Mathematics Question Thread

[spnmox]

The quadrilateral formed by (1, 2), (3, 3), (1, 7), (-1, 3) is ?

John has 10 marbles in a bag, 3 of which are green and the rest are yellow. He randomly draws 1 out of
the bag, notes its colour and then puts it back before drawing another out of the bag. What is the
probability that both marbles drawn are the same colour?

is the second one 0.58?? haha i'm so unsure with probability questions

[InnererSchweinehund]

The quadrilateral formed by (1, 2), (3, 3), (1, 7), (-1, 3) is ?

John has 10 marbles in a bag, 3 of which are green and the rest are yellow. He randomly draws 1 out of
the bag, notes its colour and then puts it back before drawing another out of the bag. What is the
probability that both marbles drawn are the same colour?

is the second one 0.58?? haha i'm so unsure with probability questions

Hi!

For the quadrilateral question, I would start by drawing a cartesian plane and plotting the points to create the shape. see first image attached
From this I got the impression that the quadrilateral is a kite, but you would look to prove that by showing that the diagonals of the kite meet at a right angle (I'm not 100% sure but I think you can do this by showing that the gradients of the two diagonals, when multiplied together equal -1)

For the probability question, I used a tree diagram to clearly step it out and show everything I needed to include
It's important to remember that this question uses replacement
see the second image attached

Hope this helps!!

[spnmox]

Hi!

For the quadrilateral question, I would start by drawing a cartesian plane and plotting the points to create the shape. see first image attached
From this I got the impression that the quadrilateral is a kite, but you would look to prove that by showing that the diagonals of the kite meet at a right angle (I'm not 100% sure but I think you can do this by showing that the gradients of the two diagonals, when multiplied together equal -1)

For the probability question, I used a tree diagram to clearly step it out and show everything I needed to include
It's important to remember that this question uses replacement
see the second image attached

Hope this helps!!

thank you so much!!

[spnmox]

 A game of poker uses a deck of 52 cards
with 4 suits (hearts, diamonds, spades
and clubs). Each suit has 13 cards,
consisting of an ace, cards numbered
from 2 to 10, a jack, queen and king.
If a person is dealt 5 cards find the
probability of getting a flush (all cards the same suit)

Sorry for weird formatting!

A squad of 8 is chosen at random from 3 baseball teams with 10 players in each team. Find the probability that Joe from the B team and Dan from the A team will be chosen.

I saw that the answer is 8/30 * 7/29. Isn't that the probability of choosing any 2 people - should we specify that it is Joe and Dan, or does it not matter?

[fun_jirachi]

Hey there!

For the first question, first note that the total number of ways of getting a 5 card hand from 52 cards is equal to (52x51x50x49x48)/5!. The numerator should make sense (tell me if it doesn't!) and dividing by 5! removes order ie. picking an Ace, 2, 3, 4, 5 of diamonds is the same as picking a 2, Ace, 3, 4, 5 of diamonds in that order. Getting a flush means you're limiting yourself to a certain suit, and then picking 5 cards in a similar way; ie. the number of ways will be (13x12x11x10x9)/5!, for much the same reasons as the total number of ways. Since there are four suits, we multiply this value by four, and put it all over the total number of ways



For the second question, that's not the probability of choosing any two people; rather, it's the probability of choosing any two PARTICULAR people. What this means is that we pick the people first, then choose the team, not choose the team and check if those people are there. By choosing Joe and Dan first, then finding the probability they're in the team, we simply make sure they're picked ie. 8/30 x 7/29, then disregard the rest of the team, since it doesn't matter who the rest of the team is, as long as Joe and Dan are in it. Essentially, the 8/30 x 7/29 have to be Joe and Dan.

Hope this helps

[spnmox]

Hey there!

For the first question, first note that the total number of ways of getting a 5 card hand from 52 cards is equal to (52x51x50x49x48)/5!. The numerator should make sense (tell me if it doesn't!) and dividing by 5! removes order ie. picking an Ace, 2, 3, 4, 5 of diamonds is the same as picking a 2, Ace, 3, 4, 5 of diamonds in that order. Getting a flush means you're limiting yourself to a certain suit, and then picking 5 cards in a similar way; ie. the number of ways will be (13x12x11x10x9)/5!, for much the same reasons as the total number of ways. Since there are four suits, we multiply this value by four, and put it all over the total number of ways



For the second question, that's not the probability of choosing any two people; rather, it's the probability of choosing any two PARTICULAR people. What this means is that we pick the people first, then choose the team, not choose the team and check if those people are there. By choosing Joe and Dan first, then finding the probability they're in the team, we simply make sure they're picked ie. 8/30 x 7/29, then disregard the rest of the team, since it doesn't matter who the rest of the team is, as long as Joe and Dan are in it. Essentially, the 8/30 x 7/29 have to be Joe and Dan.

Hope this helps

wait, so is 8/30 * 7/29 picking the other people after assuming that joe and dan are already picked?

[Grace0702]

A particle moves in a straight line so that its displacement, in meters, is given by
 Show that
Hence find expressions for the velocity and the acceleration in terms of t

[fun_jirachi]

wait, so is 8/30 * 7/29 picking the other people after assuming that joe and dan are already picked?

No, this is the probability that Dan and Joe are picked. This probability ensures they are picked in the group; note that the probability of picking the other 6 people is just 1, since we don't care who they are. I mustn't have made it clear, thanks for bringing this up!

A particle moves in a straight line so that its displacement, in meters, is given by
 Show that
Hence find expressions for the velocity and the acceleration in terms of t

Hey there!
This question is just asking you to differentiate the equation in a pretty convoluted way


Hope this helps

[spnmox]

hey guys, for the attached for ii), in the solution I don't understand why Pat winning on the second attempt involves P(LLW) -  shouldn't there be only one loss, which happens to Chandra, and the next one is just a win by Pat ie P(LW)?

[fun_jirachi]

There have to be two losses; consider what happens as the game is played out.

Pat goes first. If Pat wins, the game ends. Pat wins on the first throw.
If he loses, it's now Chandra's turn. If Chandra wins, the game ends. If Chandra loses, it's Pat's turn again, and he has the chance to win on his second throw.
Note that Pat must lose the first round for it to reach any further round. In general, each player must lose in order for the game to progress to the next round. Hence, we can't just disregard the result of Pat's first go; if he wins, the game is already won on the first round, and he only gets to win the second round if and only if both Pat and Chandra lose on their first throws.

[spnmox]

There have to be two losses; consider what happens as the game is played out.

Pat goes first. If Pat wins, the game ends. Pat wins on the first throw.
If he loses, it's now Chandra's turn. If Chandra wins, the game ends. If Chandra loses, it's Pat's turn again, and he has the chance to win on his second throw.
Note that Pat must lose the first round for it to reach any further round. In general, each player must lose in order for the game to progress to the next round. Hence, we can't just disregard the result of Pat's first go; if he wins, the game is already won on the first round, and he only gets to win the second round if and only if both Pat and Chandra lose on their first throws.

oh, haha yeah i completely forgot about pat losing first.thanks!!!

[Hawraa]

Hi guys,
I need some help with a past hsc question. It's from 2008 (Q10-b-ii) for math advance. My problem is with the differentiation part, when we have to differentiate the area A. I looked through the answers and the marking centre feedback but I still don't get it.
If S, L and Sin are constants, why they are not cancelled in the derivative?
And why they cancelled one L inside the brackets but left the other one squared?
I hope my question makes sense. If anyone can help me with this, it would be much appreciated. Thanks.

[RuiAce]

Hi guys,
I need some help with a past hsc question. It's from 2008 (Q10-b-ii) for math advance. My problem is with the differentiation part, when we have to differentiate the area A. I looked through the answers and the marking centre feedback but I still don't get it.
If S, L and Sin are constants, why they are not cancelled in the derivative?
And why they cancelled one L inside the brackets but left the other one squared?
I hope my question makes sense. If anyone can help me with this, it would be much appreciated. Thanks.

I'm afraid I'm not 100% certain about the issue, but it sounds like the differentiation itself is what you're asking about?
\[ \text{As }s \text{ and }\sin \alpha\text{ are constants, we can move them outside the derivative}\\ \text{using the rule }\frac{d}{dx}(cf(x)) = c\frac{d}{dx}f(x) \]
\begin{align*}
\frac{dA}{dx} &= \frac{d}{dx} \left[s\left(x-\ell +\frac{\ell^2}{2x} \right)\sin\alpha\right]\\
&= s\sin \alpha \times \frac{d}{dx}\left(x-\ell + \frac{\ell^2}{2x} \right)
\end{align*}
\[ \text{Now recall that the derivative is with respect to }x.\\ \text{As always, when there's a sum of terms, we differentiate term by term.}\\ \text{Firstly, }\frac{d}{dx} x = 1. \]
\[ \text{Then, }\frac{d}{dx}\ell = 0\text{, because }\ell\text{ is a constant.}\\ \text{So differentiating }\ell\text{ has the same effect as differentiating a number, like }2. \]
\[ \text{With the last term, the }\frac{\ell^2}{2}\text{ bit is also a constant.}\\ \text{So we move it out in front.}\\ \frac{d}{dx} \left( \frac{\ell^2}{2x} \right) = \frac{\ell^2}{2} \times \frac{d}{dx} \frac1x \]
\[ \text{And since }\frac1x\text{ is all we're left with, we just use the power rule to deduce that}\\ \frac{d}{dx} \frac1x = -\frac1{x^2}. \]
Which gives the final answer \( \frac{dA}{dx} = s\sin \alpha \times \left(1-\frac{\ell^2}{2x^2} \right) \)

Make sure you understand that the derivative is with respect to \(x\). Not with respect to something else like \(\ell\). The \(\ell\) inside was not cancelled because of the fact it was an \(\ell\), but rather because it was a constant when as opposed to \(x\) and hence its derivative is zero. Whereas that \( \frac{\ell^2}{2x}\) term actually has an \(x\) attached to it, and hence requires use of laws of derivatives.

[Hawraa]

And one more question please. This is from the (ATAR Notes topic tests HSC mathematics Edition 1 2017-2019. Page 35 Q9.) I think they accidentally forgot to put the answer for this one. The question is :

A strip of wire 96 cm in length is used to build a square prism. Supposing that the length of the square side is x cm, show that the surface area of the square prism is given by  S= 6x(16 - x). Hence, find the volume of the prism with the maximum surface area.

Thanks guys.

[Hawraa]

I'm afraid I'm not 100% certain about the issue, but it sounds like the differentiation itself is what you're asking about?
\[ \text{As }s \text{ and }\sin \alpha\text{ are constants, we can move them outside the derivative}\\ \text{using the rule }\frac{d}{dx}(cf(x)) = c\frac{d}{dx}f(x) \]
\begin{align*}
\frac{dA}{dx} &= \frac{d}{dx} \left[s\left(x-\ell +\frac{\ell^2}{2x} \right)\sin\alpha\right]\\
&= s\sin \alpha \times \frac{d}{dx}\left(x-\ell + \frac{\ell^2}{2x} \right)
\end{align*}
\[ \text{Now recall that the derivative is with respect to }x.\\ \text{As always, when there's a sum of terms, we differentiate term by term.}\\ \text{Firstly, }\frac{d}{dx} x = 1. \]
\[ \text{Then, }\frac{d}{dx}\ell = 0\text{, because }\ell\text{ is a constant.}\\ \text{So differentiating }\ell\text{ has the same effect as differentiating a number, like }2. \]
\[ \text{With the last term, the }\frac{\ell^2}{2}\text{ bit is also a constant.}\\ \text{So we move it out in front.}\\ \frac{d}{dx} \left( \frac{\ell^2}{2x} \right) = \frac{\ell^2}{2} \times \frac{d}{dx} \frac1x \]
\[ \text{And since }\frac1x\text{ is all we're left with, we just use the power rule to deduce that}\\ \frac{d}{dx} \frac1x = -\frac1{x^2}. \]
Which gives the final answer \( \frac{dA}{dx} = s\sin \alpha \times \left(1-\frac{\ell^2}{2x^2} \right) \)

Make sure you understand that the derivative is with respect to \(x\). Not with respect to something else like \(\ell\). The \(\ell\) inside was not cancelled because of the fact it was an \(\ell\), but rather because it was a constant when as opposed to \(x\) and hence its derivative is zero. Whereas that \( \frac{\ell^2}{2x}\) term actually has an \(x\) attached to it, and hence requires use of laws of derivatives.

Thanks so much. This actually makes sense now.  ☺️🌷