If a hen and a half can lay an egg and a half in a day and a half, how many eggs can six hens lay in seven days?\n\nHint: The answer is NOT 42. 10 points to the correct answer.
212912
A hen and a half therefore can lay one egg in one day. So, one hen can lay 2/3 of an egg in one day. So, in seven days, one hen can lay 14/3 eggs. Six hens in seven days can lay 28 eggs.\n\n28.
406005
The number is an even number that is also 3 greater than a multiple of 11, so it is 14 greater than a multiple of 22; it is also 13 greater than a multiple of 37, so it is 124 greater than a multiple of 814. This reasoning is so simple because 2, 11, and 37 are pairwise coprime (meaning that no two of those integers have a common factor; in fact, they are prime).\nMy guess is that she wasn't carrying around 938 or more eggs, so she probably had 124 eggs.
370037
chicky has made your F1 an octaploid! The F1 will be AAaa\n\nb) there are 16 combinations in the selfed F2\n\naaaa = 1\nAAAA =1\nAAaa = 6\nAaaa = 4\nAAAa = 4
647647
There are 2 ways to work this out\n\n1) You have 2 equations for 3 unknowns so you can solve for one in terms of the other\n\nlet x be the number of chickens, y be the number of goats and z the number of cows, then we know\n\n0.05X + Y + 5Z = 100 and\n\nX + Y + Z = 100\n\nTherefore \n\n0.05X + Y + 5Z = X + Y + Z \n\n0.05X -1X +5Z - Z =0\n\n4Z = 0.95 X\n\nZ = 0.95/4 X\n\nor since we have to be in whole numbers\n\nZ = 95/400 X = 19/80X\n\nSince 19 is a prime, this is the minimum number of cows there can be, and there must be 80 chickens to go along with this.\n\nThat leaves 1 Goat. Check back and we find we have 80 chickens (4 $) 1 goat (1 $) and 19 cows (95 $) and both animals and money add up to 100\n\n2) Chickens cost 0,05, and since the cost is a whole number of $, the number of chickens must be a multiple of 20 and therefore there can only be 20, 40 60 or 80 chickens\n\n20 chickens : there are 99 $ left with which to buy things and 80 aminals left to buy.\n\n40 chickens : there are 98 $ left with which to buy things and 60 aminals left to buy.\n\n60 chickens : there are 97 $ left with which to buy things and 40 aminals left to buy.\n\n80 chickens : there are 96 $ left with which to buy things and 20 aminals left to buy.\n\nLets look at the 20 chicken case\n\nY + 5Z = 98 : Y = 98 - 5Z\nY + Z = 80 : Y = 80 - Z\n98 - 5Z = 80 - Z\ntherefore 4Z = 18, Z = 18/4 = 4.5 not feasible\n\nYou can do the same with the 40 and 60 chicken cases, you will find they are also un feasible.\n\nLets take the 80 chicken case.\n\nY + 5Z = 96 : Y = 96 - 5Z\nY + Z = 20 : Y = 20 -Z\n96 - 5Z = 20 - Z\n76 = 4Z\nZ= 76/4= 19\n\nTherefore Y = 1\n\nThe same 80 chickens, 19 cows and a goat\n\nOkay?
508913
It is generally considered that there are 52 weeks an a year. 7 days a week times 52 weeks is 364 days, 1 day short of a normal year and 2 days short in a leap year.\n\nP.S. I just noticed your comment. There are years where there are 53 pay days in the year. We had a municipal district that tried to not pay their employees their 53rd check because they had already been paid  for 52 weeks. The employees went to court and got their 53rd paycheck.
153517
This is a simple problem, so a straight-forward solution is to draw the set of all possibilities on a piece of paper and count them.  (And you should read this carefully, because it is the first answer so far to get all  parts correct.)\n\nProblem A)\n\nThere are 36 ways to roll 2 die:\n\n66 65 64 63 62 61\n56 55 54 53 52 51\n46 45 44 43 42 41\n36 35 34 33 32 31\n26 25 24 23 22 21\n16 15 14 13 12 11\n\nOf these, there are 11 ways to roll at least one 5 (Event A):\n\n$$ 65 $$ $$ $$ $$\n56 55 54 53 52 51\n$$ 45 $$ $$ $$ $$\n$$ 35 $$ $$ $$ $$\n$$ 25 $$ $$ $$ $$\n$$ 15 $$ $$ $$ $$\n\nSo, P(A) = 11/36\n\n\nProblem B)\n\nOf the 36 ways to roll 2 die, there are 26 ways for Event B to occur:\n\n66 65 64 63 62 61\n56 55 54 53 52 51\n46 45 44 43 42 $$\n36 35 34 33 $$ $$\n26 25 24 $$ $$ $$\n16 15 $$ $$ $$ $$\n\nSo, P(B) = 26/36 = 13/18\n\n\nProblem C)\n\nNotice that if Event A occurs, so does Event B.  Therefore, there are 11 ways for Event (A and B) to occur:\n\n$$ 65 $$ $$ $$ $$\n56 55 54 53 52 51\n$$ 45 $$ $$ $$ $$\n$$ 35 $$ $$ $$ $$\n$$ 25 $$ $$ $$ $$\n$$ 15 $$ $$ $$ $$\n\nSo, P(A and B) = P(A) = 11/36\n\n\nProblem D)\n\nNotice that if Event B does not occur, neither does Event B.  Therefore, there are 26 ways for Event (A or B) to occur:\n\n66 65 64 63 62 61\n56 55 54 53 52 51\n46 45 44 43 42 $$\n36 35 34 33 $$ $$\n26 25 24 $$ $$ $$\n16 15 $$ $$ $$ $$\n\nSo, P(A or B) = P(B) = 13/18\n\n\nProblem E)\n\nP(A|B) means the probability that Event A occurs given that Event B also occurs.  In this instance, there are only 26 ways to roll the dice so that B occurs:\n\n66 65 64 63 62 61\n56 55 54 53 52 51\n46 45 44 43 42 \n36 35 34 33 \n26 25 24 \n16 15 \n\nOf these 26 possibilities, there are 11 ways that Event A could occur:\n\n$$ 65 $$ $$ $$ $$\n56 55 54 53 52 51\n$$ 45 $$ $$ $$ \n$$ 35 $$ $$ \n$$ 25 $$ \n$$ 15 \n\nP(A|B) = 11/26\n\nAlternatively, you can directly use the definition of conditional probability, which is:\n\nP(A|B) = P(A and B)/P(B)\n\nSo, P(A|B) = (11/36)/(26/36) = 11/26
530923
There are 6 correct numbers.\nThat means 49-6 = 43 numbers are incorrect.\nYou need to choose 4 correct numbers and 2 incorrect numbers.\nC(6,4) = 15\nC(43,2) = 903\n15*903 = 13545
396962
A few facts. Eggs that are on the market are not fertilized. Eggs that have been refrigerated will not hatch.(those from the store)\n\nIf you can get fertilized eggs, keep them warm.Chicks require a high temperature also.  Although there should be a 50 : 50 ratio of male to female, males are weeded out of the mix very early. Those are the ones that are available for Easter chicks. Chickens eat their weight every day. \n\nI do not know what you have available, but you could make a hypothesis regarding the color of the eggs and the color of the chicks.
533126
Ok, I'll help you out, but I'm not going to give you an answer because I'm not going to do the work.  Besides, you were pretty sloppy here, calling dogs kittens and #3 makes no sense if you're talking about whole oz.  So reword that ridiculous statment to say that dog B weighs EXACTLY the same as dog F.\n\nFirst thought, always:  "How many unknowns do I have"\n\nYou have seven:  A-G\n\nSecond thought, always:  "I MUST have 7 different equations or I cannot solve the problem."\n\nNote:  In this problem if you have less than 7 different independent equations, it is IMPOSSIBLE to solve the question (that is, without some other constraint.  For example if you had some problem asking you to find the total number of puppies in the kennel, and the numbers worked out you'd have 2 puppies or 0.5 puppies, you have to go with 2 because you can't have half a puppy unless you're a sicko).\n\nOk, you need 7 equations.  Start writing.\n\nA+B+C = D+E+F+G\n\nC=B+2=A+6\nF=G+12=E+3=D+13\nB=F\n\nThere are your seven equations.  I leave it to you to understand fully where those seven come from.  From here, you start substituting, until you get the first equation in terms of one single variable.  Without giving it too much thought, I would suggest "F".  Solve for that one variable, then solve for the other variables.  Since you have 7 equations and 7 unknowns, there is no "minimum" weight:  you will have the exact weight of the kittens.
43874
7\n\norder them; if you have odd number of entries, take the middle one; if even - take average of two middle ones.\n\n\n\nand it's a statistics question
475462
Answer 1:\nf(6)= 726\nAnswer 2:\nyes\n\n(If you ask it again I promise not to answer.)
266637
Grrrr.. Another homwwoek relp Q.\nAnyways i am here to earn points\n\nP + C = 50 /* Num of heads.. Each has one head*/\n\n4P + 2C =  110 /* Num of feet Pig has 4,chicken 2*/\n\nSolving these gives:\nP = 5 and C =45
196047
A)Prob. that a livestock will die 0.5%\nProb. that a livestock will not die 99.5%\nProb. that 200 livestock will not die:\n99.5%\n x 99.5% x 99.5% x 99.5% x ... (200 times), or\n= 99.5%^200\n\nB)\nProb. that a B is not received at the first stimulus 0.3\nProb. that a B is not received at the second stimulus 0.3 x 0.3\nProb. that a B is received for the first time at the third stimulus:\n0.3 x 0.3 x 0.7 = 0.063\n\n^_^
147194
Let today's number be 'x'\nDay before yesterday = x/5\nyesterday = x/5+18 and this is equal to x/3\ntherefore,\nsolving x/5+18=x/3 you get x=135\nNow putting in the values,\nday before yesterday = x/5 = 27\nyesterday = x/3 = 45\ntoday = 135\ntomorrow = 4 times yesterday = 180
412633
In other words, what is the number of amebas in 10 minutes compared to the number of ameba after 30 minutes?\n\nSince they generate 8 after 30 minutes and 2 after 10 minutes. the ratio would be 2/8 of the total.
242953
OK - let's assume that Large Chicken is A, Medium Chicken is B, and Small Chicken is C. \n\nA+B=10.6\nA+C=8.5\nB+C=6.1\n\nNow, logically: 10.6+8.5 = (A+B)+(A+C), or 2A+(B+C)\n\nSo - 10.6+8.5=2A+6.1 ~ 19.1=2A+6.1 ~ 19.1-6.1=2A ~ 13=2A ~ A=6.5\n\nSo, if A=6.5, then (10.6-6.5)=B ~ B=4.1\nand (8.5-6.5)=C ~ C=2.0\n\nTHEREFORE:\nA=6.5 - Large Chicken\nB=4.1 - Medium Chicken\nC=2.0 - Small Chicken
336216
Divide 278 by four to get the number of horses, then subract that number of horses from the heads to get the number of ostriches. MAke sense? If it doesn't email me. \n\nR u sure you have the number of hooves right?????? Cause ostriches don't have hooves but there is a remainder of 2 did u mean feet? if u meant feet then...\n\nummm...\n\nOOOO i no.... 123 horses and 0 ostriches! works don't it?