#$&* course MTH 151 04/11/2014 at 2:37 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `a** You would put all dogs in a circle, and this circle would be inside another circle consisting of all things that love to bury bones. Archie is outside this bone-burying circle and since this circle contains all dogs Archie isn't a dog. This makes the argument valid. The x for Archie has to go outside the outer circle, so it has to be outside the inner circle. Thus the x can't be in the inner circle, and Archie therefore can't be a dog. The conclusion can't be contradicted. COMMON ERROR WITH INSTRUCTOR RESPONSE: I put 'all dogs like to bury bones' in one circle and 'archie likes to bury bones' in another. INSTRUCTOR RESPONSE: You don't want to use a single circle to represent a compound statement. 'All dogs like to bury bones' and 'Archie likes to bury bones' are compound statements. SIMILAR ERROR: in one circle ,I put all dogs love to bury bones, inthe other circle I put Archie, so I knew that Archie wasn't a dog, so the statement is valid . INSTRUCTOR COMMENT: See previous comment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery 3.5.20 all chickens have a beak. All hens are chickens. Therefore all hens have beaks. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (This problem is # 19 in my book.) The smallest circle will be “hens”, inside the “chickens” circle. Then, both of these will be in the largest “have beaks” circle. All hens have beaks is valid. confidence rating #$&*: 3. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** You need to show the relationship between chickens and things with breaks. You would make a circle for everything with a beak and another circle for chickens. Since all chickens have beaks the chicken circle has to be inside the 'beaked' circle. Then you have hens. They are all chickens so the hen circle is inside the chicken circle. Since the chicken circle is already inside the beaked circle the hen circle (inside the chicken circle) is also inside the beaked circle, and you conclude that all hens have beaks. COMMON ERROR WITH INSTRUCTOR COMMENT: In the outer circle, I put chickens with beaks. Inside that circle, I made another circle for hens are chickens. INSTRUCTOR COMMENT: 'hens are chickens' is a statement, not a thing. The circles have to be defined by things. SIMILAR ERROR WITH COMMENT: Two circles: large circle of hens are chickens and a smalled circle within of hens have beaks. Valid INSTRUCTOR COMMENT: You don't put propositions into circle (e.g., 'hens are chickens' isn't a circle). You put sets of things into circles (e.g., a circle for hens and a circle for chickens, with the hens circle inside the chickens circle). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhen the diagram is drawn according to the premises, is it or is it not possible for the diagram to be drawn so that it contradicts the conclusion? If it is possible describe how. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No. This is the only was to solve this argument. confidence rating #$&*: 3. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The circle for hens must be inside the circle for chickens, which is inside the circle for beaked creatures. Therefore the circle for hens must be inside the circle for beaked creatures. No other way to draw it consistent with the conditions. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhen the diagram is drawn according to the premises, is it or is it not possible for the diagram to be drawn so that it contradicts the conclusion? If it is possible describe how. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No. This is the only was to solve this argument. confidence rating #$&*: 3. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The circle for hens must be inside the circle for chickens, which is inside the circle for beaked creatures. Therefore the circle for hens must be inside the circle for beaked creatures. No other way to draw it consistent with the conditions. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: #*&!