answer:First, let's calculate the total weight of tuna Mr. Ray has: 10 tuna * 200 pounds each = 2000 pounds of tuna Now, let's see how he can accommodate the customers with specific weight preferences: 10 customers want exactly 30-pound pieces. To satisfy these customers, Mr. Ray needs: 10 customers * 30 pounds each = 300 pounds of tuna 15 customers want exactly 20-pound pieces. To satisfy these customers, Mr. Ray needs: 15 customers * 20 pounds each = 300 pounds of tuna So far, Mr. Ray has allocated: 300 pounds (for 30-pound pieces) + 300 pounds (for 20-pound pieces) = 600 pounds of tuna This leaves Mr. Ray with: 2000 pounds (total) - 600 pounds (allocated) = 1400 pounds of tuna for the remaining customers. Each of the remaining customers wants 25 pounds of tuna. Let's see how many customers can be satisfied with the remaining 1400 pounds: 1400 pounds / 25 pounds per customer = 56 customers Now, let's add the customers with specific weight preferences to the number of customers who can be satisfied with the remaining tuna: 10 customers (30-pound pieces) + 15 customers (20-pound pieces) + 56 customers (25-pound pieces) = 81 customers Mr. Ray has 100 customers waiting, so the number of customers who will go home without any fish is: 100 customers - 81 customers = 19 customers Therefore, boxed{19} customers will go home without any fish.

answer:From the equation sin alpha cos (alpha - beta) - cos alpha sin (alpha - beta) = frac{4}{5}, we obtain sin [alpha - (alpha - beta)] = frac{4}{5}, which simplifies to sin beta = frac{4}{5}. Hence, the answer is boxed{frac{4}{5}}. This is directly derived by applying the trigonometric identity for the sine of a difference between two angles to simplify and find sin beta. This problem tests the understanding of the sine of a difference between two angles and is a basic computation question.

answer:Let t=g(t)=x^{2}-2ax+3, then the function y=log_{2}t is increasing. If the function f(x)=log_{2}(x^{2}-2ax+3) is monotonically decreasing within the interval left( frac{1}{2},1right), it is equivalent to g(t)=x^{2}-2ax+3 being monotonically decreasing within the interval left( frac{1}{2},1right) and g(1)geqslant 0, which means begin{cases} -frac{-2a}{2}=ageqslant 1 g(1)=1-2a+3geqslant 0end{cases}, solving this gives 1leqslant aleqslant 2, thus, the range of values for a is boxed{[1,2]}. By using the substitution method and combining the relationship between the monotonicity of composite functions, the conclusion can be reached. This problem mainly examines the application of the monotonicity of composite functions. Using the substitution method combined with the relationship between the monotonicity of composite functions is the key to solving this problem.

answer:To solve this problem, we start with the given sequence and repeatedly divide each term by 2 to find the next term. The sequence starts as follows: 1. 1,000,000 div 2 = 500,000 2. 500,000 div 2 = 250,000 3. 250,000 div 2 = 125,000 4. 125,000 div 2 = 62,500 5. 62,500 div 2 = 31,250 6. 31,250 div 2 = 15,625 At this point, we observe that 15,625 is the last term in the sequence that is an integer when divided by 2. The next division would be: 7. 15,625 div 2 = 7,812.5 This result is not an integer, indicating that 15,625 is the last integer in the sequence before the results stop being integers. Therefore, the last integer in this sequence is boxed{15625}.