Author Topic: Bucket's Questions (Read 47808 times) Tweet Share

bucket · « **Reply #225 on:** July 29, 2008, 12:44:54 am »

lol wtffff ahmad!!!
i think ill just expand then :p

bucket · « **Reply #226 on:** August 11, 2008, 09:33:08 pm »

the line with equation $y=4x+c$ is a tangent to the curve with equation $y=x^2-x-5$ . What is the value of c?

/0 · « **Reply #227 on:** August 11, 2008, 09:57:22 pm »

The line $y=4x+c$ has gradient 4, so you want to find where $y=x^2-x-5$ has gradient 4.

$\frac{dy}{dx} = 2x-1$

$2x-1 = 4 \Rightarrow x = \frac{5}{2}$

$\Rightarrow y = (\frac{5}{2})^2-(\frac{5}{2})-5=-\frac{5}{4}$

So the coordinates of the tangent point are $(\frac{5}{2}, - \frac{5}{4})$ , so

$- \frac{5}{4} = 4(\frac{5}{2})+c$

$c = -\frac{45}{4}$

bucket · « **Reply #228 on:** August 11, 2008, 10:02:21 pm »

love you.

bucket · « **Reply #229 on:** August 11, 2008, 10:14:46 pm »

water is draining from a cone-shaped funnel at a rate of 500cm³/min. The cone has base radius of 20cm and a height of 100cm. let h be the depth of water in the funnel at the time t minutes. the rate of decrease of h in cm/min is given by..?

Mao · « **Reply #230 on:** August 11, 2008, 10:22:31 pm »

$\frac{dh}{dt}=\boxed{\frac{dh}{dV}}\times \frac{dV}{dt}$

we want a relationship between h and V, i.e. volume of a cone:
$V=\frac{1}{3}\pi r^2 h$

by similar triangles, we find that $r=\frac{h}{5}$

$\implies V=\frac{\pi}{75}h^3$

$\implies \frac{dV}{dh}=\frac{\pi h^2}{25}$

$\implies \frac{dh}{dV}=\frac{25}{\pi h^2}$

$\frac{dh}{dt}=\frac{25}{\pi h^2}\times 500$

$\frac{dh}{dt}\right|_{h=100}=\frac{5}{4\pi}\approx 0.398 \mbox{cm min}^{-1}$

bucket · « **Reply #231 on:** August 16, 2008, 02:22:22 pm »

The probability of a target shooter hitting the bullseye on any one shot is 0.2

What is the smallest number of shots the shooter should make to ensure a probability of more than 0.95 of hitting the bullseye at least once?

In the previous parts of the question I found that $pr(X=0)=0.3277$ and hence $pr(X\geq1)=0.6723$

ell · « **Reply #232 on:** August 16, 2008, 07:34:16 pm »

Quote from: bucket on August 16, 2008, 02:22:22 pm

The probability of a target shooter hitting the bullseye on any one shot is 0.2

What is the smallest number of shots the shooter should make to ensure a probability of more than 0.95 of hitting the bullseye at least once?

$\Pr(X\geq1)>0.95$

$\implies 1 - \Pr(X<1)>0.95$

$\implies \Pr(X<1)<0.05$ (note the sign change)

Let n be the number of shots the shooter should make.

$\Pr(X<1) = \left( \begin{array}{c} n\\ 0\\ \end{array} \right)\cdot (0.2)^{0}\cdot (0.8)^{n}=(0.8)^{n}$

Now to solve $(0.8)^{n}<0.05$ , use your graphics calculator and enter $y=(0.8)^{x}$ . Go into TBLSET and make TblStart=0. Now go into TABLE and scroll down until you find the first value where Y1 is less than 0.05. It should end up being 0.04398, to which the corresponding X value is 14.

$\therefore n=14$
i.e. the smallest number of shots the shooter should make to ensure a probability of more than 0.95 at least once is 14.

bucket · « **Reply #233 on:** August 17, 2008, 03:09:31 pm »

good as!!! =]
thanks man, seriously, essential maths is shit at explanations.

bucket · « **Reply #234 on:** August 18, 2008, 04:28:44 pm »

Hey, with the previous question, what if it asked something like; ensure a probability of more than 0.95 of hitting the bullseye exactly twice?

Mao · « **Reply #235 on:** August 18, 2008, 04:34:43 pm »

$Pr(X=2)>0.95,\; \implies \binom{n}{2}\cdot (0.2)^2\cdot (0.8)^{n-2}>0.95,\; n\ge 2$

then you'd put into the y editor $\frac{x\cdot (x-1)}{2\times 1}\cdot (0.2)^2\cdot (0.8)^{x-2}$ and look through the table to see where it gets greater than 0.95.

for this case, however, the maximum probability of hitting the bullseye twice is 0.302 at 9 or 10 trials. hence the probability fo hitting the bullseye twice is NEVER greater than 0.95 (or 1/3 to be more realistic)

bucket · « **Reply #236 on:** August 18, 2008, 04:36:17 pm »

cheers.

bucket · « **Reply #237 on:** August 27, 2008, 10:50:28 pm »

The length of time, X(minutes), between the arrival of customers at an autobank has a random variable with probability density function:

$ f(x) = \left\{ \begin{array}{cc} \frac{1}{5}e^{-\frac{x}{5}} & x\geq0 \\ 0 & x<0 \\ \end{array} $

Find the probability that more than 12mins elapses between successive customers given that more than 8 minutes has passed.

I translated the question into:
Find $pr(X>12|X>8)$

is that right?
if so how do I solve it now?

I found that $pr(X>8)=0.202$

These questions that include 'given' always fuck me up :S

Collin Li · « **Reply #238 on:** August 27, 2008, 10:54:25 pm »

Since the intersection of $X>8$ and $X>12$ is $X>12$ , then the formula for conditional probability becomes:

$\mbox{Pr}(X>12|X>8) = \frac{\mbox{P}(X>12)}{\mbox{P}(X>8)}$

bucket · « **Reply #239 on:** August 27, 2008, 10:57:43 pm »

how did you find that as the intersection? :|

Search the forums now!

We have moved!

Author Topic: Bucket's Questions (Read 47808 times) Tweet Share

bucket

Re: Bucket's Questions

bucket

Re: Bucket's Questions

/0

Re: Bucket's Questions

bucket

Re: Bucket's Questions

bucket

Re: Bucket's Questions

Mao

Re: Bucket's Questions

bucket

Re: Bucket's Questions

ell

Re: Bucket's Questions

bucket

Re: Bucket's Questions

bucket

Re: Bucket's Questions

Mao

Re: Bucket's Questions

bucket

Re: Bucket's Questions

bucket

Re: Bucket's Questions

Collin Li

Re: Bucket's Questions

bucket

Re: Bucket's Questions

Recent Posts

User:
Password: