several classes (100 students)of 6th graders went on a field trip to the zoo. 1/5 of them went to c the monkeys. 1/2 of them visited the penguins. 1/4 went 2 c the bats. and the rest went to the snack bar. how many students went to the snack bar? if u hav a heart. u will help
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several classes (100 students)of 6th graders went on a field trip to the zoo. 1/5 of them went to c the monkeys. 1/2 of them visited the penguins. 1/4 went 2 c the bats. and the rest went to the snack bar. how many students went to the snack bar? if u hav a heart. u will help \n\nwell...i got 5 students went to the snack bar...\n1/5 of 100=20\n1/2 of 100=50\n1/4 of 100=25\nso \n  20+50+25=95\n  100-95=5 students\n~Brendon
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hmm.. sounds like a homework question.  tsk tsk..
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1. c\n2. d\n3. e\n4. b\n5. a\nIts been a few years since I took biology but I thought I could take a shot at your question.  Hope the answers are right:)
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In the experiment of choosing a committee of 6 from a group of 8 women and 7 men,\n\nTotal no. of possible choices \n= no. of ways in which the 6 members of the committee can be chosen from the total 15\n==> n = C(15,6)\n==> n = 15*14*13*12*11*10/6*5*4*3*2*1\n==> n = 7*13*11*5\n==> n = 5005\n\nLet "A" be the event of choosing the committee with 3 women and 3 men\n\nFor Event "A"\n\n---------- Men----Women---Total\nAvailable---8-------7------15\nTo Choose---3-------3-------6\nChoices----C(8,3)-C(7,3)--C(15,6)\n\nNo. of favorable choices\n= (no. of ways in which the 3 men can be chosen from the total 8) * (no. of ways in which the 3 wommen can be chosen from the total 7) \n==> m(a) = C(8,3) * C(7,3)\n==> m(a) = 8*7*6/3*2*1 * 7*6*5/3*2*1\n==> m(a) = 8*7 * 7*5\n==> m(a) = 1960\n\nProbabilty of choosing the committe with 3 women and 3 men \n==> Probability of occurance of event "A" = m(a)/n\n==> P(A) = 1960/5005\n==> P(A) = 392/1001\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/
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*lick* uh one\n*lick* uh two \n*lick* uh three\n*CRUNCH*\n\nso how many licks DOES it take to get to the center of a Tootsie Roll Tootsie Pop?\n\nThe world may never know.\n\n(ok i guess that doesn't help...but really it can take as many as you can muster. If you lick it 5 times and suck on it for the rest ot the time, then technically for you the answer is 5...they never really said you could ONLY lick it...it was merely implied...ok maybe that still doesn't help...it's Tuesday afternoon and I'm having a little fun. I hope you got at least a chuckle and aren't mad...)
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Total no of items\n = 7 forks + 4 spoons + 5 knives \n = 16  \n\nIn the experiment of drawing 3 items from among 16\n\nTotal no. of possible choices \n= no. of ways in which the 3 items can be drawn from the total 16\n==> n = C(16,3)\n==> n = 16 * 15 * 14/3 * 2 * 1\n==> n = 560 \n[On the assumption that the drawing is being made without replacement]\n\nLet \n"A", "B", "C" and "D" be the events of the three items drawn being 3 forks, 2 forks and a knife, 3 spoons and 1 fork, 1 knife and 1 spoon respectively\n\nFor Event "A"\n\n              Forks      Knives    Spoons     Total\nAvailable       7           5        4          16\nTo draw         3           0        0          3\nChoices       C(7,3)      C(5,0)    C(4,0)     C(16,3)\n\nNo. of favorable choices\n= No. of ways in which the 3 forks can be drawn from the total 7\n==> m(a) = C(7,3)\n==> m(a) = 7 * 6 * 5/3 * 2 * 1\n==> m(a) = 35\n\nProbability that the three items drawn are forks = m(a)/n\n==> P(A) = 35/560\n==> P(A) = 1/16\n\n\nFor Event "D"\n\n              Forks      Knives    Spoons     Total\nAvailable       7           5        4          16\nTo draw         1           1        1          3\nChoices       C(7,1)      C(5,1)    C(4,1)     C(16,3)\n\nNo. of favorable choices\n= (No. of ways in which 1 fork can be drawn from the total 7)* (No. of ways in which 1 knife can be drawn from the total 5) * (No. of ways in which 1 spoon can be drawn from the total 4)\n==> m(d) = C(7,1) * C(5,1) * C(4,1)\n==> m(d) = 7 * 5 * 4\n==> m(d) = 140\n\nProbabilty for the 3 items drawn to be 1 fork, 1 kinfe and 1 spoon = m(d)/n\n==> P(D) = 140/560\n==> P(D) = 1/4\n\nYou should be able to solve the two parts in between for Events "B" and "C".\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/probability/
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OK -I'll call seating spots  1 2 3 \n                             4 5 6  Joe I put at 5 with  V ball at 4 and C Pres @ 6. that makes 1 the Preacher's kid.  Since Sally likes Vball and aint Preacher's kid, she sits at 3 or 6. Now Frank sits opposite Valdictorian Not Joe-that means  Frank sits at 6 (pres) and Sally at 3 is Valdic-torian .By now we know 2 is cheerleader since Joe cant be-FRank likes cheerleader and this crowd is hetero. Since 2 is cheer.. then Gina sits 1 and Betty is cheer..and that leaves Ron at 4 and Joe the B ball\nSO-preachers kid is Gina @ 1spot-\nBetty @2is cheerleader-\nSally is Valdictorian @ 3 \n-Ron is V ball @4-\n-Joe is B ball @5-\n-and Frank is \nclass president @ 6\nEasier if you start with a seat grid to see it
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1. herd\n2. association\n3. space\n4. day\n5. common\n6. Mars\n7.weight\n8. calf\n9.Florida\n10.foot\n11. yellow/orange\n12.salty\n13. red\n14. water\n15.geese\n16. half a mile\n17. German\n18. Boxing\n19. 23\n\nI feel pretty good about these answers.  If you need an explanation for any of them, email me.
242807
Try this on for size...Whisper this sentence into one of the kids ear. "My father told my brother that his monkey was digging in the back yard". Now have this student whisper the same sentence to the student next to them and so on until it reaches the last student. Have that student then tell you what he was told. You will be surprised at the answer you get. This works really well with kids age 10 to 15.
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that's brilliant for a second grader!!\n1) I used to talk with my friend, but I cannot see him anymore\n2) I Was _____, and now I play American football\n (I feel ________)\n\ni'm not so sure about number 2. the ___ bits are something i don't understand. i think you meant football. the words you misplelt and i /or  can not understand are: luchadore, futbo, nonaccents. hope i could help
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I go with C. \n\na) Chickens don't fly very well.\nb)Water is denser than air, and mercury is denser than water. It's true.\nd)But, if the cat is drunk it won't land on it's feet.
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5 hour trip times 3 students standing at all time = 900 standing-minutes. 900 min divided by 19 students = 37 min 22 s.\nAt departure student 1 2 and 3 is standing. When a third of student 1's standing-time is over he/she sits down on student 4's seat, who just stood up. When two thirds of student 2's standing-time is over he/she sits down on student 5's seat, who just stood up. Student 3 stands for his fair share of time, and switch place with student 6. All students from #3 to #19 do their standingtime in a single streach. When student 17 is done he/she sits down on student 1's current seat. When student 18 is done he/she sits down on student 2's current seat, and when student 19 is done, they are at their destination!\n\nThe math teacher only has to shout "switch!" every 12min 42sec.\n\nI know you said no trickery, but she could also force the students to all stand for 5 straight hours...
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A) 2.  Since the probability of drawing a blue marble is 1/3, there must be 4 of them in the bag (4/12 = 1/3).  They also tell you that the probability of a blue or red is 1/2 = 6/12.  So 6 marbles are blue or red.  I know that 4 of them are blue.  So 2 of them must be red.\n\nB)  It says that the total group was 160.  So there must have been 160 people surveyed.\n\nC)  I don't get this part, though.  If 12 said they preferred vanilla and there were 160 people total, then 12/160 = 7.5% prefer vanilla.  But that isn't an option.  You must have copied something wrong somewhere.
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1.Let the no.of boys in the family be x+1\nTherefore,each boy has x brothers(excluding the boy himself)  and x sisters\nSo the number of girls in the family is x\nTherefore each girl has x-1 sisters(excluding herslef,of course) and as assumed x+1 brothers\nBut by the problem,\n(x-1)=(x+1)/2\n+.2x-2=x+1\n2x-x=1+2\n+.x=3\nTherefore there are 3 sosters amd 4 boys in the family and the no. of children is 7.\nso A is correct\n2>Let the no. of children who lost 1 teddybear each be x\nTherefore,the no of children who lost two teddy bears each is 32-x\nAccording to the problem\nx+2(32-x)56\n=>x+64-2x=56 \nor x=8\nTherefore,8 children lost one teddy bear each\nA is correct
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Are you kidding me?\n\nWho would find their socks?  Wash their socks?  Make the food?  Buy the food (chips and dip are NOT food groups)?\nWho would stay up till 1 o'clock finishing the science fair project?  Who would drive back and forth from home to the school to pick up clarinets and library books and permission forms?\nWHO I ask WHO?
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Wow! That is a difficult assignment! please don't turn\nthis in, you will probably funk\n\nI traveled abroad to the Galapagos islands to study the \nadaption of animals to their environment. Birds have the\nadvantage of plumage which can blend into the environment\nand protect them from predators.  I admit that this is \nabsolutely the most boring assignment I have ever done,so\nI will abandon my attempt to continue. \n\nNow, use your brain and make up your own story. Good luck
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Cool! quiz!\n1.) B, cuz A means that Michaelangelo was bronze and marble (lol) and C would have been a decent one, but it still sounds like A, only answer left:B\n2.) A, cuz 'students' is plural, meaning it should be 'were' (as in answer A) and not was (as in answer B)\n3.) B, cuz it says what Lincoln talked about, not like in A where his speech was talking.\nThis reminded about a test I was studying for earlier just before I saw this quiz...