answer:Let's assume the cost price of the product for 1 kg is 100 (for easy calculation). Since the dealer is using a counterfeit weight that is 20% less than the real weight, for every 1 kg the dealer claims to sell, he actually gives out only 0.8 kg (because 20% less than 1 kg is 0.8 kg). Now, the dealer adds 15% impurities to the product. This means for every 0.8 kg of the actual product, he adds 0.15 * 0.8 kg = 0.12 kg of impurities. So, the total weight of the product with impurities that the dealer gives out for the price of 1 kg is 0.8 kg + 0.12 kg = 0.92 kg. The cost price for 0.92 kg of the adulterated product (which the customer thinks is 1 kg) is still 100, but the actual cost to the dealer for 0.8 kg of the real product is 80 (since 0.8 kg is 80% of 1 kg and the cost price for 1 kg is 100). The dealer's profit for this transaction is the selling price minus the actual cost price, which is 100 - 80 = 20. To find the net profit percentage, we use the formula: Net Profit Percentage = (Net Profit / Cost Price) * 100 Plugging in the values, we get: Net Profit Percentage = (20 / 80) * 100 = 25% Therefore, the net profit percentage of the dealer is boxed{25%} .

answer:Since triangle ABC is a 45-45-90 triangle, the legs AC and BC are equal. Let these legs each be labeled as x. The hypotenuse AB, by the property of 45-45-90 triangles, will be xsqrt{2}. Given that the altitude CD = sqrt{2}, and in a 45-45-90 triangle, the altitude to the hypotenuse splits it into two equal segments and is also equal to frac{x}{sqrt{2}}. Setting this up: [ frac{x}{sqrt{2}} = sqrt{2} ] [ x = 2 ] Now, as AB is the hypotenuse, AB = 2sqrt{2}. The legs AC and BC are each 2 cm. To find the area of triangle ABC, compute: [ text{Area of } triangle ABC = frac{1}{2} times AC times BC = frac{1}{2} times 2 times 2 = 2 text{ square centimeters.} ] Thus, the area of triangle ABC is boxed{2} square centimeters.

answer:1. **Identifying the individual probabilities of missing the target**: - For the first hunter, the probability of missing the rabbit is: [ P(text{miss}_1) = 1 - P(text{hit}_1) = 1 - frac{3}{5} = frac{2}{5} ] - For the second hunter, the probability of missing the rabbit is: [ P(text{miss}_2) = 1 - P(text{hit}_2) = 1 - frac{3}{10} = frac{7}{10} ] - For the third hunter, the probability of missing the rabbit is: [ P(text{miss}_3) = 1 - P(text{hit}_3) = 1 - frac{1}{10} = frac{9}{10} ] 2. **Calculating the combined probability that all hunters miss the rabbit**: - The event that all hunters miss is the product of their individual probabilities of missing: [ P(text{all miss}) = P(text{miss}_1) cdot P(text{miss}_2) cdot P(text{miss}_3) ] - Substituting the individual probabilities: [ P(text{all miss}) = frac{2}{5} cdot frac{7}{10} cdot frac{9}{10} ] Performing the multiplication step-by-step: [ = frac{2}{5} times frac{7}{10} = frac{2 times 7}{5 times 10} = frac{14}{50} ] Next, multiply (frac{14}{50}) by (frac{9}{10}): [ frac{14}{50} times frac{9}{10} = frac{14 times 9}{50 times 10} = frac{126}{500} ] 3. **Calculating the probability that the rabbit is hit**: - The probability that the rabbit is hit is the complement of the event that all hunters miss: [ P(text{hit}) = 1 - P(text{all miss}) ] - Substituting the previously obtained probability (P(text{all miss})): [ P(text{hit}) = 1 - frac{126}{500} ] - Simplifying further: [ P(text{hit}) = 1 - 0.252 = 0.748 ] To conclude, the probability that the rabbit will be killed is: [ boxed{0.748} ]

answer:1. **Case 1: 5 geq x+1 and 5 geq y-5.** We have two inequalities: [ x+1 leq 5 quad text{and} quad y-5 leq 5 ] Solving these, we get: [ x leq 4 quad text{and} quad y leq 10 ] This describes the quarter-plane bounded by x = 4 and y = 10. 2. **Case 2: x+1 geq 5 and x+1 geq y-5.** We have: [ x+1 geq 5 quad text{and} quad x+1 geq y-5 ] Solving, we get: [ x geq 4 quad text{and} quad x+1 geq y-5 implies y leq x+6 ] This describes the region above the line y = x + 6 and to the right of x = 4. 3. **Case 3: y-5 geq 5 and y-5 geq x+1.** [ y-5 geq 5 quad text{and} quad y-5 geq x+1 ] Solving these, we get: [ y geq 10 quad text{and} quad y-5 geq x+1 implies y geq x+6 ] This describes the region below the line y = x + 6 and above y = 10. 4. **Conclusion:** The set S consists of regions bounded by the lines y = x + 6, x = 4, and y = 10, forming a boundary around a region in the quadrant. Thus, the correct description for S is text{a bounded region in the quadrant}. The final answer is boxed{textbf{(A)} text{a single bounded region in the quadrant}}