again, the sum is fixed regardless of ( R ).Alternatively, maybe the problem is considering something else, like the median Hamming distance or the mode. But the question specifically says average effort, which is the expected Hamming distance.Wait, perhaps the problem is not about the average over all books, but the average over all possible searches. But no, it says "the average effort over all possible books", which is the same as the expected Hamming distance from ( R ) to a randomly chosen book.Given that, and since the distribution is uniform, the expected Hamming distance is fixed at ( frac{n}{2} ). Therefore, the average effort is fixed, and there's no way to choose ( R ) to minimize it further. So, the strategy is that any ( R ) will do, as the average effort is the same.But the problem says "derive a strategy to choose ( R )", implying that there is a specific choice. Maybe I need to think differently.Wait, perhaps the problem is considering that the volunteer can choose ( R ) based on the distribution of the books. But the problem says the binary sequences are uniformly distributed, so the volunteer cannot exploit any structure in the data because it's uniform.Alternatively, maybe the problem is about minimizing the maximum Hamming distance, but the question says average effort.Wait, let me think about the definition of Hamming distance. It's the number of differing bits. So, for each bit, if ( R ) has a 0, then all sequences with 1 at that bit contribute 1 to the Hamming distance. Similarly, if ( R ) has a 1, all sequences with 0 contribute 1.Since the sequences are uniformly distributed, for each bit, exactly half of them have 0 and half have 1. Therefore, for each bit, the number of sequences that differ from ( R ) at that bit is exactly half, regardless of ( R )'s bit.Therefore, for each bit, the contribution to the average Hamming distance is ( frac{1}{2} ). So, the total average Hamming distance is ( frac{n}{2} ).Therefore, regardless of how ( R ) is chosen, the average Hamming distance remains ( frac{n}{2} ). So, the average effort is fixed, and there's no way to choose ( R ) to make it lower.So, in that case, the strategy is that any ( R ) will result in the same average effort. Therefore, the volunteer can choose any reference sequence ( R ), as it won't affect the average effort.But the problem says "derive a strategy to choose ( R )", so maybe I'm supposed to say that any ( R ) is fine, but perhaps in practice, choosing a specific ( R ) might have other benefits, like making it easier for the volunteer to remember or something. But in terms of minimizing the average effort, it doesn't matter.Alternatively, maybe the problem is considering the average Hamming distance over all pairs, but no, it's specifically about the average effort over all books, which is from ( R ) to each book.Wait, maybe I need to think about the variance. If we choose ( R ) such that it's in the middle, the variance might be minimized, but the average is still the same. So, perhaps the problem is about minimizing the maximum Hamming distance, but the question says average.Alternatively, maybe the problem is considering that the volunteer can choose ( R ) adaptively, but no, the problem says "derive a strategy to choose ( R )", so it's a fixed ( R ).Wait, perhaps I'm overcomplicating it. The key point is that for a uniform distribution over the hypercube, the average Hamming distance from any vertex is the same, which is ( frac{n}{2} ). Therefore, the average effort is fixed, and there's no way to choose ( R ) to make it lower. So, the strategy is that any ( R ) will do.But the problem says "derive a strategy", so maybe I need to explain why any ( R ) is fine. Alternatively, maybe the problem is expecting me to say that ( R ) should be the all-zero sequence or something, but that doesn't make sense because it doesn't affect the average.Alternatively, maybe the problem is considering that the volunteer can choose ( R ) such that it's the most common sequence, but since the distribution is uniform, all sequences are equally common, so that doesn't help.Wait, another thought: Maybe the problem is considering that the volunteer can choose ( R ) to minimize the sum of Hamming distances, but as I thought earlier, the sum is fixed because each bit contributes half the time. So, the sum is fixed, and thus the average is fixed.Therefore, the conclusion is that the average effort is fixed at ( frac{n}{2} ), regardless of ( R ). So, the volunteer can choose any ( R ), as it won't affect the average effort.But the problem says "derive a strategy to choose ( R )", so maybe the answer is that any ( R ) is acceptable, as the average effort is the same. Alternatively, perhaps the problem is expecting me to say that ( R ) should be chosen such that each bit is 0 or 1 with equal probability, but since ( R ) is a fixed sequence, that doesn't make sense.Wait, maybe the problem is considering that the volunteer can choose ( R ) to minimize the maximum Hamming distance, but the question is about the average. So, perhaps the answer is that any ( R ) is fine, as the average is fixed.Alternatively, maybe I'm missing a mathematical approach. Let me try to model it.Let ( R ) be a binary sequence of length ( n ). Let ( X ) be a uniformly random binary sequence of length ( n ). The Hamming distance between ( R ) and ( X ) is ( d(R, X) = sum_{i=1}^n mathbf{1}_{{R_i neq X_i}} ).The expected Hamming distance is ( E[d(R, X)] = sum_{i=1}^n E[mathbf{1}_{{R_i neq X_i}}] ).Since ( X_i ) is uniformly distributed over {0,1}, and independent of ( R_i ), we have:( E[mathbf{1}_{{R_i neq X_i}}] = P(R_i neq X_i) = frac{1}{2} ).Therefore, ( E[d(R, X)] = sum_{i=1}^n frac{1}{2} = frac{n}{2} ).So, regardless of ( R ), the expected Hamming distance is ( frac{n}{2} ). Therefore, the average effort is fixed, and there's no way to choose ( R ) to make it lower.Therefore, the strategy is that any ( R ) will result in the same average effort. So, the volunteer can choose any reference sequence ( R ), as it won't affect the average effort.But the problem says "derive a strategy to choose ( R )", so maybe I need to explain that any ( R ) is fine because of the uniform distribution. Alternatively, perhaps the problem is expecting me to say that ( R ) should be the all-zero sequence or something, but that doesn't make sense because it doesn't affect the average.Alternatively, maybe the problem is considering that the volunteer can choose ( R ) such that it's the most "central" point, but in the hypercube, all points are symmetric, so there's no central point.Therefore, the conclusion is that the average effort is fixed at ( frac{n}{2} ), and the choice of ( R ) doesn't affect it. So, the strategy is that any ( R ) is acceptable.But wait, the problem is in the context of a library, so maybe the volunteer can choose ( R ) based on the actual distribution of books, but the problem says the sequences are uniformly distributed. So, the volunteer cannot exploit any structure because it's uniform.Therefore, the answer is that any ( R ) will do, as the average effort is fixed.But the problem says "derive a strategy", so maybe I need to say that the volunteer can choose any ( R ), as the average effort is the same regardless of the choice.Alternatively, perhaps the problem is expecting me to say that ( R ) should be chosen such that each bit is 0 or 1 with equal probability, but since ( R ) is a fixed sequence, that doesn't make sense.Wait, another thought: Maybe the problem is considering that the volunteer can choose ( R ) such that it's the most "balanced" sequence, but in this case, all sequences are equally balanced because the distribution is uniform.Therefore, the strategy is that any ( R ) is acceptable, as the average effort is fixed.So, to summarize:1. The minimum length ( n ) is 8 because ( 2^8 = 256 ).2. The strategy is that any reference sequence ( R ) will result in the same average effort, which is ( frac{n}{2} = 4 ), because the distribution is uniform. Therefore, the volunteer can choose any ( R ), as it won't affect the average effort.But the problem says "derive a strategy to choose ( R )", so maybe I need to explain why any ( R ) is fine, based on the uniform distribution.Alternatively, perhaps the problem is expecting me to say that ( R ) should be chosen such that it's the most "central" point, but in this case, all points are equally central because of the uniform distribution.Therefore, the answer is that any ( R ) is acceptable, as the average effort is fixed.But to make sure, let me think about a small example. Suppose ( n = 2 ). There are 4 sequences: 00, 01, 10, 11.If ( R = 00 ), the Hamming distances are: 0, 1, 1, 2. The average is (0 + 1 + 1 + 2)/4 = 4/4 = 1.If ( R = 01 ), the distances are: 1, 0, 2, 1. The average is (1 + 0 + 2 + 1)/4 = 4/4 = 1.Similarly, for ( R = 10 ) and ( R = 11 ), the average is also 1.So, in this case, regardless of ( R ), the average Hamming distance is the same, which is ( n/2 = 1 ).Therefore, this confirms that for ( n = 2 ), the average is fixed regardless of ( R ). So, the same applies for any ( n ).Therefore, the strategy is that any ( R ) will do, as the average effort is fixed at ( n/2 ).So, putting it all together:1. The minimum length ( n ) is 8.2. The strategy is that any reference sequence ( R ) will result in the same average effort, so the volunteer can choose any ( R ).</think>