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March 29, 2024, 09:15:12 am

Author Topic: VCE Methods Question Thread!  (Read 4802809 times)  Share 

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james.358

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Re: VCE Methods Question Thread!
« Reply #18735 on: August 26, 2020, 10:22:30 am »
+6
Hey rozmaaate,

In this question, you can think of 5x as a whole. Say we let 5x = a to simplify the question, where a > 0.

2a + a < 75
3a < 75
a < 25

Subbing 5x back in, we get

5x < 25
5x < 52
Hence x < 2

Hope this makes it a bit clearer!

James
« Last Edit: August 26, 2020, 10:26:18 am by james.lhr »
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Coolgalbornin03Lo

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Re: VCE Methods Question Thread!
« Reply #18736 on: August 27, 2020, 10:07:59 am »
0
How do you solve exponential decay without sufficient information, for example in the question I have given. I cannot find k so there are too many unknown variables.
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Coolgalbornin03Lo

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Re: VCE Methods Question Thread!
« Reply #18737 on: August 27, 2020, 10:38:45 am »
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To make any decimal into and exact value (i.e 0.87676) do you just divide it by 1000........ the number of zero's depending on how many zero's the number has? My teacher said something along these lines but the SAC is tommrorw and cannot seem to contact him
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Coolgalbornin03Lo

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Re: VCE Methods Question Thread!
« Reply #18738 on: August 27, 2020, 11:40:54 am »
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How do you get exact values on CAS without using the exact function on the CAS. My teacher said not to use exact CAS function as it’s not accurate, but I can’t remember how he said to do it?

For example I got 1.39 as x- intercept so I just put 139/100 for exact. However in the solutions they got loge(4) with is equivalent to 1.39.....am I wrong? 
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keltingmeith

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Re: VCE Methods Question Thread!
« Reply #18739 on: August 27, 2020, 11:56:19 am »
+2
To make any decimal into and exact value (i.e 0.87676) do you just divide it by 1000........ the number of zero's depending on how many zero's the number has? My teacher said something along these lines but the SAC is tommrorw and cannot seem to contact him

No. An "exact value" isn't a fraction, an exact value is the EXACT thing that something is equal to. For example, if you have a 1 metre ruler, then it has an "exact value" length of 1m. If you cut it in half, it has an "exact value" length of 0.5m. You can write it as a half, but that doesn't mean 0.5 is any less exact.

How do you get exact values on CAS without using the exact function on the CAS. My teacher said not to use exact CAS function as it’s not accurate, but I can’t remember how he said to do it?

For example I got 1.39 as x- intercept so I just put 139/100 for exact. However in the solutions they got loge(4) with is equivalent to 1.39.....am I wrong? 

Yes, and this comes to confusion as to what exact means. In this case, the EXACT solution is ln(4). ln(4) is APPROXIMATELY equal to 1.39 - but that's bigger than ln(4) is EXACTLY equal. A better approximation is 1.386, and a BETTER approximation again is 1.3863, and an EVEN BETTER one again is 1.38629 and... you get the idea.

At some point in the method you've used, you've made a numerical approximation. To know where, I'd have to see your method. However, you are wrong, because the idea that fractions are exact values is wrong - just because you've turned something into a fraction, doesn't make it exact.

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Re: VCE Methods Question Thread!
« Reply #18740 on: August 27, 2020, 02:58:18 pm »
+1
For Normal Distribution, if the scenario is to do with a population OR idk time (something where it cant be negative), are we allowed to use minus infinity as the lower limit or do we use 0 in NormCDF? The answers u get are different so which one do we use?
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Re: VCE Methods Question Thread!
« Reply #18741 on: August 27, 2020, 03:12:38 pm »
+1
For Normal Distribution, if the scenario is to do with a population OR idk time (something where it cant be negative), are we allowed to use minus infinity as the lower limit or do we use 0 in NormCDF? The answers u get are different so which one do we use?

You should still be using negative infinity, or otherwise the maths doesn't work. I know it seems counterintuitive, but if the values MUST be positive, you'll often find that the proportion of actual negative numbers in the distribution is basically 0. Although constructed questions from teachers/VCAA not based on real world data or simulated data might prove otherwise lol

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Re: VCE Methods Question Thread!
« Reply #18742 on: August 27, 2020, 03:23:57 pm »
+3
I tried to find a question where we could at least see what VCAA expect. See 2016 Exam 2, question 3c, where the time taken for a laptop battery to run out is normally distributed, and we need to work out the probability that the laptop battery will run out in less than 190 minutes. The battery can not run out in negative time (or does this mean it's charging?) but the examiners report still uses Pr(X ≤ 190), rather than Pr(0 ≤ x ≤ 190). However, they give the same answer correct to 4 decimal places, which is the accuracy required by the question.

Sherlock.Holmes

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Re: VCE Methods Question Thread!
« Reply #18743 on: August 29, 2020, 08:24:50 pm »
0
Hey guys,

From the new study design for methods, do we still have to do chapter 15 from the methods Cambridge textbook?

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Re: VCE Methods Question Thread!
« Reply #18744 on: August 29, 2020, 08:33:10 pm »
0
Hey guys,

From the new study design for methods, do we still have to do chapter 15 from the methods Cambridge textbook?
What's in the 2020 probability curriculum: 13AB, Ch 14, Ch 16, understanding of the importance of shape (positive and negative skew & symmetry), centre (mean = µ = average = long term expected value), spread (std dev = √var = sigma)

So, no.
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Re: VCE Methods Question Thread!
« Reply #18745 on: August 29, 2020, 09:04:49 pm »
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understanding of the importance of shape (positive and negative skew & symmetry)

Everything else you said is essentially right, not sure about this one or where you heard it, though? Skew has never been a part of the study design

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Re: VCE Methods Question Thread!
« Reply #18746 on: August 29, 2020, 09:41:09 pm »
0
Everything else you said is essentially right, not sure about this one or where you heard it, though? Skew has never been a part of the study design
I copied and pasted from what my teacher said. Didn't look it up myself. Maybe it's just saying the importance, not that you need to know how to do it or anything? I don't know.
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Re: VCE Methods Question Thread!
« Reply #18747 on: August 30, 2020, 12:07:15 pm »
+1
Everything else you said is essentially right, not sure about this one or where you heard it, though? Skew has never been a part of the study design
Could possibly be a “learn this but not actually examinable” thing. Pretty sure this is usually a concept covered in Further, but for data in general, rather than probability, so there’s a possibility that the teacher thought it might be helpful but not exam-necessary for Methods or they simply could be confused between subjects if they teach both.
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Azila2004

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Re: VCE Methods Question Thread!
« Reply #18748 on: August 30, 2020, 06:46:33 pm »
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Hello!  ;D

I have a question on circular functions:
Let P(x1, y1) be a point on the circle with equation x^2 + y^2 = a^2. Show that the equation of the tangent at P(x1, y1) is x1x + y1y = a^2

So if we have to find the equation of the tangent line at P, wouldn't we need to just need to find the gradient and P(x1,y1)?
It has been determined that the gradient of the tangent is -x1/y1.

However, it says that the equation of the tangent is x1x + y1y = a^2. I don't really understand how a^2 comes here. Sorry for the weird format  :P Some help would really be appreciated, and please let me know if I may have wrote the question wrong.
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keltingmeith

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Re: VCE Methods Question Thread!
« Reply #18749 on: August 30, 2020, 06:52:47 pm »
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Hello!  ;D

I have a question on circular functions:
Let P(x1, y1) be a point on the circle with equation x^2 + y^2 = a^2. Show that the equation of the tangent at P(x1, y1) is x1x + y1y = a^2

So if we have to find the equation of the tangent line at P, wouldn't we need to just need to find the gradient and P(x1,y1)?
It has been determined that the gradient of the tangent is -x1/y1.

However, it says that the equation of the tangent is x1x + y1y = a^2. I don't really understand how a^2 comes here. Sorry for the weird format  :P Some help would really be appreciated, and please let me know if I may have wrote the question wrong.

Remember that all you have here is the equation of the gradient - you need to put that into the equation of the line that touches the circle at P. Why don't you try that, and come back to us if you're still struggling