Mathematics Marathon HSC 09 (1 Viewer)

[hermand]

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]

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]

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]

(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)

 

[ninetypercent]

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

log(e) x^2 + 1

 

[ninetypercent]

Show that the curve: y = x^3 + 3x^2 + 3x -2 has one stationary point

 

[untouchablecuz]

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]

Show that the curve: y = x^3 + 3x^2 + 3x -2 has one stationary point

dy/dx=3x^2+6x+3

Δ=6²-4(3)(3)=36-36=0

.'. one stationary point

 

[untouchablecuz]

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

 

[ForrestGump]

Haha you lost me on that one. i hate apps of calculus to physical world.

 

[Schoey93]

because a circle goes around forever, there is no defined beginning or end.

Correct - the question was rather quite simple... !

 

[untouchablecuz]

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]

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.

 

[ninetypercent]

does that make sense?

did you mean for which the roots are equal?

oh sorry. yeah that's what I meant

 

[emmatheron]

angle between the lines/curves:
y=x^2
y=x^3

 

[hermand]

angle between the lines/curves:
y=x^2
y=x^3

you need to specify a point.

 

[00iCon]

you need to specify a point.

the only point they meet is O(0,0).

 

[hermand]

the only point they meet is O(0,0).

no, they meet at two points.

(0,0) and (1,1)

 

[s2Vicki]

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.

 

[untouchablecuz]

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

noone?