VCE Methods Question Thread!

[shadows]

Insight 2010 exam 2....
What is fewest number of spins required in order to be 80% sure of getting a 3 at least twice?  ( 3 Marks)

This is what I did (I tried to really harsh and end up giving myself 1 out of 3 for this o.o) Will VCAA examiners be this harsh?

X~Binom (n= unknown, p=1/7)

Pr(X>or equal 2) >0.8

1- (Pr (X=1) + (Pr(x=0))>0.8

6(1/7)*(6/7)^1-n + (6/7)^n >0.2

n=21

(can you directly solve this on a cas for n?, I can not find the way) so i defined the binomcdf, graph it, and get the table of values to the first x value that gives a probability greater than 0.8.

In the marking scheme there's no inequalities...., i realised that they didnt say.. (at least.. or more than 80%)
Will i lose a mark?

apparently the answer is 20. I try it with binomcdf and i get 0.78...  ? Someone please explain?

[rhinwarr]

I think some lines in your working are wrong.

First of all, it says, the number of spins required in order to be 80%, not more than 80%, which means it should be 'exactly' 80% (equal to).

1- (Pr (X=1) + (Pr(x=0))>0.8

This should be 1 - Pr(X=1) - Pr(X=0) = 0.8

6(1/7)*(6/7)^1-n + (6/7)^n >0.2

This should then be

That should give you the correct answer of 20. I don't think you can directly solve it on the CAS (not on t-inspire anyway). I don't know what you mean by table of values?

You might lose marks for the inequalities because technically you haven't defined the problem correctly.

I tried it with binomcdf for 20 and I got 0.8014. Maybe you typed it incorrectly?

[shadows]

ugh maybe i'm not thinking straight.. actually really tired?

binomcdf(20,1/7,2,6) instead of binomcdf (20,1/7,2,20) lol


What I mean by table of values is I create the binomcdf (x,1/7,2,7)
define it.. (eg: f(x)

go and graph f(x)

you won't see anything, but press Ctrl T. opens values. Scroll down to the corresponding x value where probability is first greater than 0.8.

If you don't do it this way, may I ask what method you use?

[rhinwarr]

Oh, I never knew you could do that
How do you know to use 7 when you define f(x)?

I just typed in the equation and solved it for n using solve(...... ,n)

(can you directly solve this on a cas for n?, I can not find the way)

ooohhhh is that what you meant?

[shadows]

oops binomcdf(x,1/7,2,x) not binomcdf (x,1/7,2,7)

what calc are you using?

when I solve for n normally, it  does nothing but restates the equation... can anyone help me out on this?

Because it's a bit of a hassle using the table thing.

[BasicAcid]

I got 20 on the CAS.

binomialCDf(0,1,x,1/7) on main screen
Copy and paste it into graphs and tables
Draw the table
You should see

0 Error
1 1
2 0.9796
3 0.9446
etc...
18 0.2495
19 0.2227
20 0.1986
21 0.1767

We take the first value below 0.2 to ensure this probability (we can't take non whole numbers)
Therefore, n=20

[shadows]

Can vcaa expect us to know what (sin(x))^2 looks like?

Neap 2006 paper . Tech free.

Find the range of 1-(3sin(x)^2 ,

Are we expected to know this?  Unless we had to figure out the period, sub in end points....etc.z.

[teletubbies_95]

Hey guys!
Just to clarify the solutions given in one of the Engage Exam 2 papers.
It was an MCQ :The number of solutions to the equation -3tan (ax) + 5   = k , over the interval [ pi/2 , 3pi/2 ] , with "a" being a subset of Z and k being a subset of R

The correct answer was modulus (a )

The solution said : The period of the tangent function given is modulus ( pi/a) , and it attains all real values once within each period, hence the number of solutions is given by interval width /  period

Is that true for all cases for finding the number of solutions of tan or only this question?

Thanks !
Sorry guys ...Im not sure how to use latex! I'll learn how to do it soon!

[b^3]

Can vcaa expect us to know what (sin(x))^2 looks like?

Neap 2006 paper . Tech free.

Find the range of 1-(3sin(x)^2 ,

Are we expected to know this?  Unless we had to figure out the period, sub in end points....etc.z.

Well yes. Look at it this way. What values can put out? right? Now what happens to what values it can put out if you square it? We cannot get any negative values anymore. The maximum it can still put out is , but we can get anything from up to this. So our range of will be . So will just be a transformation of this.

You could also look at it as a composite function of and as well.

[darklight]

Hey guys,

A bit stuck on Algebra; Tech-Active Test 1, very last question (Extended Response, 3 (c)). I read the solutions and I understood up until the paragraph about deducing the relationship from the graph and noticing its 'curvature.' How do you know that h(x) = (h(x1) + h(x2))/2 when x < (x1 + x2)/2? Can someone please expand on this for me?

To mods: Can I post up the question for those who don't have the book?

[Phy124]

To mods: Can I post up the question for those who don't have the book?

You certainly may. In fact, I encourage it!

We (or possibly just I) have considered implementing a rule whereby the user must post the question in its entirety, whether it be in text or image format, rather than a reference to the said question.

After all, it benefits all those involved - those looking for a question to be answered will have it answered more promptly and those looking to answer questions, who do not have the referenced material, will be able to do so

[shadows]

tips on graphing circular functions? esp when there's annoying transformations (including horizontal shifting)

I always take forever on these questions.

[darklight]

Hey guys,

A bit stuck on Algebra; Tech-Active Test 1, very last question (Extended Response, 3 (c)). I read the solutions and I understood up until the paragraph about deducing the relationship from the graph and noticing its 'curvature.' How do you know that h(x) = (h(x1) + h(x2))/2 when x < (x1 + x2)/2? Can someone please expand on this for me?

To mods: Can I post up the question for those who don't have the book?

Here's the qn (it's part c):

Consider two adjacent x-intercepts of the graph of f(x) = Modulus of sin(ln(2x+5)), (x1,0) and (x2,0). Explain, in words, whether the stationary point between those two x-intercepts is to the left, to the right or the same as the point ((x1+x2)/2), f((x1+x2)/2)!

[revcose]

teletubbies_95: That is true for all cases where interval width / period is a natural number. If it's not a natural number, you round down, and consider whether or not there is a solution in the part of the graph where there wasn't a full period completed.

[~T]

Here's the qn (it's part c):

Consider two adjacent x-intercepts of the graph of f(x) = Modulus of sin(ln(2x+5)), (x1,0) and (x2,0). Explain, in words, whether the stationary point between those two x-intercepts is to the left, to the right or the same as the point ((x1+x2)/2), f((x1+x2)/2)!

If you plot the graph say, from -5/2 to 10, you'll see what the general shape of each peak looks like. It looks as if you had a sin graph and skewed it by slanting it to the left.

(x1+x2)/2 gives us the average of the two x intercepts, in essence, the middle value between them. Thus, with our knowledge that the graph has a peak that is sort of skewed to the left, the peak does not occur in the middle of the two intercepts - it occurs when x < (x1+x2)/2.

I hope that makes sense, it's more a matter of trying to word it well. Perhaps the solutions you were speaking of will make a little more sense now?