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question:From the sale of sleeping bags, a retailer made a gross profit of 14% of the wholesale cost. Each sleeping bag was sold for 28. What was the wholesale cost per bag?
answer:Let's denote the wholesale cost per bag as ( W ). The retailer made a gross profit of 14% of the wholesale cost, which means the profit per bag is ( 0.14W ). The selling price per bag is the sum of the wholesale cost and the profit, which is ( W + 0.14W ). We know the selling price per bag is 28, so we can set up the following equation: [ W + 0.14W = 28 ] Combining like terms, we get: [ 1W + 0.14W = 1.14W ] [ 1.14W = 28 ] Now, we can solve for ( W ) by dividing both sides of the equation by 1.14: [ W = frac{28}{1.14} ] [ W approx 24.56 ] Therefore, the wholesale cost per bag is approximately boxed{24.56} .
question:Carolyn and Paul are playing a game starting with a list of the integers 1 to n. The rules of the game are: bullet Carolyn always has the first turn. bullet Carolyn and Paul alternate turns. bullet On each of her turns, Carolyn must remove one number from the list such that this number has at least one positive divisor other than itself remaining in the list. bullet On each of his turns, Paul must remove from the list all of the positive divisors of the number that Carolyn has just removed, and he can optionally choose one additional number that is a multiple of any divisor he is removing. bullet If Carolyn cannot remove any more numbers, then Paul removes the rest of the numbers. For example, if n=8, a possible sequence of moves could be considered. Suppose that n=8 and Carolyn removes the integer 4 on her first turn. Determine the sum of the numbers that Carolyn removes.
answer:The list starts as 1, 2, 3, 4, 5, 6, 7, 8. If Carolyn removes 4, then Paul removes the divisors of 4 (that is 1 and 2). He can also choose to remove another number that is a multiple of either 1 or 2. Assume he chooses to remove 6 as it is a multiple of 2. This leaves 3, 5, 7, 8. Next, Carolyn must choose a number with at least one divisor other than itself remaining. She can remove either 8 (with divisor 4 removed, but 2 still there). After removing 8, Paul removes 2 (remaining divisor of 8). This leaves 3, 5, 7. Carolyn cannot remove any number as none has a positive divisor other than itself remaining. Thus, Paul removes 3, 5, 7. In summary, Carolyn removes 4 and 8 for a sum of 4 + 8 = boxed{12}. Paul removes 1, 2, 6, 3, 5, and 7 for a sum of 1 + 2 + 6 + 3 + 5 + 7 = 24.
question:a certain pond, 80 fish were caught, tagged, and returned to the pond. A few days later, 80 fish were caught again. The percent of tagged fish in the second catch approximates the percent of tagged fish in the pond. The approximate number of fish in the pond is 3200. How many fish in the second catch were found to have been tagged?
answer:Let's denote the total number of fish in the pond as N. According to the problem, the second catch of 80 fish is a representative sample of the entire pond. Therefore, the proportion of tagged fish in the second catch should be approximately the same as the proportion of tagged fish in the entire pond. We know that 80 fish were tagged out of the total number of fish N, so the proportion of tagged fish in the pond is 80/N. We are given that the approximate number of fish in the pond is 3200, so N is approximately 3200. The problem states that the percent of tagged fish in the second catch approximates the percent of tagged fish in the pond. Therefore, the number of tagged fish in the second catch (let's call this number T) out of the 80 fish caught should be the same proportion as 80 out of 3200. So we have the proportion: T/80 = 80/3200 Now we can solve for T: T = (80/3200) * 80 T = (1/40) * 80 T = 2 So, approximately boxed{2} fish in the second catch were found to have been tagged.
question:Let A be a moving point on the circle (x2)^{2}+(y2)^{2}=2. The maximum distance from A to the line xy4=0 is ______.
answer:The center of the circle (x-2)^{2}+(y-2)^{2}=2 is (2,2), and its radius is sqrt{2}. The distance d from (2,2) to the line is d=frac{|2-2-4|}{sqrt{2}}=2sqrt{2}. Therefore, the maximum distance from a point on the circle (x-2)^{2}+(y-2)^{2}=2 to the line x-y-4=0 is 3sqrt{2}. Thus, the answer is boxed{3sqrt{2}}. This is obtained by finding the center and radius first, then calculating the distance from the center to the line, and finally adding this distance to the radius. This problem tests the comprehensive application of knowledge about the relationship between lines and circles, as well as the formula for the distance from a point to a line, and can be considered a basic problem.