Perhaps its a bit late, but I was just wondering...
Show that v=Ce-kt satisfies the equation dv/dt=-kv, basically exponential growth and decay and easy throw away marks, but what do you actually write to get the marks? I looked up the proof in the text book, but its about ten lines long... not worth it for 2 marks.
Then make Fn = 0, determine the number of years, then take this value, multiply by 15400 and finally multiply by $10 each to get the total income. This was basically the last question in the 1998 HSC 2 unit exam.
Lol didn't realise my probability question would get 2 pages of attention. As I've said before, the answer is 7/10 and I have posted the full solution above.
Then make Fn = 0, determine the number of years, then take this value, multiply by 15400 and finally multiply by $10 each to get the total income. This was basically the last question in the 1998 HSC 2 unit exam.
I assume you mean differentiate.
(lnx + (x-1)/x)x^(x-1) by rewriting as e^ln(x(x-1))
Next (a question I made):
P(a,a^3) and Q(b,b^3) are the points of intersection of y = k^2/x and y = x^3 (k>0)
i) Explain why a and b are the roots of k/x^2 - x^3 = 0. Without finding the roots of the equation, show that a+b=0 and ab=-k
ii) Show that the line PQ has equation y-a^3 = k(x-a)
iii) Find an expression for PQ in terms of b.
iv) Hence show that the equation of PQ is y = kx.
if P and Q are the intersections, then they must satisfy both equations. they do satisfy y = x^3 ; but for y = k^2/x :
1) a^3 = k^2/a ---> a^4 = k^2 ---> a^2 = +/- k
2) b^3 = k^2/b ---> b^4 = k^2 ---> b^2 = +/- k
but both a^2 & b^2 are > 0 since 'a' and 'b' are reals. hence, there are only two possible cases:
(i) a^2 = k , and, b^2 = k
(ii) a^2 = -k , and, b^2 = -k
but k > 0 ; hence, case (i) is the only possibility.
ie. a^2 = k = b^2 ; assuming that 'a' does not = 'b', then a = -b = sqrt(k)
Therefore;
Answer to (i):
a = -b ---> a+b = 0 ; and, ab = (sqrt(k))(-sqrt(k)) = -k
Answer to (ii):
just algebra and substitution using results obtained in (i)
Answer to (iii):
equation from (ii) is y - a^3 = k(x - a) ; but since a = -b ; then equation in terms of 'b' is:
y + b^3 = k(x + b)
Answer to (iv):
from (ii), equation is y = kx - ak + a^3 ; since -k = ab = -a^2, then -ak = -a^3
ie. equation: y = kx -a^3 + a^3 = kx
thus, y = kx is the equation of PQ.
Next question:
In a game of (fair) coin toss Jack and Jill take alternating turns to toss a coin. Jill gets to go first, followed by Jack. the winner of the game is the person who tosses the first head.
(a) is the game "fair"? who has the higher probability of winning any given game? and what is that probability?
(b) Jack and Jill then decide to play in a competition in which they play a number of the coin toss game. it was agreed that for every game Jack wins in the competition Jill will compensate him $1. the competition ends, however, when Jill gets her first win.
Find the probability that Jack will have more than $99 before the competition ends.
Well maths is better than economics anyday so let's see,
a) No the game isn't fair. Jill clearly has the higher probability of winning as she tosses first.
for Jill to win, she must toss a head on first turn, or toss a head on second turn after Jack tosses a tail on his second turn etc...
i.e. P(Jill wins) = 0.5 + 0.5*0.5*0.5 + 0.55 + ...
this is a limiting sum with a = 0.5, r = 0.52
so P(Jill wins) = a/(1 - r) = 0.5/0.75 = 2/3
so P(Jack wins) = 1 - P(Jill wins) = 1/3
hence Jill has the higher probability of winning: 2/3
b) P(Jack will have more than 99 bucks) = P(Jill loses the first 100 games then wins, + 101 games + 102 games + ...) = (2/3)(1/3)100 + (2/3)(1/3)101 + (2/3)(1/3)102 + ... = (2/3)(1/3)100/(2/3) = (1/3)100 = pretty unlikely , = P(Jill loses the first 100 games) yeah i just repeated myself like an idiot
cbf don't have calculator
Question:
If x, y and 9 are the first three terms of a geometric series and y, x and 2 are the first three terms of an arithmetic sequence, find the values of x and y.
Question:
If x, y and 9 are the first three terms of a geometric series and y, x and 2 are the first three terms of an arithmetic sequence, find the values of x and y.
Question:
The region bounded by the curve y=ex + e-x, the x axis and the line x=0 and x=2 is rotated about the axis. Find the volume of the solid formed .
Originally Posted by word
Well maths is better than economics anyday so let's see,
a) No the game isn't fair. Jill clearly has the higher probability of winning as she tosses first.
for Jill to win, she must toss a head on first turn, or toss a head on second turn after Jack tosses a tail on his second turn etc...
i.e. P(Jill wins) = 0.5 + 0.5*0.5*0.5 + 0.55 + ...
this is a limiting sum with a = 0.5, r = 0.52
so P(Jill wins) = a/(1 - r) = 0.5/0.75 = 2/3
so P(Jack wins) = 1 - P(Jill wins) = 1/3
hence Jill has the higher probability of winning: 2/3
b) P(Jack will have more than 99 bucks) = P(Jill loses the first 100 games then wins, + 101 games + 102 games + ...) = (2/3)(1/3)100 + (2/3)(1/3)101 + (2/3)(1/3)102 + ... = (2/3)(1/3)100/(2/3) = (1/3)100 = pretty unlikely , = P(Jill loses the first 100 games) yeah i just repeated myself like an idiot
Question:
The region bounded by the curve y=ex + e-x, the x axis and the line x=0 and x=2 is rotated about the axis. Find the volume of the solid formed .
V=pi x integral 0->2 y2dx
pi x integral 0->2 (ex-e-x)2dx
=pi x integral 0->2 (e2x+2+e-2x)dx
=pi[e2x/2+2x-e-2x/2]0->2
=pi(e4-1)/2e4+4)
=pi(e8-1)/2e4)+4pi units3
