I would like some help remembering someththing that I learned in HS.\n\nWhat is the formula for estimating height of a person if you only have a single bone to base that on?  Knowing that bones are a certain ratio to height.\n\nAlso,\n\nWhat is the formula for knowing how many choices you have with a set number of options.  I know that if I have 3 switches, there are seven possible combinations (including all open and all closed), but what if I have 8 switches, how many combinations do I have then?\n\nNo homework involved, this has been bugging me for some time and I thought one of you "math" people could help me.
120230
The answer to the three switch problem is 8 choices, not 7. The answer figured numerically is 2^3.\n\nWith 8 switches there are 2^8 or 256 combinations possible.\n\nThere is a formula based on the length of the femur:\nhttp://ublib.buffalo.edu/libraries/projects/cases/bones/activity4.html
159709
Suppose the number of combinations of k objects out of n objects without respect to order chosen is C(n, k). If we wanted to count order, we would multiply each combination by the number of ways it could be permuted. Thus P(k)*C(n, k) is the number of ways one could choose k people out of n people with regard to order. \nBut we can easily count this number another way. The number of possible choices for the first person is n, for the second person n-1, and so on until we reach n - k + 1. Thus, we get the equation P(k)*C(n, k) = P(n, k) = n*(n - 1)*...*(n - k + 1) = n!/(n - k)!. Solving for C(n, k), we get C(n, k) = n!/[P(k)(n - k)!] = n!/[k!(n - k)!]. \nNow you should be able to derive it whenever you get confused.
625535
I may be barking completely up the wrong tree here. Are you comparing one ratio with another?\n\nIn the first, there are 45 parts, in the second there are 19.\n\nMost math problems have a glimmer of sense about them, in that when you hit upon the right way to solve them or what they are really asking, the calculation is quite easy.  I was expecting the number of total parts in each expression to be an easy multiple of each other, which they are not.\n\nIf you are expected to apply the first ratio to the 18 side of the second, that looks a little more reasonable (45 and 18 are both multiples of 9.)\n\nDoes that help without seeming to? (Also I understand you to mean 'factor' and 'easy')
522279
The h(t) function sounds like it tells the height at time t, so you should let h(t)=0, and calculate t, looking for a positive value.\n\nThe answer is approximately 5.1516, presumably in seconds.
549782
alright. Electron configurations. I had the exact same problem three weeks ago. I was SO confused. But i got it in the end.\n\nLet me explain.\nThere are two types of electron configurations. One is the simple electron shell configuration. Let's say 2-8-16 and the other is called spdf notation like 1s2 2s2(i can' t type the two after the s like an exponent so bear with me)\n\nNow the first one is simple. Shell notation. An atoms outer shells are divided into shells that each have a maximum limit for the number of electrons that they can contain. the first can hold 2, the second 8 and so on. They all follow a rule. It is always 2 time the shell number squared. lets say level 1. Square it. You get 1. Multiply it by 2. You get 2. Therefore 2 is the maximum # that level one can hold. You said 15. So fill each level so that each contains the max. It would be 2-8-5. (The last one is five so that the all ad up to 15). Got it?\n\nNext, we have spdf notation. This takes a lot of time to explain so i'll just give you a site to go to. \n\nhttp://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson36.htm\n\nGood luck!
490248
1) Don't know\n2) what the hell is a "25-gon"?\n\n3)360/8(360 deg. in an octogon. divided by the # of sides. And voila. you get your answer)\n4) 360/24(same as above)
539023
I would start by asking the kids questions.\n\nHow many days are in a year?\n\nSo, if we have 80,000 people how could we estimate how many people will be at each party? At least 1 kid will be smart enough to tell you to divide it. Be sure that write it on the board as you go.\n\nWhen you come up with 219 then ask the kids if they really think that the birthdays will be divided equally. If that is true, then if there are 24 kids in your class... there should be 2 birthdays in each month. (Maybe as a starter for the lesson you could have some sort of birthday poster and could quickly write the names of the kids under each month that their birthday falls. You may have some months with 5 birthdays, and others with 1 or none). If they say that they think that the birthdays will be evenly then use your birthday chart and explain that their birthdays are random; that it does not work out to exactly 2 birthdays in each month.\n\nThen you can explain to them that with 80,000 people there are likely to be a few birthdays on more days than others. They will get it; especially when you use their own data to help explain it.\n\nIf time allows you could do another problem that is similar, but maybe using something like the day they got their first pet (I don't know; something...) anyway, because some people may have never had a pet you could ask if it is possible that some people never come to celebrate. Or maybe use vet patients for the problem because then we could assume that all patients would qualify.
172695
You use combinations to determine how many different combinations you can make out of a number of different objects at a certain number. For example, You have 10 people to be chosen for a 3-person commitee. How many different 3-person commitees can be formed? This means that out of 10 people, you need to choose 3 different people, and how many different combinations out of that. Think of combinations this way, as it alters the definition of it: In fighting games like Mortal Kombat and Tekken, you use "combo's" to fight successfully. You can do multiple different combo's to defeat your opponent. Now, Combo's is just a short term for combinations.\n\nBasic probability fraction is the chances of something happening, not how many different combinations can happen.\n\nHope this helps.\n\nPS Penguins!!!
224823
Count the total number of groups:\n9 choose 5 = 9*8*7*6/(4*3*2*1) = 9*2*7 = 126.\n\nb) Count the number of groups where 3 women where chosen and 2 men.\n(4 choose 3) * (5 choose 2) = 4 * 10 = 40.\n\nThe prob that a random committee has exactly 3 women = 40/126\n\nc) If there were no women then the number of ways this can happen is:\n(4 choose 0) * (5 choose 5).  As you could probably already see, there is only 1 way to have a committe of 5 men and 0 women.  The prob = 1/126.\n\nf) >=3 women is the same as saying 3 or 4 women as 4 women is the most you can have.\n\n(4 choose 3)*(5 choose 2) + (4 choose 4) * (5 choose 1)=\n4*10 + 1*5 = 45.  Prob = 45/126.
550549
ok, im going to assume that you know (rho)(g)(h) = pressure\n\nlets label 4 points:\nA is at the bottom of the 1.28cm mark\nB is at the top of the 1.28cm mark\nC is at the bottom of the hl mark\nD is at the top of the hl mark\n\nNotice 2 things:\n1. the pressures at A on the right and the left must be the same\n2.if you removed the same amount of blue from both sides, hl would still be the same measurement (so lets remove it for calulations)\n\nnow the blue on the left side is of a height of 1.28+hl and the green on the right is 1.28\n\nat the top they are equal, so:\n2.25*g*(1.28+hl)=8.41*g*(1.28)\n2.25(1.28+hl)=8.41*1.28         g cancels from both sides\n2.25*1.28+2.25*hl=8.41*1.28\n2.88 + 2.25hl = 10.765\n2.25hl = 10.765 - 2.88=7.885\nhl = 7.885/2.25 = 3.50435.....cm
39850
how do you din the height of a trapezoid if one of the non parallel sides is straight, and the other is slanted. Which makes the two parallel sides different lengths? Also, how would you find the width of a trapezoid in this situation?\n\nIf the length of the shorter parallel side is A and the length of the longer parallel side is B and the length of the straight side is C and the length of the slanted side is D then:\n\nthe height is:\nh = C (the length of the straight side.\n\nThe width depends on where you measure the width.\n\nI am imagining that your trapezoid looks like this:\n.__\n/__|
182848
I can't believe that this question is being asked in the sixth grade.  Maybe half of my college students could answer this.\n\nThe best way to answer this question is to make a table.  Make the rows the results of the first die and the columns the results of the second die.  Then put the product of the two roll inside of the table.\n            1   2   3   4   5   6\n          ------------------\n        1| 1   2   3   4   5   6\n        2| 2   4   6   8  10  12\n        3| 3   6   9  12  15  18\n        4| 4   8  12  16  20  24\n        5| 5  10  15  20  25  30\n        6| 6  12  18  24  30  36\n\n(Sorry I can't get the columns to even out!)\n\nNotice that there are 36 possible rolls of the two dice.  To get the two probabilities, just count the number of times you get the event in the table and divide by 36.\n\nProbability of an even product = 27/36\nProbability that the product is less than or equal to six = 14/36\n\nI see a lot of incorrect answers up there.  I teach statistics at the college level.  Trust me, this is right.
420160
Let us call the first 2 "2 sub 1", the second "2 sub 2" ... same thing for the 6's.\n\nLet us say that "2 sub 1" is distinct from "2 sub 2", and so on.\n\nWe have 8 choices for the first number, seven choices for the second number, etc. This means that there are 8*7*6*5*4*3*2*1 choices, or 8!. \n\nNow if you look at any number, take all of the 2's out. You see that you can rearrange the 2's in 3! different ways (including the original arrangement) using the procedure in the paragraph above. If we take out the 2's rearrange them, and put them back in we have created the a new arrangement. So for each arrangement of numbers, we have counted it 3! times. So we divide by 3!. The same thing for the sixes, we divide by 2!.\n\nThe result it (8!)/((3!)(2!)) = 3360 choices.
184539
Although this seems to be a problem about Gregor Mendel and his tall and dwarf pea plants, once it is realised that the probability of any child being short is 1/2, it simplifies to being the simpler and more familiar one of tossing a coin, heads or tails, n times\n\nIt would have produced more interesting results perhaps if the parents were Aa and Aa and three children out of four were therefore tall and we had looked at grandchildren (offspring of children who are one quarter AA, one half Aa and one quarter aa, eg if they intermarry with another such distribution among their partners.) However ...\n\n(i) is imprecisely worded. How big is the family?\n\nIf there is one child, the chance of that child being short is  1/2\n\nIf there are two children, the chance that both are tall is 1/4 and the chance that at least one child is short is 3/4. (Of one being short is 1/2 and of both being short is a quarter.)\n\nIf there are three children, the chance that all three are tall is 1/8 and the chance that at least one child is short is 7/8 (of one being short: 3/8, of two being short also 3/8, of all three being short: 1/8).\n\nWith 4 children, the chance of at least 1 being short is 15/16, of 1  being short is 4/16 or 1/4; of 2 being short is 6/16 or 3/8; of 3   being short is 4/16 or 1/4; of all 4 being short is 1/16;\n\nOne just reads across successive lines of Pascal's Triangle.\n\n(ii) the answer to this is buried in the above: 3/8ths\n\n(iii) again, imprecisely phrased. Does "such families" refer back to (i) or to (ii)? Taking it to mean families of four children, there is one chance in 16 of 4 tall children, 4 chances in 16 of 3 tall children, 6 chances in 16 of 2 tall children, 4 chances in 16 of 1 tall child and 1 chance in 16 of 0 tall children.\n\nIn 16 such families, there are 64 children and unsurprisingly 32 are tall\n(1 x 4) + (4 x 3) + (6 x 2) + (4 x 1) + 0\n\nwith an average number of 2 tall children per family, and as you would expect an average number of 2 short children per family, too.\n\n(iv) 256 = 16 squared and so the expectation is 16 families out of 256 will have 4 tall children, given 1 family in 16 has this particular expectation.
28773
Here in Slovakia, we use a simple formula to calculate percentages.  And yes, there are 3 kinds of percentage problems, but the formula applies to all of them, you just have to play with it a little bit.\nOkay, problem no. 1: there are 35 kids (total = T) in a class, 40% (P as in percentage) of which are boys.  How many (value = V) boys are there?\nHere's the formula:\nV = T/100*P\n(V = 35/100*40 = 14)\n\nProblem no. 2: there are 14 boys and 21 girls (T = 14 + 21)in the class.  What's the percentage(P) of boys (V)?\nSame formula as before, really:\nP = V/(T/100)\n(P = 14/(35/100) = 40%)\n\nProblem no. 3: how much did your shoes originally cost (T) if your 10% discount (P) meant $2 (V) off?\nSame formula:\nT = V*100/P\n(T = 2*100/10 = $20)\n\nOnce you know which is P, V, and T, you're fine.  You don't necessarily have to remember the formulas.
588784
What if there is only one child and one adult?\nWhat if there is at least two adults and only adults?
418677
I don't understand your question.  Do you mean using the associative law?\n\nabc*de = \ncba*de =\nbca*de =\nacb*de =\ncab*de = \nbac*de \nNow all of these are multiplied by the two ways to postiion de\n\nThat's 3! = 6\ntimes  2! = 2\n\nfor a total of 12.  \n\nFor a 5th grader I'm sure that you're just supposed to list all 12.  For convenience, note that I've put each letter in the middle and switched the other two.  For neatness, I'd recommende just putting another column to the right of each of these 6 with de and ed .\nWhen doing counting like this, called combinotorics, it's helpful to arrange things neatly.