course Mth 151 *********************************************
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: B and A’= since “A” consist of at least one even digit, then “A’” will have to consist of zero even numbers. Since it looks like “B” consists of odd numbers I’m guessing “B and A’” have in common that their main numbers are odd numbers B and A= “A” means there has to be at least one even number, and “B” means there is at least one odd number, I guess they both have in common that they both to have at least one of each to make it True confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aOf the numbers in B, 8, 89, 104, 4321 each have at least one even digit and so are common to both sets. 3 is odd, both of the digits in the number 35 are odd, as are all three digits in the number 357. Both of these numbers are therefore in A ' . &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I see that I didn’t pay enough attention to B set, so what I stated “B” had to have a negative in False. ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q002. I have in a room 8 people with dark hair brown, 2 people with bright red hair, and 9 people with light brown or blonde hair. Nobody has more than one hair color. Is it possible that there are exactly 17 people in the room?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 8+2+9= 19 So there cannot just be 17 people there has to be 19 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf we assume that dark brown, light brown or blonde, and bright red hair are mutually exclusive (i.e., someone can't be both one category and another, much less all three), then we have at least 8 + 2 + 9 = 19 people in the room, and it is not possible that we have exactly 17. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating #$&* ********************************************* Question: `q003. I have in a room 6 people with dark hair and 10 people with blue eyes. There are only 14 people in the room. But 10 + 6 = 16, which is more than 14. How can this be?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because there has to be a combination of the two, and we can assume this is possible because it didn’t state that we couldn’t. and since you add 10+6 to equal 16, but there can only be 14 people there has to be 2 people that has both dark hair and blue eyes confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe key here is that there is nothing mutully exclusive about these categories-a person can have blue eyes as well as dark hair. So if there are 2 people in the room who have dark hair and blue eyes, which is certainly possible, then when we add 10 + 6 = 16 those two people would be counted twice, once among the 6 blue-eyed people and once among the 10 dark-haired people. So the 16 we get would be 2 too high. To get the correct number we would have to subtract the 2 people who were counted twice to get 16 - 2 = 14 people. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q004. In a set of 100 child's blocks 60 blocks are cubical and 40 blocks are cylindrical. 30 of the blocks are red and 20 of the red blocks are cubical. How many of the cylindrical blocks are red?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since you only have 30 red blocks you have to subtract the 20 from the 30 which will give you 10, so there can only be 10 red cylindrical Confidence rating #$&*
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Given Solution: `aOf the 30 red blocks 20 are cubical, so the rest must be cylindrical. This leaves 10 red cylindrical blocks. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* "