answer：First, I need to determine the possible sums that can be rolled with two 12-sided dice. The smallest sum is 2 (1+1), and the largest sum is 24 (12+12).Next, I identify all the combinations of two numbers between 1 and 12 that add up to 15. These combinations are:- 3 and 12- 4 and 11- 5 and 10- 6 and 9- 7 and 8- 8 and 7- 9 and 6- 10 and 5- 11 and 4- 12 and 3There are 10 such combinations. Since each die is distinct, each combination represents two distinct outcomes (e.g., 3 on the first die and 12 on the second die is different from 12 on the first die and 3 on the second die). Therefore, there are a total of 20 favorable outcomes.The total number of possible outcomes when rolling two 12-sided dice is 12 * 12 = 144.Finally, the probability of rolling a sum of 15 is the number of favorable outcomes divided by the total number of possible outcomes:Probability = 20 / 144 = 5 / 36

answer：Alright, so I need to prove that for any natural number ( n ), the number of odd binomial coefficients ( mathrm{C}_{n}^{k} ) (where ( 0 leq k leq n )) is a power of 2. Hmm, okay. I remember that binomial coefficients can be even or odd, and there's something about their parity related to binary representations. Maybe Lucas' Theorem is involved here? Let me think.First, I recall that the binomial coefficient ( mathrm{C}_{n}^{k} ) is odd if and only if there is no carryover when adding ( k ) and ( n - k ) in binary. That sounds familiar. So, if I write ( n ) and ( k ) in binary, and for each bit position, the bit of ( k ) is less than or equal to the corresponding bit of ( n ), then ( mathrm{C}_{n}^{k} ) is odd.Okay, so maybe the number of such ( k ) values is related to the binary representation of ( n ). Let's say ( n ) has a binary representation with ( m ) ones. Then, the number of ( k ) such that ( mathrm{C}_{n}^{k} ) is odd would be ( 2^m ). Is that right? Because for each 1 in the binary representation, we have a choice in the corresponding bit of ( k ).Wait, let me test this with an example. Take ( n = 4 ), which is ( 100 ) in binary. The number of ones is 1, so the number of odd coefficients should be ( 2^1 = 2 ). Checking the binomial coefficients for ( n = 4 ): ( 1, 4, 6, 4, 1 ). The odd ones are ( 1 ) and ( 1 ), so that's 2. Okay, that works.Another example: ( n = 5 ), which is ( 101 ) in binary. The number of ones is 2, so the number of odd coefficients should be ( 2^2 = 4 ). The binomial coefficients are ( 1, 5, 10, 10, 5, 1 ). The odd ones are ( 1, 5, 5, 1 ), which is 4. Good.Wait, what about ( n = 3 ), which is ( 11 ) in binary. The number of ones is 2, so ( 2^2 = 4 ). The coefficients are ( 1, 3, 3, 1 ), all odd. That's 4, which matches. Okay, seems consistent.So, if I can show that the number of ( k ) such that ( mathrm{C}_{n}^{k} ) is odd is equal to ( 2^m ), where ( m ) is the number of ones in the binary representation of ( n ), then I'm done. Because ( 2^m ) is obviously a power of 2.How do I formalize this? Maybe using induction on the number of bits in ( n ). Let's try that.Base case: ( n = 1 ), which is ( 1 ) in binary. The number of ones is 1, so the number of odd coefficients should be ( 2^1 = 2 ). The coefficients are ( 1, 1 ), which are both odd. So, base case holds.Inductive step: Assume that for all numbers with up to ( m ) bits, the number of odd binomial coefficients is ( 2^{text{number of ones}} ). Now consider a number ( n ) with ( m + 1 ) bits. Let me write ( n ) as ( 2^k + m ), where ( m < 2^k ). Hmm, not sure if that's the right way.Alternatively, maybe consider the binary representation of ( n ). If the least significant bit is 1, then ( n ) is odd, and the number of odd coefficients is double the number for ( n - 1 ). Wait, is that true?Wait, no. Maybe it's better to think in terms of the binary digits. Each time you have a 1 in the binary representation, it gives you a choice in the corresponding bit of ( k ). So, if ( n ) has ( m ) ones, then for each of those ( m ) positions, ( k ) can have either 0 or 1, giving ( 2^m ) possibilities.But how do I make this rigorous? Maybe using Lucas' Theorem, which states that ( mathrm{C}_{n}^{k} ) is odd if and only if each bit in the binary representation of ( k ) is less than or equal to the corresponding bit in ( n ). So, the number of such ( k ) is exactly the product over each bit of ( n ) of (1 + number of choices for that bit). Since each bit that is 1 in ( n ) gives 2 choices (0 or 1) for ( k ), and each bit that is 0 gives only 1 choice (0). So, the total number is ( 2^m ), where ( m ) is the number of ones in ( n ).Therefore, the number of odd binomial coefficients is ( 2^m ), which is a power of 2.Wait, let me check another example to be sure. Take ( n = 6 ), which is ( 110 ) in binary. Number of ones is 2, so number of odd coefficients should be ( 4 ). The coefficients are ( 1, 6, 15, 20, 15, 6, 1 ). The odd ones are ( 1, 15, 15, 1 ), which is 4. Yep, that works.Another one: ( n = 7 ), which is ( 111 ) in binary. Number of ones is 3, so number of odd coefficients should be ( 8 ). The coefficients are ( 1, 7, 21, 35, 35, 21, 7, 1 ). All are odd, which is 8. Perfect.Okay, I think I've got it. The key idea is that the number of odd binomial coefficients corresponds to the number of ways to choose bits in ( k ) such that they don't exceed the bits in ( n ), which is exactly ( 2^m ) where ( m ) is the number of ones in ( n )'s binary representation. Therefore, the number of odd binomial coefficients is always a power of 2.

answer：Okay, so I have this problem here: Given that -1 < α < β < 1, which of the following inequalities always holds? The options are A: -2 < α - β < 0, B: -2 < α - β < -1, C: -1 < α - β < 0, and D: -1 < α - β < 1.Hmm, let me try to figure this out step by step. First, I know that α and β are both between -1 and 1, and α is less than β. So, α is to the left of β on the number line between -1 and 1.I need to find the range of α - β. Since α < β, subtracting β from α will give a negative number. So, α - β is definitely less than 0. That helps narrow it down because options C and D have upper bounds greater than 0, but since α - β is negative, those might not be the right choices. But let me not jump to conclusions yet.Let me think about the maximum and minimum possible values of α - β. Since both α and β are between -1 and 1, the smallest α can be is just above -1, and the largest β can be is just below 1. So, if α is close to -1 and β is close to 1, then α - β would be approximately -1 - 1 = -2. But since α is greater than -1 and β is less than 1, α - β will be greater than -2. So, the lower bound is -2.On the upper side, since α < β, the maximum α - β can be is when α is as large as possible and β is as small as possible. The largest α can be is just below 1, and the smallest β can be is just above α. So, if α is close to 1 and β is just a little bit larger than α, then α - β would be a small negative number, approaching 0. Therefore, the upper bound is 0.Putting that together, α - β must be between -2 and 0. So, the inequality -2 < α - β < 0 should always hold. That would be option A.Wait, let me double-check. If α is -0.5 and β is 0.5, then α - β is -1, which is within -2 and 0. If α is -0.9 and β is 0.9, then α - β is -1.8, which is still within -2 and 0. If α is 0.9 and β is 0.99, then α - β is -0.09, which is also within -2 and 0. So, it seems like no matter what values α and β take within the given range, α - β will always be between -2 and 0.But what about the other options? Option B says -2 < α - β < -1. That would mean α - β is always less than -1, but in my example where α is 0.9 and β is 0.99, α - β is -0.09, which is greater than -1. So, option B isn't always true.Option C says -1 < α - β < 0. But in my first example, α - β was -1, which is not greater than -1. So, option C isn't always true either.Option D says -1 < α - β < 1. Well, we already established that α - β is always less than 0, so the upper bound of 1 is irrelevant, but the lower bound is -1. However, as in the example where α is -0.9 and β is 0.9, α - β is -1.8, which is less than -1. So, option D isn't always true either.Therefore, the only inequality that always holds is option A: -2 < α - β < 0.

answer：First, I need to determine the fraction of students who are female at Barkely University. Let's denote this fraction as ( F ). Consequently, the fraction of students who are male will be ( 1 - F ).I know that ( frac{5}{6} ) of the female students are on the honor roll and ( frac{2}{3} ) of the male students are on the honor roll. The overall fraction of students on the honor roll is given as ( 0.7333333333333333 ).To find ( F ), I can set up an equation that represents the total fraction of students on the honor roll:[frac{5}{6}F + frac{2}{3}(1 - F) = 0.7333333333333333]Next, I'll simplify this equation. First, expand the terms:[frac{5}{6}F + frac{2}{3} - frac{2}{3}F = 0.7333333333333333]To combine the terms involving ( F ), I'll find a common denominator. The common denominator for 6 and 3 is 6, so I'll convert ( frac{2}{3}F ) to ( frac{4}{6}F ):[frac{5}{6}F + frac{4}{6} - frac{4}{6}F = 0.7333333333333333]Now, combine the ( F ) terms:[left(frac{5}{6} - frac{4}{6}right)F + frac{4}{6} = 0.7333333333333333][frac{1}{6}F + frac{4}{6} = 0.7333333333333333]Subtract ( frac{4}{6} ) from both sides to isolate ( F ):[frac{1}{6}F = 0.7333333333333333 - frac{4}{6}][frac{1}{6}F = 0.7333333333333333 - 0.6666666666666666][frac{1}{6}F = 0.0666666666666667]Finally, multiply both sides by 6 to solve for ( F ):[F = 0.0666666666666667 times 6][F = 0.4]Therefore, the fraction of students who are female is ( 0.4 ) or ( frac{2}{5} ).

answer：Alright, so I have this problem here: I need to determine the sum of all prime numbers ( p ) for which there exists no integer solution in ( x ) to the congruence ( 5(10x + 2) equiv 3 pmod{p} ). Hmm, okay. Let me try to break this down step by step.First, I should probably simplify the given congruence to make it easier to work with. The original congruence is ( 5(10x + 2) equiv 3 pmod{p} ). Let me expand this out. Multiplying 5 by ( 10x + 2 ) gives me ( 50x + 10 ). So, the congruence becomes:[ 50x + 10 equiv 3 pmod{p} ]Now, I can subtract 10 from both sides to isolate the term with ( x ):[ 50x equiv 3 - 10 pmod{p} ][ 50x equiv -7 pmod{p} ]Hmm, so ( 50x equiv -7 pmod{p} ). I can also write this as:[ 50x equiv p - 7 pmod{p} ]But I'm not sure if that helps directly. Maybe I should think about whether this congruence has a solution for ( x ). In modular arithmetic, a linear congruence of the form ( ax equiv b pmod{m} ) has a solution if and only if the greatest common divisor of ( a ) and ( m ) divides ( b ). In this case, ( a ) is 50, ( m ) is ( p ), and ( b ) is ( -7 ).So, the condition for a solution to exist is that ( gcd(50, p) ) divides ( -7 ). Since ( p ) is a prime number, ( gcd(50, p) ) can only be 1 or ( p ) itself, depending on whether ( p ) divides 50 or not.Let me factorize 50 to find its prime factors. 50 is equal to ( 2 times 5^2 ). So, the prime factors of 50 are 2 and 5. Therefore, if ( p ) is 2 or 5, then ( gcd(50, p) ) will be ( p ) itself (since 2 and 5 divide 50). For any other prime ( p ), ( gcd(50, p) ) will be 1.Now, going back to the condition for the existence of a solution. If ( gcd(50, p) = 1 ), then 1 divides ( -7 ), which is always true. Therefore, for any prime ( p ) that is not 2 or 5, the congruence ( 50x equiv -7 pmod{p} ) will have a solution. That means the only primes where there might be no solution are 2 and 5.But wait, I should check whether these primes actually result in no solution or not. Let me test ( p = 2 ) first.For ( p = 2 ), the congruence becomes:[ 50x equiv -7 pmod{2} ]But modulo 2, 50 is congruent to 0, because 50 is even. Similarly, -7 modulo 2 is 1, since -7 divided by 2 is -4 with a remainder of 1. So the congruence simplifies to:[ 0 times x equiv 1 pmod{2} ][ 0 equiv 1 pmod{2} ]Which is clearly not true. Therefore, there is no solution when ( p = 2 ).Now, let's check ( p = 5 ).For ( p = 5 ), the congruence becomes:[ 50x equiv -7 pmod{5} ]Again, 50 modulo 5 is 0, since 50 is a multiple of 5. Similarly, -7 modulo 5 is 3, because -7 + 10 = 3. So the congruence simplifies to:[ 0 times x equiv 3 pmod{5} ][ 0 equiv 3 pmod{5} ]Which is also not true. Therefore, there is no solution when ( p = 5 ) either.So, both primes 2 and 5 result in the congruence having no solution. Therefore, these are the primes we are interested in.Now, the problem asks for the sum of all such primes. So, I need to add 2 and 5 together.[ 2 + 5 = 7 ]Therefore, the sum of all prime numbers ( p ) for which there exists no integer solution in ( x ) to the given congruence is 7.Wait, let me just double-check my reasoning to make sure I didn't miss anything. I considered that for primes other than 2 and 5, the gcd(50, p) is 1, which means the congruence will have a solution. For primes 2 and 5, the gcd is not 1, so the congruence might not have a solution. Then, I checked both cases and confirmed that indeed, there are no solutions for ( p = 2 ) and ( p = 5 ). So, adding them together gives 7. Yeah, that seems correct.I don't think I missed any other primes because 50 only has prime factors 2 and 5. Any other prime would not divide 50, so the gcd would be 1, ensuring a solution exists. Therefore, my conclusion is that the sum is 7.**Final Answer**The sum of all such prime numbers is boxed{7}.

answer：First, calculate Paul's earnings from his main job before taxes and fees:Main job earnings = 15.75/hour × 45 hours = 708.75Next, calculate the taxes on his main job earnings. The first 40 hours are taxed at 22%, and the remaining 5 hours are taxed at 25%:Tax on first 40 hours = 40 hours × 15.75/hour × 22% = 138.60Tax on next 5 hours = 5 hours × 15.75/hour × 25% = 19.6875Total tax on main job = 138.60 + 19.6875 = 158.2875After tax earnings from main job = 708.75 - 158.2875 = 550.4625Now, calculate the 10% fee on his main job earnings:Fee on main job = 708.75 × 10% = 70.875After fee earnings from main job = 550.4625 - 70.875 = 479.5875Next, calculate Paul's earnings from his part-time job before taxes:Part-time job earnings = 13.25/hour × 20 hours = 265Now, calculate the taxes on his part-time job earnings:Tax on part-time job = 265 × 18% = 47.70After tax earnings from part-time job = 265 - 47.70 = 217.30Add Paul's after-tax earnings from both jobs:Total after-tax earnings = 479.5875 (main job) + 217.30 (part-time job) = 696.8875Calculate how much Paul spends on gummy bears, his phone bill, and how much he saves for a vacation:Spending on gummy bears = 696.8875 × 20% = 139.3775Spending on phone bill = 696.8875 × 5% = 34.844375Savings for vacation = (696.8875 - 139.3775 - 34.844375) × 30% = 157.3331625Finally, calculate the remaining amount for other essential expenses:Remaining amount = 696.8875 - 139.3775 - 34.844375 - 157.3331625 = 365.3324625Therefore, Paul has boxed{365.33} left for other essential expenses.