VCE Specialist 3/4 Question Thread!

[dm9195]

This is definitely a method that VCAA will accept and is the first method I used. The other method I used was just calculating the mean and sd for 5X. So mean= 5x200, variance= 5^2x10^2. But this doesn't work, it only works if you don't do 5^2 in the variance formula but the formula says that you should so I'm confused.

Yeah I see what you're saying... That is strange. On reflection, 0.3023 SEEMS to be more likely (and I'm almost certain that you have to square the variance coefficient so there's no getting around that). What is wrong with that initial calculation of 0.1318 then? Maybe it has something to do with the low sample size contributing to a lack of true normality? I'm not sure. Do you have solutions for that particular question through which we may gain some insight?

[greenironbat7]

Yeah I see what you're saying... That is strange. On reflection, 0.3023 SEEMS to be more likely (and I'm almost certain that you have to square the variance coefficient so there's no getting around that). What is wrong with that initial calculation of 0.1318 then? Maybe it has something to do with the low sample size contributing to a lack of true normality? I'm not sure. Do you have solutions for that particular question through which we may gain some insight?

Each tangerine is independent of the one before. So you use Var(x1+x2....+x5)=5Var(x)
It's not T=5X, where T is total mass. It's T=X1+X2....+X5

[dm9195]

Each tangerine is independent of the one before. So you use Var(x1+x2....+x5)=5Var(x)
It's not T=5X, where T is total mass. It's T=X1+X2....+X5

Ahh that makes much more sense! Thanks!

[guest123]

Each tangerine is independent of the one before. So you use Var(x1+x2....+x5)=5Var(x)
It's not T=5X, where T is total mass. It's T=X1+X2....+X5

When would you use 5X? Thanks

[keltingmeith]

I used a similar approach to the one outlined in this Khan Academy vid (https://www.khanacademy.org/math/statistics-probability/sampling-distributions-library/sample-means/v/sampling-distribution-example-problem) and got 0.1318 by estimating the sampling distribution of the sample mean (ie. pr(175<X<195)| X~N(200,sqrt(100/5))). I'm not too sure if that's what VCAA wants us to do (I doubt it because I haven't seen it in any text book) but that number sorta seems intuitively right to me. Let me know what you think

Fun fact: the method Khan academy uses is identical to finding var(X1+X2+...+X5).

When would you use 5X? Thanks

Only when each following event is identical to the first. Example - you flip a coin, 1 for heads, 0 for tails. Then, you flip 4 more coins, and each coin gives the exact same result as the first, as opposed to being random themselves.

A more realistic example: you toss a coin - if you land heads, you win $200, if you land tails, you win nothing. In this case, the random variable X~Bi(1,0.5) describes the coin toss, and Y=200X describes the amount of money you win.

If you're having multiple, independent, events, ALWAYS use the sum formula.

[guest123]

Fun fact: the method Khan academy uses is identical to finding var(X1+X2+...+X5).

Only when each following event is identical to the first. Example - you flip a coin, 1 for heads, 0 for tails. Then, you flip 4 more coins, and each coin gives the exact same result as the first, as opposed to being random themselves.

A more realistic example: you toss a coin - if you land heads, you win $200, if you land tails, you win nothing. In this case, the random variable X~Bi(1,0.5) describes the coin toss, and Y=200X describes the amount of money you win.

If you're having multiple, independent, events, ALWAYS use the sum formula.

You also get the correct answer by using a sample mean. So I think this is the safer method

[YellowTongue]

Okay, but what process do I use to calculate the time?

Bump

Original question:

If given the equation for the mass of a  substance in terms of time, how do you find the time that the concentration will be a particular value?

E.g. If x=(t-20)^3/800, find how long it takes for the concentration of the solution to reach 0.2 grams per litre. There is initially 10 grams of the substancein 40 litres of water.

[haxor4chan]

How many marks do you have to get wrong in both exam 1 and 2 to get a 40 or above mark? Thankss 

[lzxnl]

Bump

Original question:

If given the equation for the mass of a  substance in terms of time, how do you find the time that the concentration will be a particular value?

E.g. If x=(t-20)^3/800, find how long it takes for the concentration of the solution to reach 0.2 grams per litre. There is initially 10 grams of the substancein 40 litres of water.

Think about this logically. The volume isn't changing, right? So, if you want 0.2 g/L, and you have 40 L, that's 8 g in total. So x = 8. Now solve for t.

How many marks do you have to get wrong in both exam 1 and 2 to get a 40 or above mark? Thankss 

Seriously depends on the year. Look up past years' assessment reports. You have to get, on average, above the A+ cutoff for all graded assessments (of course, you can smash exam 2 and mess up exam 1, but I mean on average). On exam 2, that's normally 20% of the marks that you can lose (in my time at least; dunno if they've changed that as it's been 4 years).

[Gogo14]

For reciprocal functions where you have y=f(x)/g(x) fully simplified. How can there be a horizontal asympote at y=0 and have an x intercept at the same time?
According to maths quest

[keltingmeith]

For reciprocal functions where you have y=f(x)/g(x) fully simplified. How can there be a horizontal asympote at y=0 and have an x intercept at the same time?
According to maths quest

This is an interesting one - you see, asymptotes only hold as you go to infinity. Consider the graph of y=x*e^(-x). This OBVIOUSLY has an asymptote at y=0 (just graph it if you don't believe me!), however, y=0 <===> x*e^(-x)=0, which is obviously true for x=0. SO, we have an x-intercept at x=0, as well as an asymptote at y=0. In fact, y=(x-a)e^(-x) ALWAYS has an asymptote at y=0, but x-intercept at x=a. You should try graphing these on your calculator so you can see for yourself:

y=(x-1)*e^(-x)
y=(x-2)*e^(-x)
y=x*e^(-x)
y=(x+1)*e^(-x)
y=(x+2)*e^(-x)

[Buddster]

For reciprocal functions where you have y=f(x)/g(x) fully simplified. How can there be a horizontal asympote at y=0 and have an x intercept at the same time?
According to maths quest

y can be equal to zero if f(x)=0, and g(x)=/=0
for example, see y=x/(x^2 + 1)

edit:

This is an interesting one - you see, asymptotes only hold as you go to infinity. Consider the graph of y=x*e^(-x). This OBVIOUSLY has an asymptote at y=0 (just graph it if you don't believe me!), however, y=0 <===> x*e^(-x)=0

ooooh that is a sexy example

[Gogo14]

But I always though that an asymptote is a line which the function reaches but never touches......or is my understanding wrong?

[keltingmeith]

But I always though that an asymptote is a line which the function reaches but never touches......or is my understanding wrong?

VCE tends to teach a pretty flawed understanding of asymptotes, tbh. An asymptote's behaviour actually depends on if it's horizontal/oblique, or vertical.

In the case of a vertical asymptote, the function will head towards plus/minus infinity as x approaches the value of that asymptote.

Otherwise, an asymptote is the curve that the function will approach (but never reach/touch) as x goes to positive or negative infinity. The important part being you APPROACH the curve of the asymptote. Let's say that our asymptote is g(x) - this means that as x->infinity, f(x)=g(x) AS WELL AS f'(x)=g'(x). Importantly, in the case of non-vertical asymptotes, you can ALWAYS hit a point on the asymptote, your approaching behaviour is ONLY true in limiting cases.

Another cool example: f(x)=sin(x)/x. In this case, the function actually CONSTANTLY hits the line y=0 (or g(x)=0, using the notation from earlier). HOWEVER, as x->infinity (remember - it's only this limiting case that's important!!), f(x)->0=g(x), AND f'(x)->0=g'(x).

[Sine]

VCE tends to teach a pretty flawed understanding of asymptotes, tbh. An asymptote's behaviour actually depends on if it's horizontal/oblique, or vertical.

In the case of a vertical asymptote, the function will head towards plus/minus infinity as x approaches the value of that asymptote.

Otherwise, an asymptote is the curve that the function will approach (but never reach/touch) as x goes to positive or negative infinity. The important part being you APPROACH the curve of the asymptote. Let's say that our asymptote is g(x) - this means that as x->infinity, f(x)=g(x) AS WELL AS f'(x)=g'(x). Importantly, in the case of non-vertical asymptotes, you can ALWAYS hit a point on the asymptote, your approaching behaviour is ONLY true in limiting cases.

Another cool example: f(x)=sin(x)/x. In this case, the function actually CONSTANTLY hits the line y=0 (or g(x)=0, using the notation from earlier). HOWEVER, as x->infinity (remember - it's only this limiting case that's important!!), f(x)->0=g(x), AND f'(x)->0=g'(x).

hmmmm is this an VCE explanation or the (real) explanation?

Asymptotes are allowed to cut the graph and you are allowed to draw it in cutting through in VCE exams.

Asymptotes : The curve approaches,not touching not curving away.

To find vertical asymptotes solve the denominator to  = 0* or find logs and equate the terms inside to 0. Horizontal asymptotes

Horizontal/oblique asymptotes let x approach positive or neg infinity.

Note*: Be careful as sometimes percieved asymptotes are really just a point of discontinuity if factors in the numerator and denominator can cancel out.