🧠:<think>First, I need to address Sub-problem 1. The journalist has two independent opportunities to secure an interview with the President. The probability of success in the first time slot is 0.3, and in the second time slot, it's 0.5. To find the probability of securing an interview in either slot, I'll use the formula for the probability of the union of two independent events: P(A ∪ B) = P(A) + P(B) - P(A) * P(B). Plugging in the values, this becomes 0.3 + 0.5 - (0.3 * 0.5) = 0.3 + 0.5 - 0.15 = 0.65. So, there's a 65% chance she'll secure the interview on any given day.Next, for Sub-problem 2, the journalist attempts to secure an interview each day for a week, making 7 independent attempts. The probability of not securing an interview on a single day is 1 - 0.65 = 0.35. To find the probability of not securing any interviews throughout the week, I'll raise this daily failure probability to the power of 7: 0.35^7 ≈ 0.0007. Therefore, the probability of securing at least one interview during the week is 1 - 0.0007 ≈ 0.9993, or 99.93%.</think>

🧠:<think>Okay, so I have this problem here about a Chinese entrepreneur who wants to establish trade agreements with three countries: A, B, and C. The potential trade volumes with each country are proportional to their GDPs and inversely proportional to their geographical distances from China. Let me try to break this down step by step. First, part 1 asks me to express the potential trade volumes ( V_A ), ( V_B ), and ( V_C ) in terms of the given variables: proportionality constant ( k ), GDPs ( G_A ), ( G_B ), ( G_C ), and distances ( D_A ), ( D_B ), ( D_C ). Then, since the total trade volume is expected to be 10 billion USD, I need to find ( k ) in terms of these variables.Alright, so if the trade volume is proportional to GDP and inversely proportional to distance, that should translate to a formula like ( V = k times frac{G}{D} ). So, for each country, it would be:( V_A = k times frac{G_A}{D_A} )( V_B = k times frac{G_B}{D_B} )( V_C = k times frac{G_C}{D_C} )That makes sense. So, each trade volume is directly proportional to the GDP and inversely proportional to the distance, scaled by the constant ( k ).Now, the total trade volume is the sum of these three, which is given as 10 billion USD. So:( V_A + V_B + V_C = 10 ) billion USDSubstituting the expressions for each ( V ):( k times frac{G_A}{D_A} + k times frac{G_B}{D_B} + k times frac{G_C}{D_C} = 10 )I can factor out the ( k ):( k left( frac{G_A}{D_A} + frac{G_B}{D_B} + frac{G_C}{D_C} right) = 10 )So, to solve for ( k ), I can divide both sides by the sum of ( frac{G_A}{D_A} + frac{G_B}{D_B} + frac{G_C}{D_C} ):( k = frac{10}{left( frac{G_A}{D_A} + frac{G_B}{D_B} + frac{G_C}{D_C} right)} )Hmm, that seems straightforward. So, ( k ) is 10 divided by the sum of each country's GDP divided by their respective distances.Okay, moving on to part 2. The entrepreneur can invest in infrastructure improvements, which reduce the effective distance by a factor ( alpha ) for each country. So, the new distances become ( D'_A = alpha D_A ), ( D'_B = alpha D_B ), and ( D'_C = alpha D_C ). Therefore, the new trade volumes ( V'_A ), ( V'_B ), ( V'_C ) would be:( V'_A = k times frac{G_A}{D'_A} = k times frac{G_A}{alpha D_A} = frac{1}{alpha} times k times frac{G_A}{D_A} = frac{1}{alpha} V_A )Similarly,( V'_B = frac{1}{alpha} V_B )( V'_C = frac{1}{alpha} V_C )So, each new trade volume is just the original divided by ( alpha ). Therefore, the total new trade volume ( V' ) is:( V'_A + V'_B + V'_C = frac{1}{alpha} (V_A + V_B + V_C) )We know from part 1 that ( V_A + V_B + V_C = 10 ) billion USD. So,( V' = frac{1}{alpha} times 10 )But the entrepreneur wants the total trade volume to be 15 billion USD after the improvements. So,( frac{10}{alpha} = 15 )Solving for ( alpha ):Multiply both sides by ( alpha ):( 10 = 15 alpha )Then, divide both sides by 15:( alpha = frac{10}{15} = frac{2}{3} )So, the reduction factor ( alpha ) needs to be ( frac{2}{3} ). That means the effective distance is reduced to two-thirds of the original distance.Wait, let me double-check that. If the distance is reduced by a factor ( alpha ), does that mean the new distance is ( alpha times ) original distance? So, if ( alpha = frac{2}{3} ), the distance becomes two-thirds of what it was. Since trade volume is inversely proportional to distance, reducing distance would increase trade volume. So, if distance is multiplied by ( alpha ), the trade volume is multiplied by ( frac{1}{alpha} ).So, starting with a total of 10 billion, after improvement, it becomes ( frac{10}{alpha} ). We set that equal to 15:( frac{10}{alpha} = 15 implies alpha = frac{10}{15} = frac{2}{3} ). Yeah, that seems correct.Alternatively, I can think about it in terms of the proportionality. If the distance is reduced by a factor ( alpha ), then the trade volume increases by ( frac{1}{alpha} ). So, to go from 10 to 15, which is an increase by a factor of ( frac{15}{10} = 1.5 ), so ( frac{1}{alpha} = 1.5 implies alpha = frac{2}{3} ). Yep, that's consistent.So, I think that's solid.Let me recap:1. Expressed each trade volume as ( V = k times frac{G}{D} ), summed them up, set equal to 10, solved for ( k ).2. With the distance reduced by factor ( alpha ), each trade volume becomes ( frac{1}{alpha} ) times original, so total becomes ( frac{10}{alpha} ). Set equal to 15, solved for ( alpha = frac{2}{3} ).I don't see any mistakes in the reasoning. The key was recognizing that the trade volume is inversely proportional to distance, so reducing distance increases volume proportionally. Another way to think about part 2 is to express the new trade volumes in terms of ( k ) and the new distances, then sum them up and solve for ( alpha ). Let me try that approach to confirm.So, ( V'_A = k times frac{G_A}{alpha D_A} ), similarly for B and C. So, total new trade volume is:( V' = k times left( frac{G_A}{alpha D_A} + frac{G_B}{alpha D_B} + frac{G_C}{alpha D_C} right) = frac{k}{alpha} times left( frac{G_A}{D_A} + frac{G_B}{D_B} + frac{G_C}{D_C} right) )But from part 1, we know that ( k times left( frac{G_A}{D_A} + frac{G_B}{D_B} + frac{G_C}{D_C} right) = 10 ). Therefore,( V' = frac{10}{alpha} )Set this equal to 15:( frac{10}{alpha} = 15 implies alpha = frac{10}{15} = frac{2}{3} )Same result. So, that's consistent. I think that's thorough enough. I considered two different approaches to part 2 and both led to the same answer. So, I feel confident that ( alpha = frac{2}{3} ) is correct.Final Answer1. The proportionality constant ( k ) is ( boxed{dfrac{10}{frac{G_A}{D_A} + frac{G_B}{D_B} + frac{G_C}{D_C}}} ).2. The necessary reduction factor ( alpha ) is ( boxed{dfrac{2}{3}} ).</think>

🧠:<think>First, I need to convert the base-7 number "31452" to its decimal equivalent. I'll do this by multiplying each digit by 7 raised to the power of its position, starting from the rightmost digit.Next, for the arithmetic progression problem, I'll start by converting the first term "52" (base-7) and the common difference "13" (base-7) into decimal. Once I have these values in base-10, I can use the formula for the sum of an arithmetic progression, which is ( S_n = frac{n}{2} times (2a + (n-1)d) ), where ( n ) is the number of terms, ( a ) is the first term, and ( d ) is the common difference. Plugging in the values, I'll calculate the sum of the 20 terms.</think>

🧠:<think>First, I need to determine how much time Dr. Turner spends on discussing books and seeking tax advice. She spends 20% of her time on books and 5% on tax advice. With a 60-hour workweek, that's 12 hours on books and 3 hours on tax advice. This leaves 45 hours for teaching, research, and administrative duties.Next, I'll define the time spent on administrative duties as ( x ) hours. According to the problem, she spends 3 times as much on research as on administrative duties, which is ( 3x ) hours, and twice as much on teaching as on administrative duties, which is ( 2x ) hours. Adding these together gives the equation ( x + 3x + 2x = 45 ) hours. Solving for ( x ), I find that ( x = 9 ) hours. Therefore, she spends 9 hours on administrative duties, 27 hours on research, and 18 hours on teaching.For the second part, Dr. Turner dedicates 15% of her research time to writing the book. Since she spends 27 hours on research, 15% of that is 4.05 hours per week. At a rate of 2000 words per week during this time, after 10 weeks, she will have written 20,000 words.</think>