Maths Methods 3/4 Help Thread 2011

[luken93]

1) Set up a matrix with the 4 probabilities, multiply it by [1,0]

[gossamer]

Here's 2 conditional probability Q's.
1-Pr(Nathan kicks more than 4 goals on a wet day)=0.3
Pr(Nathan kicks more than 4 goals on a dry day)=0.6
Pr(that it'll be wet on the day of next game)=0.7
What is the probability that Nathan will kick more than 4 goals in next game?

Pr(>4goals|W)=0.3 Pr(W)= 0.7, so Pr(>4 goals and W) = 0.21
Pr(>4goals|D)=0.6 Pr(D)=0.3 (1-0.7), so Pr(>4 goals and D)= 0.18
Therefore, Pr(>4 goals) = 0.21 + 0.18 = 0.39

I guess for these questions it's just understanding the information given to you -- when you read the question you need to understand that Pr(Nathan kicks more than 4 goals on a wet day)=0.3 is the probability given it's a wet day.

For the second one, Pr (person with disease yields a positive result)=0.95 is the probabily that the result is positive given the person has a disease, and so on.

[happycat]

Thanx for the help gossamer

[#1procrastinator]

Why do you have to change (rationalise?) x/(√(x+4) - 2) before finding the limit as x->0? Don't you only have to do that when you get 0 in the denominator?

[b^3]

if you plug in lim->0 you will get 0 in the denominator as root(4)=2 and 2-2=0

[#1procrastinator]

Ah, it has to be LIM->0? I was told to just plug in the value (0) and see what you get...obviously 0

[xZero]

Ah, it has to be LIM->0? I was told to just plug in the value (0) and see what you get...obviously 0

if you sub x=0 in, you'll get 0/0 which is undefined

[b^3]

Ah, it has to be LIM->0? I was told to just plug in the value (0) and see what you get...obviously 0

if you sub x=0 in, you'll get 0/0 which is undefined

Yeh sorry, my bad wording. Thats what I meant, I have 2 many numb3rs g01ng thr0ugh my h3ad r1ght n0w.

[epinephrine]

WOW is it just me or are kilbaha's exams quite good !   

Edit: probably harder than Derrick Ha's !!!

Wait this is the wrong place to post this comment.

[#1procrastinator]

Ah, it has to be LIM->0? I was told to just plug in the value (0) and see what you get...obviously 0

if you sub x=0 in, you'll get 0/0 which is undefined

LOL shit what the f how the hell did I miss that, I was adding 2 to the square root of 4 instead of subtracting haha

Thanks

[cltf]

How do I do these?

Express in the form of "" where "a" is real and "b" is [0, pi/2]
Then find g^-1

given that express y in terms of X

and finally E(2-3X) is the same as 2-3E(X) yes?

[brightsky]

g(t) = sin(t) + cos(t)
sqrt(2)/2 g(t) = sqrt(2)/2 sin(t) + sqrt(2)/2 cos(t)
sqrt(2)/2 g(t) = cos(pi/4) sin(t) + sin(pi/4) cos(t)
sqrt(2)/2 g(t) = sin(t + pi/4)
g(t) = sqrt(2) sin(t + pi/4) [after rationalization]

[cltf]

g(t) = sin(t) + cos(t)
sqrt(2)/2 g(t) = sqrt(2)/2 sin(t) + sqrt(2)/2 cos(t)
sqrt(2)/2 g(t) = cos(pi/4) sin(t) + sin(pi/4) cos(t)
sqrt(2)/2 g(t) = sin(t + pi/4)
g(t) = sqrt(2) sin(t + pi/4) [after rationalization]

Where does the rt(2)/2 come from and why?

[brightsky]

your second question is a bit ambiguous but:

if it's x = (y^2 + 1)/(y^2 + 2)
then
x(y^2 + 2) = y^2 + 1
xy^2 + 2x = y^2 + 1
xy^2 - y^2 = 1 - 2x
y^2(x - 1) = 1 - 2x
y^2 = (1-2x)/(x-1)
y = +- sqrt((1-2x)/(x-1))

if it's x = y^2 + (1/y^2) + 2
multiply everything by y^2:
xy^2 = y^4 + 1 + 2y^2
y^4 + (2-x) y^2 + 1 = 0
let y^2 = u
u^2 + (2-x)u + 1 = 0
use quadratic formula and then find y.

[brightsky]

g(t) = sin(t) + cos(t)
sqrt(2)/2 g(t) = sqrt(2)/2 sin(t) + sqrt(2)/2 cos(t)
sqrt(2)/2 g(t) = cos(pi/4) sin(t) + sin(pi/4) cos(t)
sqrt(2)/2 g(t) = sin(t + pi/4)
g(t) = sqrt(2) sin(t + pi/4) [after rationalization]

Where does the rt(2)/2 come from and why?

you multiply both sides by sqrt(2)/2. this allows you to exploit the compound angle formula for sine on the RHS.