course Mth 152
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:34:41 `q002. List the different orders in which the letters a, b and c could be arranged (examples are 'acb' and 'cba'). Explain how you know that your list contains every possible order.
......!!!!!!!!...................................
RESPONSE --> These are the different orders in which the letters a, b, and c could be arranged: abc,acb,bca,bac,cba,cab I know that these are every possible answer (assuming that I can't use one of the letters more than once) because there are only 2 possible ways with each letter being in the front. For example, if a is the first letter, then the 2nd letter can only be b or c (only 2 ways for each one). confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:35:09 The only reliable way to get all possible orders is to have a system which we are sure the list every order without missing any. Perhaps the simplest way to construct all possible orders is to list then alphabetically. We start with abc. There is only one other order that starts with a, and it is obtained by switching the last two letters to get acb. The next alphabetical order must start with b. The first possible listing starting with b must follow b with a, leaving c for last. The orders therefore bac. The only other order starting with b is bca. The next order must start with c, which will be followed by a to give us cab. The next order is obtained by switching the last two letters to get cba. This exhausts all possibilities for combinations of the three letters a, b and c. Our combinations are, in alphabetical order, abc, acb, bac, bca, cab, cba.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:43:34 `q003. One collection consists of the letters a, c, d and f. Another collection consists of the letters a, b, d and g. List the letters common to both collections. List the letters which appear in at least one of the collections. List the letters in the first half of the alphabet which do not appear in either of the collections.
......!!!!!!!!...................................
RESPONSE --> Having two lists of letters with a,c,d and f in one and a, b, d and g in the second I will go through the first list one letter at a time to see if that letter is in the second list. a - is in the second list c - is not in the second list d - is in the second list f - is not in the second list I do not need to look at remaining letters in the second list, because if they were not in the first list they are not relevant to respond to the first request of listing the letters commom to both collections ... a and d only. To list the letters which appear in at least one of the collections I will list the letters already listed from the work done on the first exercise and then go through second list. a - was in first list b - was not in first list d - was in first list g - was not in first list Adding letters from first list to letters in second list that were unused, I get the group of letters that appear in at least one of the collections to be a, c, d, f, b and g. To list the letters in the first half of the alphabet which do not appear in either of the collections I'll have to go through each letter .. a, b, c, d, e, f, g, h, i, j, k, l and m. The letters not used in either collection are e, h, i, j, k ,l and m. confidence assessment: 3
.................................................
......!!!!!!!!...................................
17:44:07 To letters a and d each appear in both collections. No other letter does. The letters a, c, d, and f appear in the first collection, so they all in at least one of the collections. In addition to letters b and g appear in the second collection. Therefore letters a, b, c, d, f and g all appear in at least one of the collections. We consider the letters in the first half of the alphabet, in alphabetical order. a, b, c and d all appear in at least one of the collections, but the letter e does not. The letters f and g also appear in at least one of the collections, but none of the other letters of the alphabet do. The first half of the alphabet ends at m, so the list of letters in the first half of the alphabet which do not occur in at least one of the collections is e, h, i, j, k, l, m.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:48:59 `q004. Give the next element in each of the following patterns and explain how you obtained each: 2, 3, 5, 8, 12, ... 3, 6, 12, 24, ... 1, 3, 4, 7, 11, 18, ...
......!!!!!!!!...................................
RESPONSE --> The next element in each of the following patterns is ... 2, 3, 5, 8, 12 .... next element is 17. Difference between values goes up by 1 each time. Difference between 2 and 3 is 1, between 3 and 5 is 2, between 5 and 8 is 3 etc. 3, 6, 12, 24 ... element is 48. Next value is calculated by doubling prior element. 3 plus itself is 6, 6 plus itself is 12, 12 plus itself is 24 etc. 1, 3, 4, 7, 11, 18 ... next element is 29. Next value is calculated by added two prior values together. 3 + 1 = 4, 4+3 = 7, 7+4 = 11 etc. confidence assessment: 3
.................................................
......!!!!!!!!...................................
17:49:26 The pattern of the sequence 2, 3, 5, 8, 12, ... can be seen by subtracting each number from its successor. 3-2 = 1, 5-3 = 2, 8-5 = 3, 12-8 = 4. The sequence of differences is therefore 1, 2, 3, 4, ... . The next difference will be 5, indicating that the next number must be 12 + 5 = 17. The pattern of the sequence 3, 6, 12, 24, ... can be discovered by dividing each number into its successor. We obtain 6/3 = 2, 12/6 = 2, 24/12 = 2. This shows us that we are doubling each number to get the next. It follows that the next number in the sequence will be the double of 24, or 48. The pattern of the sequence 1, 3, 4, 7, 11, 18, ... is a little obvious. Starting with the third number in the sequence, each number is the sum of the two numbers proceeding. That is, 1 + 3 = 4, 3 + 4 = 7, 4 + 7 = 11, and 7 + 11 = 18. It follows that the next member should be 11 + 18 = 29.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:52:21 `q005. The number 18 can be 'broken down' into the product 9 * 2, which can then be broken down into the product 3 * 3 * 2, which cannot be broken down any further . Alternatively 18 could be broken down into 6 * 3, which can then be broken down into 2 * 3 * 3. Show how the numbers 28 and 34 can be broken down until they can't be broken down any further. Show that there at least two different ways to break down 28, but that when the breakdown is complete both ways end up giving you the same numbers.
......!!!!!!!!...................................
RESPONSE --> Breaking down the number 28 ... 14 * 2 7 * 2 * 2 or 4 * 7 2 * 2 * 7 Both of these break downs get you to the same numbers. Breaking down the number 34 2 * 17 confidence assessment: 3
.................................................
......!!!!!!!!...................................
17:52:51 A good system is to begin by attempting to divide the smallest possible number into the given number. In the case of 34 we see that the number can be divided by 2 give 34 = 2 * 17. It is clear that the factor 2 cannot be further broken down, and is easy to see that 17 cannot be further broken down. So the complete breakdown of 34 is 2 * 17. To breakdown 28 we can again divide by 2 to get 28 = 2 * 14. The number 2 cannot be further broken down, but 14 can be divided by 2 to give 14 = 2 * 7, which cannot be further broken down. Thus we have 28 = 2 * 2 * 7. The number 28 could also the broken down initially into 4 * 7. The 4 can be further broken down into 2 * 2, so again we get 28 = 2 * 2 * 7. It turns out that the breakdown of a given number always ends up with exactly same numbers, no matter what the initial breakdown.
......!!!!!!!!...................................
RESPONSE --> Should have explained that exercise more self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:57:07 `q006. Give the average of the numbers in the following list: 3, 4, 6, 6, 7, 7, 9. By how much does each number differ from the average?
......!!!!!!!!...................................
RESPONSE --> The average of the numbers 3, 4, 6, 6, 7, 7, 9. Adding the numbers 3 + 4 + 6 + 6 + 7 + 7 + 9 up you get 42. Then divide the total amount by the number of units in the equation which is 42 divided by 7 ... 42 / 7 = 6. This equation shows that the average of these numbers is 6. To find how much each number differs from the average, we would subtract each number from the average number. 6-3=3 6-4=2 6-6=0 6-6=0 6-7=-1 6-7=-1 6-9=-3 confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:57:51 To average least 7 numbers we add them in divide by 7. We get a total of 3 + 4 + 6 + 6 + 7 + 7 + 9 = 42, which we then divide by 7 to get the average 42 / 7 = 6. We see that 3 differs from the average of 6 by 3, 4 differs from the average of 6 by 2, 6 differs from the average of 6 by 0, 7 differs from the average of 6 by 1, and 9 differs from the average of 6 by 3. A common error is to write the entire sequence of calculations on one line, as 3 + 4 + 6 + 6 + 7 + 7 + 9 = 42 / 7 = 6. This is a really terrible habit. The = sign indicates equality, and if one thing is equal to another, and this other today third thing, then the first thing must be equal to the third thing. This would mean that 3 + 4 + 6 + 6 + 7 + 7 + 9 would have to be equal to 6. This is clearly not the case. It is a serious error to use the = sign for anything but equality, and it should certainly not be used to indicate a sequence of calculations.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
18:01:28 `q007. Which of the following list of numbers is more spread out, 7, 8, 10, 10, 11, 13 or 894, 897, 902, 908, 910, 912? On what basis did you justify your answer?
......!!!!!!!!...................................
RESPONSE --> To see which list of numbers is more spread out I take the highest value and subtract the lowest value from it. On the first list of 7, 8, 10, 10, 11, 13 ... I substract 7 from 13 and get a difference of 6. On the second list of 894, 897, 902, 908, 910, 912 .... I subtract 894 from 912 and get a difference of 18. The second list is more spread out. confidence assessment: 1
.................................................
......!!!!!!!!...................................
18:01:44 The first set of numbers ranges from 7 to 13, a difference of only 6. The second set ranges from 894 to 912, a difference of 18. So it appears pretty clear that the second set has more variation the first. We might also look at the spacing between numbers, which in the first set is 1, 2, 0, 1, 2 and in the second set is 3, 5, 6, 2, 2. The spacing in the second set is clearly greater than the spacing in the first. There are other more sophisticated measures of the spread of a distribution of numbers, which you may encounter in your course.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
18:08:43 `q008. 12 is 9 more than 3 and also 4 times 3. We therefore say that 12 differs from 3 by 9, and that the ratio of 12 to 3 is 4. What is the ratio of 36 to 4 and by how much does 36 differ from 4? If 288 is in the same ratio to a certain number as 36 is to 4, what is that number?
......!!!!!!!!...................................
RESPONSE --> 36 - 4 is 32 and 36/4 is 9. 36 differs from 4 by 32, and the ratio of 36 to 4 is 9. If 288 is in the same ratio to a certain number as 36 is to 4, to find out that number we would have to know that the ratio of 36 to 4 is 9. We would then divide 288 by 9 ... 288/9=32. The ratio of 288 to 32 is 9. confidence assessment: 2
.................................................
......!!!!!!!!...................................
18:09:02 Just as the ratio of 12 to 3 is 12 / 3 = 4, the ratio of 36 to 4 is 36 / 4 = 9. 36 differs from 4 by 36 - 4 = 32. Since the ratio of 36 to 4 is 9, the number 288 will be in the same ratio to a number which is 1/9 as great, or 288 / 9 = 32. Putting this another way, the question asks for a 'certain number', and 288 is in the same ratio to that number as 36 to 4. 36 is 9 times as great as 4, so 288 is 9 times as great as the desired number. The desired number is therefore 288/9 = 32.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
18:14:21 `q009. A triangle has sides 3, 4 and 5. Another triangle has the identical shape of the first but is larger. Its shorter sides are 12 and 16. What is the length of its longest side?
......!!!!!!!!...................................
RESPONSE --> To find the length of the triangles longest side, we would first have to figure out how much bigger the second triangle is than the first and then solve for y (y being the length of the second triangles longest side). Here are the sides of the first triangle 3,4,5 Here are the sides of the second triange 12,16,y The first side of the second triangle is 4 times larger than the first ( 12/3 = 4). The second side of the triangle if also 4 times larger than the first (16/4 = 4). So the third and largest side of the second triangle would also have to be 4 times larger than the first (y/5 = 4 which calculates to y=20) confidence assessment: 3
.................................................
......!!!!!!!!...................................
18:14:27 ** You need to first see that that each side of the larger triangle is 4 times the length of the corresponding side of the smaller. This can be seen in many ways, one of the most reliable is to check out the short-side ratios, which are 12/3 = 4 and 16/4 = 4. Since we have a 4-to-1 ratio for each set of corresponding sides, the side of the larger triangle that corresponds to the side of length 5 is 4 * 5 = 20. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
™×iëýŠ¡º¿•±¢|éɼŒRàÖÌ· assignment #004 004. Liberal Arts Mathematics qa initial problems 01-12-2008