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May 11, 2021, 07:48:00 am

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rukayabal

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« Reply #18780 on: September 15, 2020, 09:03:45 pm »
0
That exam question is from the current study design, so yes.
Just adding on to what S_R_K has said - these questions also aren't too difficult, don't be scared just because they're technically in the realm of combinatorics. Remember - this is just a binomial distribution problem, all you need to do is apply the binomial distribution material as you were taught it in class.

Thank you

emonerd

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« Reply #18781 on: September 15, 2020, 09:14:04 pm »
0
For this graph here, would you say it's strictly increasing over the interval (0,1) and (3,5) OR [0,1] and [3,5]? I'm just a bit confused about whether we include cusps and endpoints.

thanks
2019 -Biology
2020- Chem, Methods, Spec, English and Literature

keltingmeith

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« Reply #18782 on: September 15, 2020, 09:24:07 pm »
+3
For this graph here, would you say it's strictly increasing over the interval (0,1) and (3,5) OR [0,1] and [3,5]? I'm just a bit confused about whether we include cusps and endpoints.

thanks

Note that definition of strictly increasing/decreasing has NOTHING to do with the derivative - and as a result, you DO include end-points.

A function is strictly increasing over an interval if for every b>a on that interval, f(b)>f(a). Let's consider the point x=1 - in this case, b=1, and f(b)=1. Pick whatever you want for a. If a=0.5, f(a)=0.5<f(b). We can include 0.5 in that interval. In fact, no matter how close a gets to x=1 (or the limit as a goes to 1), f(b) is always going to be bigger. "But keltingmeith", you say, "this explanation means it would be an OPEN interval, not a closed one!" And the answer to that is it doesn't make sense to check for when a=b - because of course f(b) is not going to be bigger than f(b), they must be the same. As a result, the mathematical community basically just decided that it's okay to include end-points, and so that's what people do. Kind of frustrating explanation, I'm sure, but it is what it is.

TheEagle

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« Reply #18783 on: September 15, 2020, 09:25:11 pm »
0
For this graph here, would you say it's strictly increasing over the interval (0,1) and (3,5) OR [0,1] and [3,5]? I'm just a bit confused about whether we include cusps and endpoints.

thanks

As a rule of thumb, include endpoints for strictly increasing/decreasing. However, if is is asking for just increasing/decreasing, then do not include the endpoints.

Hence, in this case, include the endpoints, i.e. [0,1] and [3,5].

If you want the proof, I believe Tal from Edrolo has linked a bulletin which was released by VCAA back in 2011 explaining the difference, though, the proof itself isn't really needed.

rukayabal

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« Reply #18784 on: September 15, 2020, 09:26:24 pm »
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Hi,
Does anyone know the correct working out for question 4)aii as I keep getting it wrong.
Thankyou

keltingmeith

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« Reply #18785 on: September 15, 2020, 09:31:21 pm »
0
Hi,
Does anyone know the correct working out for question 4)aii as I keep getting it wrong.
Thankyou

Well, what have you tried?

As a rule of thumb, include endpoints for strictly increasing/decreasing. However, if is is asking for just increasing/decreasing, then do not include the endpoints.

Hence, in this case, include the endpoints, i.e. [0,1] and [3,5].

If you want the proof, I believe Tal from Edrolo has linked a bulletin which was released by VCAA back in 2011 explaining the difference, though, the proof itself isn't really needed.

I can't actually find this article, reckon you could share me? Particularly because the technical, mathematical definition for increasing would indicate you include the endpoints (though this wouldn't be the first time the mathematical community and VCAA disagreed - fucking looking at you, cis(theta))

rukayabal

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« Reply #18786 on: September 15, 2020, 09:42:58 pm »
0
Well, what have you tried?]

sorry here is the working out. I know its very wrong.
« Last Edit: September 15, 2020, 09:45:02 pm by rukayabal »

emonerd

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« Reply #18787 on: September 15, 2020, 09:51:26 pm »
0
Note that definition of strictly increasing/decreasing has NOTHING to do with the derivative - and as a result, you DO include end-points.

A function is strictly increasing over an interval if for every b>a on that interval, f(b)>f(a). Let's consider the point x=1 - in this case, b=1, and f(b)=1. Pick whatever you want for a. If a=0.5, f(a)=0.5<f(b). We can include 0.5 in that interval. In fact, no matter how close a gets to x=1 (or the limit as a goes to 1), f(b) is always going to be bigger. "But keltingmeith", you say, "this explanation means it would be an OPEN interval, not a closed one!" And the answer to that is it doesn't make sense to check for when a=b - because of course f(b) is not going to be bigger than f(b), they must be the same. As a result, the mathematical community basically just decided that it's okay to include end-points, and so that's what people do. Kind of frustrating explanation, I'm sure, but it is what it is.
As a rule of thumb, include endpoints for strictly increasing/decreasing. However, if is is asking for just increasing/decreasing, then do not include the endpoints.

Hence, in this case, include the endpoints, i.e. [0,1] and [3,5].

If you want the proof, I believe Tal from Edrolo has linked a bulletin which was released by VCAA back in 2011 explaining the difference, though, the proof itself isn't really needed.

Awesome, thank you so much guys!!! )
2019 -Biology
2020- Chem, Methods, Spec, English and Literature

keltingmeith

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« Reply #18788 on: September 15, 2020, 09:55:10 pm »
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sorry here is the working out. I know its very wrong.

Okay, so what you're doing here is noticing the word "given" is likely indicating conditional probability and blindly applying some probabilities that don't make sense to apply. You need to think about what's going on here:

Let's go step-by-step. Firstly, if it's conditional probability, let's write out the full equation:

$\mathbb{P}(X|Y)=\frac{\mathbb{P}(X\cap Y)}{\mathbb{P}(Y)}$

Next, we need to figure out what each event is. So, Y must be the event that the first draw is a red tomato. X would be the event that the second draw is a tomato and the third draw is a green tomato. Finally, the intersection would be that he draws in the order of red, red, green.

Do you think you might be able to combine these two pieces of information to come to the right answer?

TheEagle

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« Reply #18789 on: September 15, 2020, 09:59:42 pm »
-1
Hi,
Does anyone know the correct working out for question 4)aii as I keep getting it wrong.
Thankyou

For part ii, they've already told you that he will definitely select a red tomato in the first grab, so you're only working out the probability of him selecting two green tomatoes in the next two selections. Since there are 3 green tomatoes and 12 tomatoes in total (he already has taken one out, namely, a red one), the probability of selecting a green one now is 3/12. When he selects again, the probability will be 2/11.

i.e. 3/12 * 2/11 = 6/132 = 1/22

TheEagle

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« Reply #18790 on: September 15, 2020, 10:08:06 pm »
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Okay, so what you're doing here is noticing the word "given" is likely indicating conditional probability and blindly applying some probabilities that don't make sense to apply. You need to think about what's going on here:

Let's go step-by-step. Firstly, if it's conditional probability, let's write out the full equation:

$\mathbb{P}(X|Y)=\frac{\mathbb{P}(X\cap Y)}{\mathbb{P}(Y)}$

Finally, the intersection would be that he draws in the order of red, red, green.

Do you think you might be able to combine these two pieces of information to come to the right answer?

Why red, red, green? It's asking for 2 green tomatoes after the first selection of the red tomato.

keltingmeith

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« Reply #18791 on: September 15, 2020, 10:11:53 pm »
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Why red, red, green? It's asking for 2 green tomatoes after the first selection of the red tomato.

Oops, should be red, green, green - simple typo

rukayabal

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« Reply #18792 on: September 16, 2020, 09:47:36 am »
0

For part ii, they've already told you that he will definitely select a red tomato in the first grab, so you're only working out the probability of him selecting two green tomatoes in the next two selections. Since there are 3 green tomatoes and 12 tomatoes in total (he already has taken one out, namely, a red one), the probability of selecting a green one now is 3/12. When he selects again, the probability will be 2/11
i.e. 3/12 * 2/11 = 6/132 = 1/22

Okay, so what you're doing here is noticing the word "given" is likely indicating conditional probability and blindly applying some probabilities that don't make sense to apply. You need to think about what's going on here:

Let's go step-by-step. Firstly, if it's conditional probability, let's write out the full equation:

$\mathbb{P}(X|Y)=\frac{\mathbb{P}(X\cap Y)}{\mathbb{P}(Y)}$

Next, we need to figure out what each event is. So, Y must be the event that the first draw is a red tomato. X would be the event that the second draw is a tomato and the third draw is a green tomato. Finally, the intersection would be that he draws in the order of red, red, green.

Do you think you might be able to combine these two pieces of information to come to the right answer?

I get it now. Thanks:)

a weaponized ikea chair

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« Reply #18793 on: September 18, 2020, 06:29:30 pm »
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Question c

I do not understand how to solve c and the solution does not give any explanation. I can figure out that for x<2, f'(x)>0, but I do not know how to work out the next bit.
« Last Edit: September 18, 2020, 06:32:34 pm by a weaponized ikea chair »

keltingmeith

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