Author Topic: Mathematics Question Thread (Read 1628151 times) Tweet Share

RuiAce · « **Reply #3180 on:** December 11, 2017, 08:07:01 pm »

Quote from: sssona09 on December 11, 2017, 07:59:02 pm

wondering why the answer to this isn't (9/14 x 9/14)
the answer is 17/42

the question is:
students studying at least one of the languages, french and Japanese, attend a meeting. Of the 28 students present, 18 study french and 22 study Japanese.
(i)whats the probability that two randomly chosen students both study french?
(ii) what is the probability that a randomly chosen student studies both languages?

thank you

$\text{Because once you choose a student, you can't choose them again;}\\ \text{they are excluded for the choice of the second student chosen.}\\ \text{The correct answer is }\frac{18}{28}\times \frac{17}{27}.$

itssona · « **Reply #3181 on:** December 11, 2017, 08:58:20 pm »

got another
david has invented a game in which he throws two die repeatedly until the sum of the two numbers shown is either 7 or 9. If the sum is 9, david wins. If the sum is 7, he loses. If it's another number, he continues to throw until it's a 7 or 9.
(I) probability he wins on first throw is 1/9 (I found out)
(ii) prob that a second throw is needed is 26/36 (I found out)
(iii) what is the probability tgat david wins on his first, second or third throw? leave answer in unsimplified form
(iv) probability that david wins the game?

RuiAce · « **Reply #3182 on:** December 11, 2017, 09:08:26 pm »

Quote from: sssona09 on December 11, 2017, 08:58:20 pm

got another
david has invented a game in which he throws two die repeatedly until the sum of the two numbers shown is either 7 or 9. If the sum is 9, david wins. If the sum is 7, he loses. If it's another number, he continues to throw until it's a 7 or 9.
(I) probability he wins on first throw is 1/9 (I found out)
(ii) prob that a second throw is needed is 26/36 (I found out)
(iii) what is the probability tgat david wins on his first, second or third throw? leave answer in unsimplified form
(iv) probability that david wins the game?

$\text{Continuing on from part ii), if a second throw is needed}\\ \text{and in fact, we }\textbf{do}\text{ win on the second throw,}\\ \text{borrowing computations from part i) this probability is }\boxed{\frac{1}{9}\times \frac{13}{18}}$
$\text{But if instead, a third throw is needed, then we must've thrown}\\ \text{a number, not 7 or 9, on the }\textbf{second throw as well}.\\ \text{Borrowing computations from part i), the probability a third throw is needed is }\frac{13}{18}\times \frac{13}{18} = \left(\frac{13}{18}\right)^2$
$\text{Given now that a third throw is needed,}\\ \text{the probability (that we need to further multiply) he wins on that third throw will, again, be }\frac19.\\ \text{Hence, the probability he wins on the third throw is }\boxed{\frac{1}{9}\times\left(\frac{13}{18}\right)^2 }$
$\text{So the probability David wins on the first, second }\textbf{or}\text{ third, will be}\\ \boxed{\frac19 + \frac19\left(\frac{13}{18}\right) + \frac19\left(\frac{13}{18}\right)^2}$
____________________________________________________________________________________
$\text{The probability that David wins the game is essentially the sum of all probabilities}\\ \text{consisting of him specifically winning on the first go, the second go,}\\ \text{the third, the fourth, the fifth and so on.}$
$\text{But we can see that in a very similar manner}\\ \text{The probability he wins on the fourth turn is }\frac19\left(\frac{13}{18}\right)^3\\ \text{The probability he wins on the fifth turn is }\frac19\left(\frac{13}{18}\right)^4\\ \text{and so on.}$
$\text{Hence, the probability he will ultimately win is}\\ \frac19+\frac19\left(\frac{13}{18}\right)+\frac19\left(\frac{13}{18}\right)^2+\frac19\left(\frac{13}{18}\right)^3+\frac19\left(\frac{13}{18}\right)^4+\dots\\ \text{which is, in fact, }\textbf{a geometric series}\\ \text{with common ratio }r=\frac{13}{18}.$
$\text{So using our formula for the limiting sum}\\ \text{our answer is }\frac{\frac19}{1-\frac{13}{18}}$

itssona · « **Reply #3183 on:** December 11, 2017, 10:33:45 pm »

stumped for part iii in here

A box contains
twelve chocolates all of exactly the same appearance. Four of the chocolates are hard and eight are soft. Kim eats three chocolates chosen randomly from the box.
Find probability that:
I) the first chocolate kim eats is hard
ii) kim eats three hard chocolates
iii) kim eats exactly one hard chocolate
thank youuu

RuiAce · « **Reply #3184 on:** December 11, 2017, 10:37:32 pm »

Quote from: sssona09 on December 11, 2017, 10:33:45 pm

stumped for part iii in here

A box contains
twelve chocolates all of exactly the same appearance. Four of the chocolates are hard and eight are soft. Kim eats three chocolates chosen randomly from the box.
Find probability that:
I) the first chocolate kim eats is hard
ii) kim eats three hard chocolates
iii) kim eats exactly one hard chocolate
thank youuu

Hint: There's more than one way in which he can eat exactly 1 hard chocolate, since he could eat the hard chocolate first, second or third.

Calley123 · « **Reply #3185 on:** December 14, 2017, 06:56:31 pm »

Sketch the curve y= (x^2+1)/e^2
showing any stationary points and inflexions.

I know how to do the question but i'm not getting all the points
Answers: its a graph answer -Something at ( 1,2/e) )-got this one AND ( 3, 10/e^3) ---not getting this answer
-thanks

RuiAce · « **Reply #3186 on:** December 14, 2017, 07:15:09 pm »

Quote from: siri4533 on December 14, 2017, 06:56:31 pm

Sketch the curve y= (x^2+1)/e^2
showing any stationary points and inflexions.

I know how to do the question but i'm not getting all the points
Answers: its a graph answer -Something at ( 1,2/e) )-got this one AND ( 3, 10/e^3) ---not getting this answer
-thanks

$\text{Are you sure you typed that correctly?}\\ y=\frac{x^2+1}{e^2}\text{ is just a parabola.}$

Calley123 · « **Reply #3187 on:** December 14, 2017, 07:37:45 pm »

Quote from: RuiAce on December 14, 2017, 07:15:09 pm

$\text{Are you sure you typed that correctly?}\\ y=\frac{x^2+1}{e^2}\text{ is just a parabola.}$

Sorry !!
- meant to say e^x at the bottom.

RuiAce · « **Reply #3188 on:** December 14, 2017, 07:40:41 pm »

Quote from: siri4533 on December 14, 2017, 07:37:45 pm

Sorry !!
- meant to say e^x at the bottom.

I had a quick look on GeoGebra and for $y=\frac{x^2+1}{e^x} $ there appears to be a point of inflexion at $\left(3,\frac{10}{e^3}\right)$. This comes out from setting $ \frac{d^2y}{dx^2}=0$.

Personally, I reckon using the quotient rule on that is disgusting and it'd be better to differentiate with $ y=e^{-x}(x^2+1)$. (At least, using the quotient rule the second time will be disgusting.) If that doesn't work, tell me what your $y^{\prime\prime}$ was and I'll investigate

Note: The one at $ \left(1,\frac{2}{e}\right) $ is a horizontal point of inflexion, and thus satisfies both $\frac{dy}{dx}=0$ and $\frac{d^2y}{dx^2}=0$.

Calley123 · « **Reply #3189 on:** December 14, 2017, 08:39:10 pm »

I got both the points using the 2nd derivative !!! So both points are inflexions and there is no stationary points ??

I don't understand why I couldn't have gotten both points through the first derivative ( only gave me x=1) though.

- Thanks

RuiAce · « **Reply #3190 on:** December 14, 2017, 09:25:43 pm »

Quote from: siri4533 on December 14, 2017, 08:39:10 pm

I got both the points using the 2nd derivative !!! So both points are inflexions and there is no stationary points ??

I don't understand why I couldn't have gotten both points through the first derivative ( only gave me x=1) though.

- Thanks

$\text{Because setting the first derivative to zero gives you the }\textbf{stationary points.}\\ \text{A stationary point }\textbf{does not necessarily have to be, but CAN also be}\text{ a point of inflexion.}$
$\text{A point that is not a stationary point but is a point of inflexion is called exactly that,}\\ \text{ a point of inflexion.}\\ \text{But a point that is both a stationary point AND a point of inflexion is called}\\ \text{a }\textbf{horizontal point of inflexion.}$
__________________________________________________
$\text{In your case, at }x=1\text{ we had }\frac{dy}{dx}=0\textbf{ and }\frac{d^2y}{dx^2}=0\\ \text{So }\left(1,\frac{2}{e}\right)\text{ is both a stationary point }\textbf{and}\text{ a point of inflexion.}\\ \text{Thus, it is a }\textbf{horizontal point of inflexion}.$
$\text{Whereas at }x=3\text{, we had }\frac{d^2y}{dx^2}=0\text{ but }\frac{dy}{dx}\text{ was not zero.}\\ \text{So }\left(3,\frac{10}{e^3}\right)\text{ is }\textbf{only}\text{ a point of inflexion.}$
Of course, in saying all that stuff about points of inflexion you do have to test both sides of $\frac{d^2y}{dx^2}$. But I'll just assume you already knew that.

Calley123 · « **Reply #3191 on:** December 14, 2017, 10:15:11 pm »

Quote from: RuiAce on December 14, 2017, 09:25:43 pm

$\text{Because setting the first derivative to zero gives you the }\textbf{stationary points.}\\ \text{A stationary point }\textbf{does not necessarily have to be, but CAN also be}\text{ a point of inflexion.}$
$\text{A point that is not a stationary point but is a point of inflexion is called exactly that,}\\ \text{ a point of inflexion.}\\ \text{But a point that is both a stationary point AND a point of inflexion is called}\\ \text{a }\textbf{horizontal point of inflexion.}$
__________________________________________________
$\text{In your case, at }x=1\text{ we had }\frac{dy}{dx}=0\textbf{ and }\frac{d^2y}{dx^2}=0\\ \text{So }\left(1,\frac{2}{e}\right)\text{ is both a stationary point }\textbf{and}\text{ a point of inflexion.}\\ \text{Thus, it is a }\textbf{horizontal point of inflexion}.$
$\text{Whereas at }x=3\text{, we had }\frac{d^2y}{dx^2}=0\text{ but }\frac{dy}{dx}\text{ was not zero.}\\ \text{So }\left(3,\frac{10}{e^3}\right)\text{ is }\textbf{only}\text{ a point of inflexion.}$
Of course, in saying all that stuff about points of inflexion you do have to test both sides of $\frac{d^2y}{dx^2}$. But I'll just assume you already knew that.

Thank you!!!

gilliesb18 · « **Reply #3192 on:** December 15, 2017, 09:55:40 am »

Helloo,

Just need help on maxima/minima, stationary points question:
The formula for the surface area of a cylinder is given by S=2πr(r+h). Show that if the cylinder holds a volume of 54πm^3, the surface area is given by the equation S=2πr^2 + 108π/r. Hence find the radius that gives the minimum surface area.

Thanks heaps!

Natasha.97 · « **Reply #3193 on:** December 15, 2017, 10:43:03 am »

Quote from: gilliesb18 on December 15, 2017, 09:55:40 am

Helloo,

Just need help on maxima/minima, stationary points question:
The formula for the surface area of a cylinder is given by S=2πr(r+h). Show that if the cylinder holds a volume of 54πm^3, the surface area is given by the equation S=2πr^2 + 108π/r. Hence find the radius that gives the minimum surface area.

Thanks heaps!

Hi!
a) We know that V = πr²h. Making h the subject will give you 54/r². Substituting h back into S=2πr(r+h) will give you S=2πr² + 108π/r.

Not 100% sure on this:
b) We've already taken into account the given volume in part a. Deriving S will give 4πr - 108π/r². Equating it to 0 and simplifying will result in a radius of 3m.

Hope this helps

gilliesb18 · « **Reply #3194 on:** December 15, 2017, 10:47:46 am »

Oh ak!! I made r the subject which was probably the dumbest thing to do!!!
Yes that makes sense. And I just checked the back of the book, and 3m is the right answer.

Thanks heaps!!

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