VCE Methods Question Thread!

[Splash8]

Not sure how to work this out, any help would be greatly appreciated

In an industrial process, a large trough is being filled with recycled water at a constant rate of 500 litres per minute. The trough is 6.0m long and the ends are isosceles trapeziums, with parallel sides of 2.0m and 3.0m, and a perpendicular height of 1.5m

b)  Suppose that t minutes after the trough starts filling, the depth of water is h metres. Find, in terms of h, the volume of water in the trough at time t minutes.


I know that the chain rule need to be used to develop a rule, I am just not sure how to find the rules I need. (If that makes sense  )

Thanks

[Splash8]

, d= depth of water in metres at a port entrance, t is in hours.

A certain ship needs the depth at the port entrance to be more than 5m. The ship can be loaded and unloaded, and in and out of the port in 9 hours. Assumng that the ship enters the port just as the depth at the entrance passes 5 metres, will the ship be able to exit 9 hours later? How long will it have to spare, or by how many minutes will it miss out?

I've managed to find whether it will be able to exit 9 hours later, but I'm not sure how to figure out the amount of time to spare it has.

Thanks :>

I will try to attach my working out for this question, but I'll try to explain it.
 Firstly I let h (or d(t)) equal 5m, and then I solved the equation for t to find the time at which the depth was greater than 5m. Then I drew a graph as a visual aid and calculated the period. Next, I subtracted t from half the value of the period to find when the depth of the water fell below 5m. Then I converted these two values of t to hours and minutes to find the time. Then I worked out when the boat would have to leave if it was 9 hours after the depth rose above 5m. Then I calculated the spare time by subtracting this 9hour time from the time at which the depth fell below 5m.

I hope this helps

[Splash8]

Hopefully this attachment is readable

[09Ti08]

Hi!
If an exam question defines a function with a restricted domain such as:



Do transformations (i.e. dilations) always affect a restricted domain?

eg.

Spoiler

If
What is the amplitude and domain of this new function?

Short answer: yes
Long answer:
Amplitude=2/3, now if you want to find the domain, just pretend that you have X=1/2*(x-pi/2), then the problem becomes finding the domain of f(X). What do you want the domain of X to be? You want it to be exactly the same as that of x, any other input of X outside the domain will be invalid. I hope this doesn't confuse you. Now you just need to solve two inequalities which satisfy: 0<X<2*pi

Not sure how to work this out, any help would be greatly appreciated

In an industrial process, a large trough is being filled with recycled water at a constant rate of 500 litres per minute. The trough is 6.0m long and the ends are isosceles trapeziums, with parallel sides of 2.0m and 3.0m, and a perpendicular height of 1.5m

b)  Suppose that t minutes after the trough starts filling, the depth of water is h metres. Find, in terms of h, the volume of water in the trough at time t minutes.


I know that the chain rule need to be used to develop a rule, I am just not sure how to find the rules I need. (If that makes sense  )

Thanks

Umm, do you have the answer there? I think we can do this using basic trigonometry, even though the answer is a bit messy. This is how I think about it: you draw a cross section of the trough which looks like a trapezium, draw a vertical height. Then you work out the angle between that line and one side of the trapezium. Pick an arbitrary value for h now, you can work out the area of that cross section and finally multiply it by the length.

[joey7]

Hey guys quick question,

A few weeks ago I had a sac and in one of the questions we had to draw a graph of a restricted domain (2,3) on a provided axes there was no zero marked on the origin so I labelled where the y axis is as 2 and drew the graph. I though this would be correct as in further maths, a trick they like to pull is start the y axis at some random number and then ask for the y intercept which many students mistake for the intercept pictured but I was marked incorrect. Are you able to draw graphs this way or was I actually wrong?

[BasicAcid]

Hey guys quick question,

A few weeks ago I had a sac and in one of the questions we had to draw a graph of a restricted domain (2,3) on a provided axes there was no zero marked on the origin so I labelled where the y axis is as 2 and drew the graph. I though this would be correct as in further maths, a trick they like to pull is start the y axis at some random number and then ask for the y intercept which many students mistake for the intercept pictured but I was marked incorrect. Are you able to draw graphs this way or was I actually wrong?

Well I'm not sure if it's "actually" wrong but I'd mark you wrong.

Usually in these questions (drawing draphs with restricted domains) marks are allocated for intercepts, shape and endpoints.
I'm pretty sure you're supposed to draw the entire axis then restrict the graph by labelling the endpoints with the correct coordinates and circle (coloured in or empty).

[Eugenet17]

I will try to attach my working out for this question, but I'll try to explain it.
 Firstly I let h (or d(t)) equal 5m, and then I solved the equation for t to find the time at which the depth was greater than 5m. Then I drew a graph as a visual aid and calculated the period. Next, I subtracted t from half the value of the period to find when the depth of the water fell below 5m. Then I converted these two values of t to hours and minutes to find the time. Then I worked out when the boat would have to leave if it was 9 hours after the depth rose above 5m. Then I calculated the spare time by subtracting this 9hour time from the time at which the depth fell below 5m.

I hope this helps

thanks buddy!

[fleet street]

Hi everyone,

When taking the mean and standard deviation of a discrete probability distribution, do they have to be rounded to the nearest value in the distribution? I thought they didn't, but after this multiple choice question, I'm not sure. I'd just like to have clarification on this.

Thanks!

[abcdqd]

yes, it's a discrete distribution, so the x-values can only be 2,3,4,5,6,7 or 8

[chemdeath]

Need help with question 2) and therefore the rest as they kinda lead on from there. I just don't know how to approach these types of questions without getting mixed up with formulas. Any help will be appreciated 

[Zealous]

Need help with question 2) and therefore the rest as they kinda lead on from there. I just don't know how to approach these types of questions without getting mixed up with formulas. Any help will be appreciated 

I would start by listing the different ways that will satisfy the criteria of the question.

a: So far part a, they've specified that the first roll has to be a six, and the second roll is not a six.
So this will be:
, the probability of getting a six, then immediately after not getting a six.

b: For part b, there are now two ways you can roll one six, and not roll a six.
You can roll a 6 and not roll a six on the second roll, or not roll a six and roll a 6. (and generally means multiply, or generally means add)

Probability of rolling a six, then not rolling a six plus the probability of not rolling a six then rolling a six, which satisfies what the question is asking.

c: How many ways, using two dice, can the sum equal 6?
You can have two 3's, a 1 and a 5 or a 4 and a 2. Note, you could have a 5 then a 1, or a 1 then a 5 and thus the probability is doubled.
.

d: Conditional probability. Find the interesection between the two events. If the first is the sum of rolls is six, the second event being one of the rolls is a three, the intersection will be when the sum of a roll is six and one of the rolls is a three (which in this case can only happen once, if both dies are three), then set up conditional probability like usual.

Let me know if i made any errors.. it's kinda late and I'm tired =p

[chemdeath]

Thanks heaps sushi!!!! 

c: How many ways, using two dice, can the sum equal 6?
You can have two 3's, a 1 and a 5 or a 4 and a 2. Note, you could have a 5 then a 1, or a 1 then a 5 and thus the probability is doubled.
.

This question doesn't work the answer is meant to be 0.19, might be me doing it wrong though

And can I please also get help on 3b) and 4b) Thanks

[09Ti08]

Sushi's method is correct and i did get 0.19: 2*0.1*0.1+2*0.2*0.2+0.3*0.3=0.19

Question 3:
A) for the biased die: Pr(X=6)=0.1
For a fair die: Pr(X=6)=1/6
Pr(X=6)=0.1*1/3+2*1/6*1/3=13/90
B) 0.1/(13/90)=9/13

Question 4:
A) i assume that this question has nothing to do with question 3? Then the answer is 0.1
B) when a six is rolled: the probability that Sarah has to pay $(10+x) is 0.1. Otherwise, the probability that she gets $x is 0.9 (x: fee for playing the game)
Then we need to solve: -(10+x)*0.1+0.9*x=0<=>x=$1.25

[chemdeath]

Thanks!!

3
B) 0.1/(13/90)=9/13

This is supposed to be 3/13 according to the answers? 

Question 4:
A) i assume that this question has nothing to do with question 3? Then the answer is 0.1
B) when a six is rolled: the probability that Sarah has to pay $(10+x) is 0.1. Otherwise, the probability that she gets $x is 0.9 (x: fee for playing the game)
Then we need to solve: -(10+x)*0.1+0.9*x=0<=>x=$1.25

And this is based on question 1,2,3. A) is 13/90 and I get that but for b) Im having trouble 

[09Ti08]

Ok, sorry, my bad.

Question 3:
I forgot that the probability of getting a 6 for the biased coin is 0.1*1/3
So the answer is (0.1*1/3)/(13/90)=3/13

Question 4:
Solve: -(10+x)*13/90+(1-13/90)*x=0<=>x=$2.03