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Mathematics Marathon HSC 09 (6 Viewers)

hermand

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1/64.


Q.It is estimated that 85% of students in Australia own a mobile phone.
(i) Two students are selected at random. What is the probability that neither
of them owns a mobile phone?
(ii) Based on a recent survey, 20% of the students who own a mobile phone
have used their mobile phone during class time. A student is selected at
random. What is the probability that the student owns a mobile phone
and has used it during class time?
 

ninetypercent

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Q.It is estimated that 85% of students in Australia own a mobile phone.
(i) Two students are selected at random. What is the probability that neither
of them owns a mobile phone?

0.15 x 0.15 = 0.0225 => 2.25%

(ii) Based on a recent survey, 20% of the students who own a mobile phone
have used their mobile phone during class time. A student is selected at
random. What is the probability that the student owns a mobile phone
and has used it during class time?

0.85 x 0.2 = 0.17 => 17%





Solve tan2x = cot40
 

untouchablecuz

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1/64.


Q.It is estimated that 85% of students in Australia own a mobile phone.
(i) Two students are selected at random. What is the probability that neither
of them owns a mobile phone?
(ii) Based on a recent survey, 20% of the students who own a mobile phone
have used their mobile phone during class time. A student is selected at
random. What is the probability that the student owns a mobile phone
and has used it during class time?
i) (0.15)*(0.15)=9/400
i) (0.85)*(0.2)=17/100

amiright?
 

ForrestGump

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(1/4)^3
=1/64
Now find the most likely combination(in any order) with 6 such lights.
1/64.
Q.It is estimated that 85% of students in Australia own a mobile phone.
(i) Two students are selected at random. What is the probability that neither
of them owns a mobile phone?
(ii) Based on a recent survey, 20% of the students who own a mobile phone
have used their mobile phone during class time. A student is selected at
random. What is the probability that the student owns a mobile phone
and has used it during class time?

Okay for the first question, the most likely combination:
P(GGG) = (3/4)^3 = 27/64
P(2xG, 1xR) = P(GGR) + P(GRG) + P(RGG)
= 3(3/4.3/4.1.4) = 9/64
So the most likely is getting 3 green lights

And for the Second Question:
i) P(NN) = P(N) x P(N) = (0.15)^2 = 0.0225
ii) P(Has phone and used in class) = 17/100

Okay how about:

Find the primitive function of 2x/(x2+1)
 

untouchablecuz

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Q.It is estimated that 85% of students in Australia own a mobile phone.
(i) Two students are selected at random. What is the probability that neither
of them owns a mobile phone?

0.15 x 0.15 = 0.0225 => 2.25%

(ii) Based on a recent survey, 20% of the students who own a mobile phone
have used their mobile phone during class time. A student is selected at
random. What is the probability that the student owns a mobile phone
and has used it during class time?

0.85 x 0.2 = 0.17 => 17%





Solve tan2x = cot40
tan2x=cot40=tan(5π/18)

.'. 2x=πk+(5π/18) => x=0.5(πk+(5π/18)) i.e. x=(π/36)(18k+5)
 

untouchablecuz

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i) find ∫tan(x)dx by writing the integral in the form f'(x)/f(x)

ii) a(t)=t3+sec2t

with initial conditions x(0)=6 m and v(0)=4 ms-1

using i), find x as a function of t
 
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untouchablecuz

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the question is pseudo-mathematics with the aim of making you seem intelligent. it doesn't make sense to say that a function equals infinity; a function describes a relation that produces an output value for an input value
 

eldore44

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Okay for the first question, the most likely combination:
P(GGG) = (3/4)^3 = 27/64
P(2xG, 1xR) = P(GGR) + P(GRG) + P(RGG)
= 3(3/4.3/4.1.4) = 9/64
So the most likely is getting 3 green lights
the q was actually for 6 lights. but doesnt matter.
2green and 1 red =(3.3.3.1)/(4.4.4)=27/64
and therefore it is equally likely as three greens.
 

s2Vicki

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the angle between 2 curves is 3 unit but not hard.

NEW QUESTION:

differentiate sin(^3)x (so that is sin cubed x)
3 sin(^2)x cos x

Question:

Find the value of n for which the equation (n+2)x^2 + 3x - 5 = 0 has one root triple the other.
 
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