lots of qs help me (2 Viewers)

totallybord

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hi,
can someone help me with these questions?
1) from eight letters of the word FREQUENT, 3 are taken at random and placed in a line. How many different letter sequences are possible?
2) What are the end points of a parabola?
3) What is the probability of throwing a 6 with up to 3 throws of a dice? Does the up to 3 throws bit affect anything?
4) what is the inverse function of 2x/(x^2+1) ?
5) I cant seem to get the following answer: Prove the tngent at y=tanx and y=cosx are perpendicular.
6) There is a row of 5 coloured lights each with its own switch. When switched on , each lghts i equally likely to show green, red or amber. If all lights are switched on, What is the probabblity that a)first 3 lights from leftare grren but not the 4th. b) first 3 lights from left are same colour bt not the 4th (answer to a) x3 right) and c) exactly 3 lights are green? d) if all lights are switched on 5 times, find, the probability that exactly 3 lights will be the same colour on 2 or 3 occassions.

Thank you to whoever answers:D:D
 

vds700

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totallybord said:
hi,
can someone help me with these questions?
1) from eight letters of the word FREQUENT, 3 are taken at random and placed in a line. How many different letter sequences are possible?
2) What are the end points of a parabola?
3) What is the probability of throwing a 6 with up to 3 throws of a dice? Does the up to 3 throws bit affect anything?
4) what is the inverse function of 2x/(x^2+1) ?
5) I cant seem to get the following answer: Prove the tngent at y=tanx and y=cosx are perpendicular.
6) There is a row of 5 coloured lights each with its own switch. When switched on , each lghts i equally likely to show green, red or amber. If all lights are switched on, What is the probabblity that a)first 3 lights from leftare grren but not the 4th. b) first 3 lights from left are same colour bt not the 4th (answer to a) x3 right) and c) exactly 3 lights are green? d) if all lights are switched on 5 times, find, the probability that exactly 3 lights will be the same colour on 2 or 3 occassions.

Thank you to whoever answers:D:D
Prove the tngent at y=tanx and y=cosx are perpendicular.

Did this today. At the point of intersection, tanx = cosx

y=tanx, y' = sec^2 x,

y=cosx, y' = -sinx

product of gradients = -sinx/cos^2 x = -tanx/cosx = -tanx/tanx (from above)
=-1
Therefore tangents are perpendicular.
 

Azreil

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1) 8P3/2! = 336/2 = 168
2) Trick question? Doesn't have endpoints?
3) I'm not sure how to interpret this question. My thinking goes like this:
5/6 * 5/6 * 1/6 + 5/6 * 1/6 + 1/6 = 91/216
P(6 on 3rd throw) + P(6 on 2nd throw) + P(6 on 3rd throw)
But I'm not sure if they keep throwing the dice after they get a 6.
4) y = 2x/(x^2+1)
D(f) = x >= 1
R(f) = y all real

x = 2y/(y^2+1)
2y = x(y^2+1)
2y = xy^2+x
2y-xy^2 = x
x-2y+xy^2 = 0

I can't get any further than this but I've always sucked at these questions. Make y subject and then you have your answer. Make sure you include the range and domain of the inverse as well.
5) Answered above
6) There is a row of 5 coloured lights each with its own switch. When switched on, each lghts i equally likely to show green, red or amber. If all lights are switched on, What is the probabblity that
a) 1/3 * 1/3 * 1/3 * 2/3 = 2/81
b) 1 * 1/3 * 1/3 * 2/3 = 2/27
c) 9 * (1/3 * 1/3 * 1/3 * 2/3 * 2/3) = 4/27 (?)
d) Yeah no. Not working through that right now sorry.
 

ronnknee

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4) what is the inverse function of 2x/(x^2+1) ?

Since the original function is y = f(x) = 2x / (x2 + 1)
The equation for the inverse function is x = 2y / (y2 + 1)
Now we rearrange the equation so that y is the subject

x = 2y / y2 + 1
xy2 + x - 2y = 0

Notice how it is a quadratic equation in terms of y
Therefore, we use the quadratic formula

y = [2 +- root (4 - 4x2)] / 2x
y = [2 +- 2 root (1 - x2)] / 2x
y = [1 +- root (1 - x2)] / x

Therefore the inverse functions are

f(x) = [1 + root (1 - x2)] / x
f(x) = [1 - root (1 - x2)] / x
where -1 < x < 1 and x cannot be 0
 

ronnknee

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6) There is a row of 5 coloured lights each with its own switch. When switched on , each lghts i equally likely to show green, red or amber. If all lights are switched on, What is the probabblity that

c) exactly 3 lights are green?

Let P(E) be the probability of 3 lights exactly green
There are 5C3 ways of 3 lights having the exact colour
For one way of three lights same, (1/3)3 . (2/3)2
There are three ways of having the same colour, but we only want green

Therefore
P(E) = 5C3 . (1/3)3 . (2/3)2 . 1/3
P(E) = 40/729


d) if all lights are switched on 5 times, find, the probability that exactly 3 lights will be the same colour on 2 or 3 occassions.


Similarly let P(F) be the probability of exactly 3 lights will be the same colour on 2 or 3 occassions

There are 5C3 + 5C2 ways of having 3 or 2 occasions

Let P(G) be the probabilty of each occasion having 3 lights of exact colour
For each occasion, there are 5C3 ways of 3 lights having the exact colour
For one way of three lights same, (1/3)3 . (2/3)2

Therefore P(G) = 5C3 . (1/3)3 . (2/3)2

Therefore
P(F) = {5C3 . P(G)3 . [1 - P(G)]2} + {5C2 . P(G)2 . [1 - P(G)]3}
 
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ronnknee

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The end points of a parabola are the points where a chord/secant touches the parabola
 

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