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donchichi
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Given x2 - y2 = 77 and x + y = 11 find x?
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 |
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donchichi
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A boot that costs x dollars is marked down by 20 percent off its original price on Monday, then an additional 10 percent on Tuesday. What is the price of the jacket, in terms of x?
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 |
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holamiday (m)
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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.  |
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holamiday (m)
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4sinxsinx-3=0 find the value of x hint:sin 60=(root 3)/2
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bawomolo (m)
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4sinxsinx-3=0
sin^2x=3/4
sinx=square root of 3/4
sinx=.866
x=60degrees or pi/3
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donchichi
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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. 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 the whole equation. And even when u divide botgh sides by a-a, you're actually dividing both sides by 0. |
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bawomolo (m)
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a(a-a)=(a+a)(a-a)[difference of two squares]
divide both sides by (a-a) division by zero is a logical fallacy. you can't divide both sides by (a-a). sorry to bust your bubble bro |
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RichyBlacK (m)
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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.
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top_kin (m)
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My brain bin sharp one time before o!  All these equations seem to me like magic now. Men, make i go jack o! |
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Imani (f)
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@donchichi Are you a mathematician?  You are good. I take off my hat to you. Post "easier" questions so at least so i can contribute now. My brain bin sharp one time before o! All these equations seem to me like magic now. Men, make i go jack o! No be only you with "cobwebs" for brain  . To be fair, if you don't have to use this level of maths, either at uni or in your line of work, then what is the point of learning them? It makes me question why we have to learn certain subjects which have no "continuity" in the real world and is just for knowledge purposes. Mathematics is different, you have to have numerical skills in most aspects of life. |
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pahtahkee
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WOW!!! |
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pahtahkee
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***Running. . . . I have to pick up my Engineering Mathematics and GRE textbook again. This is good!***
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RichyBlacK (m)
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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?
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RichyBlacK (m)
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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?
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dinozzo
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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.
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  |
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donchichi
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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.
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 Dinozzo, I'm sure your answer is correct because 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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huxley (m)
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In seven years time the combined ages of myself and my two sons will be 89. What will it be in 9 years time?
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huxley (m)
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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?
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donchichi
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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?
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 |
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donchichi
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@donchichi Are you a mathematician? You are good. I take off my hat to you. Post "easier" questions so at least so i can contribute now. Na my mama cause am!!! |
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donchichi
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In seven years time the combined ages of myself and my two sons will be 89. What will it be in 9 years time?
Let your current age be x and that of your two sons be y and z. In 7 years 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 years; (x+9) + (y + 9) + (z + 9); x + y +z +27. Remember x + y + z = 68. So x + y +z +27 = 68 + 27 = 95 |
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huxley (m)
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Let your current age be x and that of your two sons be y and z.
In 7 years 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 years;
(x+9) + (y + 9) + (z + 9);
x + y +z +27.
Remember
x + y + z = 68.
So x + y +z +27 = 68 + 27 = 95
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. |
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donchichi
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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?
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 = 4Put y = 4 in eqn 4 x-2(4) = 2; x = 10.So Ryan(y) = 4, Brian(x) = 10 Abeg make i go baff!!! |
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dinozzo
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Dinozzo,
I'm sure your answer is correct because 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.
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. |
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holamiday (m)
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still solving.hold on.
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huxley (m)
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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?
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RichyBlacK (m)
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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 Correct! That was a two-state Markov process, and you were solving for the stationary distribution. |
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RichyBlacK (m)
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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
Correct! 6C 2/total sample size = 15/36 |
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RichyBlacK (m)
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Dinozzo,
I'm sure your answer is correct because 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.
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. |
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bawomolo (m)
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you guys are going to deep into probability and statistics lol, beyond the realm of algebra
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RichyBlacK (m)
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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?
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 |
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engineerd (m)
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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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