VCE Methods Question Thread!

[cosine]

Help please:


Question 1:
The height of a tide, h meters, at the entrance to a port is given by where t is the number of hours after midnight and a is a positive integer. The number of times the height is 10m within the first day is?

A. 3
B. 6
C. 6/a
D. 3a
E. 6a

For this one, I just don't know what to do... The value of  a can drastically change the amount of times it reaches 10 m.

Question 2:
Two dice are rolled. The probability distribution for X, the difference between the numbers on the dice, is given by:

x:              0       1          2        3     4     5
Pr(X=x)  6/36  10/36   8/36     a      b   2/36

The probability that X is less than 2 given that X is less than 4 is:

A. 8/18
B. 8/15
C. 12/17
D. 17/18
E. 2/3

For this one, I cannot work out a or b because I can only formulate one equation. So I used and I know that for X to be less than 2, it is ALWAYS less than 4, so I know have: . But now I have no idea how to work it out.

Many thanks.

[The Mathemagician]

Question 1
- The first step is to know what changing the value of a does to the equation. In this case it changes the period of the function.
- The next step is knowing how the function behaves (how many times does it cross 10 m per period)
- Also determine how many cycles are in 24 hours.

Solution

For any generic sine curve , the period is In this case , so the period is .
The function has an amplitude of 3 and is shifted upwards by 8 so the range is [5,11]. If you draw the curve out you'll notice that it crosses 10 twice within a period.
We now need to know how many cycles occur within 24 hours. Let c be the number of cycles, so now we solve , which gives .
Now for each cycle it crosses 10 m twice, so the number of times it crosses is which is answer E.

Question 2
This should really say that they are 6 sided dice numbered 1-6, but assuming so, one possible solution would be to write out all permutations of the dice and check. (If you really want to, you can verify all the values as an exercise, although in an exam you obviously don't have the time). Then you should be able to calculate the required probability. As you only need Pr(X<4) you only need to calculate the value of a (a difference of 3).

Solution

In this case the permutations that give a difference of 3 are 1,4 ; 2,5 ; 3,6 ; 4,1 ; 5,2 ; 6,3. The total number of permutations for the dice is 36, so the probability of having a difference of 3 is 6/36. Now the probability of the difference being less than 4 is (6+10+8+6)/36 = 30/36, thus the required probability is which gives answer B.

[warya]

I have a number of trials binomial q and do not have the answer

n=? p=3/7 what is the minimum number of matches that must be played to ensure probability of 95% of winning at least one match

I have done this: Pr(X=0)=0.05
nC0*(3/7)^0*(4/7)^n=0.05
(4/7)^n=0.05
n=at least 6 matches
Am I correct? 

[keltingmeith]

I have a number of trials binomial q and do not have the answer

n=? p=3/7 what is the minimum number of matches that must be played to ensure probability of 95% of winning at least one match

I have done this: Pr(X=0)=0.05
nC0=0.05
(4/7)^n=0.05
n=at least 6 matches
Am I correct? 

Your working is a little confusing. I believe instead of what you do have for the second and third line, you meant:

(nC0)*(4/7)^n=0.05

Otherwise, you're on the right track, just one bit short. You actually want to find the value of n such that 1-P(X=0)=0.05, because we want to win at least one match. You've found the value of n such that we lose every match.

[warya]

Your working is a little confusing. I believe instead of what you do have for the second and third line, you meant:

(nC0)*(4/7)^n=0.05

Otherwise, you're on the right track, just one bit short. You actually want to find the value of n such that 1-P(X=0)=0.05, because we want to win at least one match. You've found the value of n such that we lose every match.

So do I solve for n such that Pr(X=0)=0.95? But isn't that the probability that we win at least one match?

[keltingmeith]

So do I solve for n such that Pr(X=0)=0.95? But isn't that the probability that we win at least one match?

No. P(X=0) is the probability that there are no successes (that is, every match is lost).

[keltingmeith]

I get that I need to find n where x is greater than 1=0.95 but when I rearrange 1-P(x=0)=0.05 I get P(X=0)=0.95 so I'm confused about what equation I have to use

Can I have another hint?

Oh shit, my bad. I misread the question, I thought you were looking for the probability that you win at least one match with probability 0.05.

Don't worry, you were right all along, all good. Just your writing was a little unclear, as I said at first.

[warya]

Haha thats okay thanks heaps!

[HopefulLawStudent]

What do they mean when they say to use a "continuous function"?

[The Mathemagician]

The manufacturer cannot (in reality) manufacture something like 0.5, 0.23 or 1/666 of a unit. So the domain should really just be the set of non-negative integers. and the resulting function would be a set of dots when plotted on a graph. Obviously, this is not differentiable as this function isn't continuous anywhere.

However for the purposes of simplicity (and the fact that the question probably wants you to use calculus), you can assume that it is somehow possible to manufacture something like 0.1 of a unit so the function is defined for all x greater than 0 (i.e the domain is from 0 to infinity).

The question is asking for average cost through so you can't simply find the minimum of the function stated in the question. You will need to find an equation that describes the average cost of a unit and then find the minimum of that. I would also round off to the nearest integer at the end.

If you haven't done the probability section yet, once you've learned about discrete random variables and continuous random variables it should be clearer what the difference is.

[cosine]

How can I do this question? Is this on the study design? Because I have never witnessed such a conceptual question. Thanks

[nerdgasm]

There are two ways that I think this question could be approached: geometric or algebraic.

For the geometric method, sketch the line 2x + y - 10 = 0 on the Cartesian axes. Call this line 'Line A'. You want the point on this line closest to the origin. Now, imagine drawing a line from the origin that is perpendicular to Line A. This line will hit a point on Line A. Let's call that point 'Point A'. It turns out that Point A is in fact the point you want! Why? Let's think in terms of parallel and perpendicular components, and Pythagoras' Theorem (or vector projections if you do Specialist Maths). Point A has some 'perpendicular distance' from the origin. This perpendicular distance is of course the distance of Point A from the origin. Every other point on Line A has both 'perpendicular distance' and 'parallel distance'. In other words, if you choose any other point on Line A, you could draw a right angled triangle using the origin, Point A, and your chosen point. Then, by Pythagoras' Theorem, we have (distance from origin to chosen point)^2 = (distance from origin to Point A)^2 + (distance from Point A to chosen point)^2. It logically follows that the distance from the origin to the chosen point is larger than the distance from the origin to point A. Hence, point A is the closest point on the line to the origin.

Therefore, all you need to do is to work out what the equation of the line from the origin perpendicular to Line A is. (Hint: use the gradient of Line A and the fact that the line must pass through the origin). You can then find the point where the two lines intersect; this is Point A. Once you know the coordinates of Point A, you can easily work out the distance.

For the algebraic approach, consider that the distance of any point (x, y) from the origin is given by sqrt(x^2 + y^2). Therefore, we may restate the question as: Find the minimum value of sqrt(x^2 + y^2) given the condition 2x + y - 10 = 0. If we rearrange the original equation for the line in terms of y and square it, we get an equation for y^2 that is a quadratic in x. Adding x^2 to this equation will give us an equation for y^2 + x^2 that is (a different) quadratic in x. Because the minimum value of sqrt(x^2 + y^2) will also be the minimum value of x^2 + y^2 (subject to our initial condition), then all we have to do is find the minimum value of this quadratic. By using calculus or completing the square, you should be able to find the minimum x-value. Then all you have to do is substitute this back into the original equation to get the y-value, and then you can work out the distance.

As for whether this question is on the study design, I believe it has been placed on at least one VCAA exam (and I remember the guy from my school's VCE revision lectures going through this style of question). Whether it is applicable to a student completing Maths Methods in 2015, I do not know.

[cosine]

Thank you Nerdgasm, still a bit unclear though

Can someone help with the attached question.

[keltingmeith]

Thank you Nerdgasm, still a bit unclear though

Can someone help with the attached question.

Sometimes, you can think of probabilities as a proportion. For example, if a particular coin has a probability 0.666 of getting a heads, then over a long period of time, we expect to see 66.6% of our coins come up as heads.

[Floatzel98]

I need help finding the constant k (attached)