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Re: Algebra Cafe: Solve Math Problems Here by donchichi: 8:39am On Apr 04, 2008 |
kobe: Using Difference of two sqaures (x+y)(x-y) = 77 Since x+y = 11 x-y = 77/11 =7 So we have x+y = 11 -----------1 x-y = 7 -------------2 From 2, x = y +7. So by substitution. y + 7 + y = 11 2y = 4; y =2 And so , x = y + 7 = 9 x=9, y =2 |
Re: Algebra Cafe: Solve Math Problems Here by donchichi: 8:45am On Apr 04, 2008 |
kobe: After Monday, The price of the boot is x - 0.2x. After Tuesday, the price is 10% less the price after monday. So that gives (x-0.2x) - 0.1(x-0.2x) = 0.8x - 0.08x = 0.72x |
Re: Algebra Cafe: Solve Math Problems Here by Nobody: 2:00pm On Apr 04, 2008 |
do you know that 1 can be proved to be equal to 2 as follows a(a-a)='a'squared-'a'squared a(a-a)=(a+a)(a-a)[difference of two squares] divide both sides by (a-a) you get a=a+a which is the same as a=2a divide both sides by a you get 1=2 though 1 can never be equal to 2 in ordinary everyday mathematics, this proving is clean correct and shows in a flawlessly that 1=2. counter my proof if you dare. ![]() |
Re: Algebra Cafe: Solve Math Problems Here by Nobody: 2:19pm On Apr 04, 2008 |
4sinxsinx-3=0 find the value of x hint:sin 60=(root 3)/2 |
Re: Algebra Cafe: Solve Math Problems Here by bawomolo(m): 2:53pm On Apr 04, 2008 |
4sinxsinx-3=0 sin^2x=3/4 sinx=square root of 3/4 sinx=.866 x=60degrees or pi/3 |
Re: Algebra Cafe: Solve Math Problems Here by donchichi: 3:02pm On Apr 04, 2008 |
holamiday: Nice Try But at a(a-a)=(a+a)(a-a)[difference of two squares], it yields 0 when a-a is executed which Zero's teh whole equation. And even when u divide botgh sides by a-a, u r actually dividing both sides by 0. |
Re: Algebra Cafe: Solve Math Problems Here by bawomolo(m): 3:03pm On Apr 04, 2008 |
a(a-a)=(a+a)(a-a)[difference of two squares] division by zero is a logical fallacy. you can't divide both sides by (a-a). sorry to bust your bubble bro |
Re: Algebra Cafe: Solve Math Problems Here by RichyBlacK(m): 3:13pm On Apr 04, 2008 |
Three of every four trucks on the road are followed by a car, while only one of every five cars is followed by a truck. What fraction of vehicles on the road are trucks. |
Re: Algebra Cafe: Solve Math Problems Here by topkin(m): 6:36pm On Apr 04, 2008 |
My brain bin sharp one time before o! ![]() ![]() |
Re: Algebra Cafe: Solve Math Problems Here by Imani(f): 9:13am On Apr 05, 2008 |
@donchichi Are you a mathematician? ![]() ![]() top_kin: No be only you with "cobwebs" for brain ![]() |
Re: Algebra Cafe: Solve Math Problems Here by pahtahkee: 9:35am On Apr 05, 2008 |
WOW!!! ![]() |
Re: Algebra Cafe: Solve Math Problems Here by pahtahkee: 9:42am On Apr 05, 2008 |
***Running. . . . I have to pick up my Engineering Mathematics and GRE textbook again. This is good!*** |
Re: Algebra Cafe: Solve Math Problems Here by RichyBlacK(m): 10:11am On Apr 05, 2008 |
A red die and a green die are cast. What is the probability that the number on the red die is larger than the number on the green die? |
Re: Algebra Cafe: Solve Math Problems Here by RichyBlacK(m): 10:20am On Apr 05, 2008 |
A student is taking a multiple-choice test in which each question has four possible answers. She knows the answers to 50% of the questions, can narrow the choices down to two 30% of the time, and does not know anything about 20% of the questions. What is the probability she will correctly answer a question at random from the test? |
Re: Algebra Cafe: Solve Math Problems Here by dinozzo(m): 5:30pm On Apr 05, 2008 |
RichyBlacK:has something to do with probability i think three of every four trucks on the road followed by cars, probabilities will be = PTC and PTT PTC = 3/4 and PTT = 1/4 one of every five cars is followed by a truck, probabilities here will be I think = PCC i.e another car following a car and PCT, a truck following a car PCC = 4/5 and PCT = 1/5 using the chapman equation; PT = PTT x PT + PCT x PC PC = 1-PT PT = 1/4PT + 1/5(1-PT) PT = 1/4PT - 1/5PT + 1/5 19/20PT = 1/5 PT = 4/19 ![]() |
Re: Algebra Cafe: Solve Math Problems Here by donchichi: 5:49pm On Apr 05, 2008 |
RichyBlacK: dinozzo: Dinozzo, I'm sure ur answer is correct cuz its wot i got. However, the picture is my illustration. if the numbers increase then i guess it would be wise to use the chapman equation. Please explain wot the PTC and PT and PCC and the rest acronyms mean. i was able to draw out the the scenario and i hope richyblack would acknolege my answer. it is just a capture of the question at any one time. Clearly, there are 19 cars on the road and for one in five cars there is a truck behinfd it(count from left), and for three in four trucks there is a car behind each of them. So from the pciture, the fraction of trucks on the road is 4/19. But then again if there were more than 4 trucks and 5 cars on the road , i would recommend using dinozzo's solution.
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Re: Algebra Cafe: Solve Math Problems Here by huxley(m): 5:54pm On Apr 05, 2008 |
In seven years time the combined ages of myself and my two sons will be 89. What will it be in 9 years time? |
Re: Algebra Cafe: Solve Math Problems Here by huxley(m): 5:57pm On Apr 05, 2008 |
Brian and Ryan are brothers. Three years ago Brian was seven times as old as Ryan. Two years ago he was four times as old. Last year he as three times as old and in two years time he will be twice as old. How old are Brian and Ryan now? |
Re: Algebra Cafe: Solve Math Problems Here by donchichi: 6:01pm On Apr 05, 2008 |
RichyBlacK: Total number of possible events is 36. Evens that satisfy the condition above are; (Red on the left - green on the right) 2-1 5-3 3-1 5-4 3-2 6-1 4-1 6-2 4-2 6-3 4-3 6-4 5-1 6-5 5-2 Total = 15. Therefore probability is 15/36 |
Re: Algebra Cafe: Solve Math Problems Here by donchichi: 6:02pm On Apr 05, 2008 |
Imani: Na my mama cause am!!! |
Re: Algebra Cafe: Solve Math Problems Here by donchichi: 6:12pm On Apr 05, 2008 |
huxley: Let your current age be x and that of your two sons be y and z. In 7 yrs time, (x+7) + (y+7) +(z+7) = 89 x + y +z +21 = 89 x + y + z = 89 - 21 = 68. The 68 value is the combined age of you and your two sons right now. So in 9 yrs; (x+9) + (y + 9) + (z + 9); x + y +z +27. Remember x + y + z = 68. So x + y +z +27 = 68 + 27 = 95 |
Re: Algebra Cafe: Solve Math Problems Here by huxley(m): 6:24pm On Apr 05, 2008 |
donchichi: Correct. Quicker solution: In 9 years time, they while be two years older than they will be in 7 years time. Since there are three people, 2 x 3 = 6. So in nine years time, the sum total will be 6 greater than the sum in 7 years. thus 89 + 6 = 95. |
Re: Algebra Cafe: Solve Math Problems Here by donchichi: 6:35pm On Apr 05, 2008 |
huxley: Let brian's age be x, and ryan's age be y. Three years ago; (x- 3) = 7(y-3) x-3 = 7y-21 7y - x = 18 -------------------------- 1 Two years ago x-2 = 4(y-2) x-2 = 4y -8 4y - x = 6 --------------------------- 2 Last yr; x - 1 = 3(y-1) x-1 = 3y -3 3y - x = 2 --------------------------- 3 Two yr time; x+2 = 2(y+2) x+2 = 2y + 4 x - 2y = 2 --------------------------- 4 Pick any two eqn's and solve simultaneously. I pick 3 and 4. Adding eqn's 3 and 4 gives; (3y+(-2y) + ((-x) + x) = 2 + 2= 4 y = 4 Put y = 4 in eqn 4 x-2(4) = 2; x = 10. So Ryan(y) = 4, Brian(x) = 10 Abeg make i go baff!!! |
Re: Algebra Cafe: Solve Math Problems Here by dinozzo(m): 7:19pm On Apr 05, 2008 |
donchichi:PTC, PTT, PCT, PCC are the transitional probabilities for the states T trucks and C cars, like using a markov chain for two given states. |
Re: Algebra Cafe: Solve Math Problems Here by Nobody: 8:42pm On Apr 05, 2008 |
still solving.hold on. |
Re: Algebra Cafe: Solve Math Problems Here by huxley(m): 8:54pm On Apr 05, 2008 |
Out of 100 ladies surveyed 85 had a white handbag 75 had black shoes 60 carried an umbrella 90 wore a ring How many ladies at least must have had all four items? |
Re: Algebra Cafe: Solve Math Problems Here by RichyBlacK(m): 11:01pm On Apr 05, 2008 |
dinozzo: Correct! That was a two-state Markov process, and you were solving for the stationary distribution. |
Re: Algebra Cafe: Solve Math Problems Here by RichyBlacK(m): 11:16pm On Apr 05, 2008 |
donchichi: Correct! 6C2/total sample size = 15/36 |
Re: Algebra Cafe: Solve Math Problems Here by RichyBlacK(m): 11:32pm On Apr 05, 2008 |
donchichi: donchichi, You're correct. You reduced a stochastic problem to a deterministic one. This is a safe technique since the set of deterministic models is a subset of the set of stochastic models. In other words, every stochastic problem has a deterministic subset. You already identified the problem, it's not always easy to find a deterministic replica of a stochastic model, especially when the number of possible states of the model is large. |
Re: Algebra Cafe: Solve Math Problems Here by bawomolo(m): 12:17am On Apr 06, 2008 |
you guys are going to deep into probability and statistics lol, beyond the realm of algebra |
Re: Algebra Cafe: Solve Math Problems Here by RichyBlacK(m): 1:12am On Apr 06, 2008 |
huxley: The maximum having all four items = 60 and theoretical minimum = 0. So, number must be between 0 and 65. I propose a geometric solution: x---------------------100--------------------------------x x--------------75-----------------------x x--------------60----------------x x------35-------x Answer = 35 |
Re: Algebra Cafe: Solve Math Problems Here by Nobody: 6:32am On Apr 07, 2008 |
with all u fools showing off your math skills, my question is, how has this helped nigeria in Science and technology. All that is emphasized is theory and theory, its about time u let these mathimatical skills let it be what it merely is which is a foundation to produce creative minds. The next step is the giant and distinguishing step, have you been able to invent anything/ or have you been able to solve any practical problems, those of you that are engineers, have you improved on any design or have you come up with something new, ![]() ![]() |
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