course Mth 151 I had severe difficulty in this section, to the point where I saved this assignment for last. So I apologize about this embarassing display of work. g¢¡ˆéÏ‘Š·‰þŒ‘ááì€}×ʸ¨¼ÿËÅþŸzassignment #008
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07:23:59 1.3.6 9 and 11 yr old hosses; sum of ages 122. How many 9- and 11-year-old horses are there?
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RESPONSE --> I could not figure this problem out, as I am horrible with these word problems, and actually could not arrive an answer. confidence assessment: 0
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07:24:55 ** If there was one 11-year-old horse the sum of the remaining ages would have to be 122 - 11 = 111, which isn't divisible by 9. If there were two 11-year-old horses the sum of the remaining ages would have to be 122 - 2 * 11 = 100, which isn't divisible by 9. If there were three 11-year-old horses the sum of the remaining ages would have to be 122 - 3 * 11 = 89, which isn't divisible by 9. If there were four 11-year-old horses the sum of the remaining ages would have to be 122 - 4 * 11 = 78, which isn't divisible by 9. If there were five 11-year-old horses the sum of the remaining ages would have to be 122 - 5 * 11 = 67, which isn't divisible by 9. The pattern is 122 - 11 = 111, not divisible by 9 122 - 2 * 11 = 100, not divisible by 9 122 - 3 * 11 = 89, not divisible by 9 122 - 4 * 11 = 78, not divisible by 9 122 - 5 * 11 = 67, not divisible by 9 122 - 6 * 11 = 56, not divisible by 9 122 - 7 * 11 = 45, which is finally divisible by 9. Since 45 / 9 = 5, we have 5 horses age 9 and 7 horses age 11. **
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RESPONSE --> I found this whole section to be the most difficult section yet, and my future answers in this assignment is something I'm not incredibly proud of. self critique assessment: 0
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07:26:16 Query 1.3.10 divide clock into segments each with same total
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RESPONSE --> I added all the numbers on the clock to get 78. With at least three different sections, that gives us 24. So I believe the answer would be 24 but I am not certain. confidence assessment: 1
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07:27:11 ** The total of all numbers on the clock is 78. So the numbers in the three sections have to each add up to 1/3 * 78 = 26. This works if we can divide the clock into sections including 11, 12, 1, 2; 3, 4, 9, 10; 5, 6, 7, 8. The numbers in each section add up to 26. To divide the clock into such sections the lines would be horizontal, the first from just beneath 11 to just beneath 2 and the second from just above 5 to just above 8. Horizontal lines are the trick. You might have to draw this to see how it works. **
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RESPONSE --> I didn't go that deep into the problem, but I'm glad I got the answer, although it wasn't the correct way to solve the problem. self critique assessment: 2
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07:29:06 Query 1.3.18 M-F 32 acorns each day, half of all acorns eaten, 35 acorns left after Friday
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RESPONSE --> I multiplied 32 by the days of the week, which was 32 times 7. That gave me 224. If half of the acorns were eaten, that leaves 112. This is as far as I got into the problem. confidence assessment: 1
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07:29:19 ** You have to work this one backwards. If they were left with 35 on Friday they had 70 at the beginning (after bringing in the 32) on Friday, so they had 70 - 32 = 38 at the end on Thursday. So after bringing in the 32 they had 2 * 38 = 76 at the beginning of Thursday, which means they had 76 - 32 = 44 before the 32 were added. So they had 44 Wednesday night ... etc. **
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RESPONSE --> self critique assessment: 0
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07:30:36 Query 1.3.30 Frog in well, 4 ft jump, 3 ft back.
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RESPONSE --> It would take 20 days for the frog. This is one of the problems I feel confident about. He reaches the 20 ft mark after the 20 days. confidence assessment: 2
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07:31:40 ** COMMON ERROR: 20 days CORRECTION: The frog reaches the 20-foot mark before 20 days. On the first day the frog jumps to 4 ft then slides back to 1 ft. On the second day the frog therefore jumps to 5 ft before sliding back to 2 ft. On the third day the frog jumps to 6 ft, on the fourth to 7 ft., etc. Continuing the pattern, on the 17th day jumps to 20 feet and hops away. The maximum height is always 3 feet more than the number of the day, and when max height is the top of the well the frog will go on its way. **
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RESPONSE --> Of course the one I'm confident about is plagued with a common error. I have no idea what my problem is with this problems. It could be because I find things too vague, and the sheer amount of possibilities overwhelms me to the point where I can't narrow any possibilities down to find an answer. self critique assessment: 2
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07:32:17 Query 1.3.48 How many ways to pay 15 cents?
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RESPONSE --> 1 dime, 1 nickel 1 dime, 5 pennies 2 nickels, 5 pennies 3 nickels 15 pennies 1 nickel 10 pennies It took me a while to reach this answer, but that makes 6 ways to pay 15 cents. confidence assessment: 3
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07:32:30 ** To illustrate one possible reasoning process, you can reason this out in such a way as to be completely sure as follows: The number of pennies must be 0, 5, 10 or 15. If you don't use any pennies you have to use a dime and a nickle. If you use exactly 5 pennies then the other 10 cents comes from either a dime or two nickles. If you use exactly 10 pennies you have to use a nickle. Or you can use 15 pennies. Listing these ways: 1 dime, 1 nickel 1 dime, 5 pennies 2 nickels, 5 pennies 3 nickels 15 pennies 1 nickel 10 pennies **
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RESPONSE --> OK, thankfully I got one right. self critique assessment: 3
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07:33:35 Query 1.3.52 Given 8 coins, how do you find the unbalanced one in 3 weighings
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RESPONSE --> I would imagine you would have to separate the eight coins first to find the unbalanced one. Maybe in two piles of four. That way if you weigh those two piles, you can find the one that's unbalanced. confidence assessment: 2
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07:34:21 ** Divide the coins into two piles of 4. One pile will tip the balance. Divide that pile into piles of 2. One pile will tip the balance. Weigh the 2 remaining coins. You'll be able to see which coin is heavier. **
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RESPONSE --> I didn't narrow it down that much, I was unsure of how many unbalanced coins there were, what coins they were. It was very vague for me. self critique assessment: 2
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