emaths probability
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amy either walks or cycles to school each morning . the probability that she walks is 2/3 . when she walks , the probability that she is late is 1/9 . when she cycles , the probability that she is late is 4/9 .
find the probability that she is late.
i cant seems to get ans right
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P(cycle) = 1 - 2/3 = 1/3
P(cycle and late) = 1/3 x 4/9 = 4/27
P(walk and late) = 2/3 x 1/9 = 2/27
P(walk and late OR cycle and late) = 4/27 + 2/27 = 2/9 -
yeah but the answer is 6/27 ..
answer wrong?
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2/9 and 6/27 same ah... u must simpify...
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Originally posted by bratorbread:
yeah but the answer is 6/27 ..
answer wrong?
reduce to simplest terms
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oh yeah! thx haha didnt notice.
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@ bread
u seem very tensed up are u ?
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yeah 2 mre weeks till o's
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but some out of topic stuff
qns seems faulted, why walk less likely to be late than cycle
In topic
In recent syllabus change , no longer need to present ans in P(E) where e is event, now just write Probability: ...(blah blah blah)... can liao
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its one of the questions from 2006 tys haha
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Originally posted by bratorbread:
its one of the questions from 2006 tys haha
no wonder so familiar x.x
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wow
queen of sgforum help u solve leh..
means the question very easy.
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haha.
some other questions which i dun get it...
given tat (3x-1)(x+p)=3x^2 +qx -2
find p,
find q.
the force F , between 2 particles are inversely proportional to the square of the distance between them.
the force is 36 units when the dist. between the particles are r metres.
find the force when the distance is 3r metres.
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Originally posted by bratorbread:
haha.
some other questions which i dun get it...
given tat (3x-1)(x+p)=3x^2 +qx -2
find p,
find q.
the force F , between 2 particles are inversely proportional to the square of the distance between them.
the force is 36 units when the dist. between the particles are r metres.
find the force when the distance is 3r metres.
Expand (3x-1)(x+p) then just compare coefficients with the right hand side equation to solve for p and q.
F = k/(dist^2)
36=k/(r^2)
k=36r^2F = (36r^2) / ((3r)^2)
= 36r^2 / 9r^2
= 4 -
Originally posted by bratorbread:
haha.
some other questions which i dun get it...
given tat (3x-1)(x+p)=3x^2 +qx -2
find p,
find q.
the force F , between 2 particles are inversely proportional to the square of the distance between them.
the force is 36 units when the dist. between the particles are r metres.
find the force when the distance is 3r metres.
Hi bratorbread,
For the first question, are the two sides equivalent?
For the second question, what they mean is F = k/(D^2), where k is a constant, and D is the distance.
What we have is that 36 = k/(r^2).
Hence, when D = 3r, we will have F = k / (3r)^2 = k / 9(r^2) = 36/9 = 4 units.
Cheers.
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when i expand it out i get 3x^2 + 3xp -x -p = 3x^2 + qx -2
i still dun get the values of p and q
idk the question did not state tat both sides are equivalent or not
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Originally posted by bratorbread:
when i expand it out i get 3x^2 + 3xp -x -p = 3x^2 + qx -2
i still dun get the values of p and q
idk the question did not state tat both sides are equivalent or not
From the equation, u can already get p since p = 2 by comparing coefficient of the constant on both side.
With p u better be able to get q if not u will really need alot of help with ur maths
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Originally posted by bratorbread:
when i expand it out i get 3x^2 + 3xp -x -p = 3x^2 + qx -2
i still dun get the values of p and q
idk the question did not state tat both sides are equivalent or not
Hi bratorbread,
Did they used = or = for the two polynomials?
Cheers
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Originally posted by TrueHeart:
Hi bratorbread,
Did they used = or = for the two polynomials?
Cheers
Such questions are always equivalent if not it can't be solved
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just = .
when p = 2 den
3x^2 + 5x -2 = 3x^2 + qx -2
so q = 5 ?
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Originally posted by bratorbread:
just = .
when p = 2 den
3x^2 + 5x -2 = 3x^2 + qx -2
so q = 5 ?
yup :)
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okie i get it . thanks :)