VCE Methods Question Thread!

[WhoTookMyUsername]

nSolve(...) should be able to do it

what is "nsolve" (what does it stand for + what is it's function _ when / where to use?)?
also what is "fsolve" ?
thanks

[pi]

I think it means "numerical solve", not sure exactly what it's definition is but it works in cases as on the previous page.

No idea on "fsolve", never used that before (and my CAS is dead so I can't test it) :/

[paulsterio]

use nSolve where you are trying to solve something that cannot be done algebraically.

[WhoTookMyUsername]

thanks
not sure where i got fsolve from... it doesn't exist on CAS lol
after playing around a lilttle bit w, CAS
equation on previous page cannot be solved with nsolve,
but if you replace > with = then it gives the answer of 6.786...
although if you just solve normally (=) it gives both 6.786 and 9.019

so
1) how do you solve the previous equation with CAS XD? is it possible? or do you have to change < / > to = and then "qualitively" / "logically" think about whether it shoudl be > or <?(for probability)
2) does nsolve give only the first positive reasonable answer? the solve function appears to be able to do what the nsolve function can (and more) for the prevous question (and for others?
thanks

[paulsterio]

1) I don't know if that's the ONLY way to do it, but yes, you can do it that way (it would be what I would do in an exam situation)

2) Yes, nSolve uses a method known as numerical bisection, i.e. halving the interval, you can google it if you want to understand more, but yeah, nSolve will just give numerical values.

[HERculina]

On the spreadsheet, how do you enter 1 to 100 in colum A using fill down.
And then after that, how do you use the cumulative binomial distribution function  [can't find this] obtained from the Distribution submenu in column B to fill down until a probability of obtaining 49 or less goals is less than 0.01 is obtained.

Or alternatively, has anyone done question 7 of exercise 15D in essentials book?

[Homer]

(2007 Exam 2, Short answer question 2a)

The concentration of insects in the gorge is a continuous function of time. The concentration C, insects per
square metre, is
(hybrid function)
f(x)=1000(cos(pi(t-8)/2)+2)^2-1000 t=[8,16]                and         M t=[0,8) and (16,24]

where t is the number of hours after midnight and m is a real constant.
a. What is the value of m?

[Hutchoo]

When t = 8, C(t) = 8000
when t = 16, C(t) = 8000.
These are the two end points shown in the upper function (of the hybrid function). For the hybrid function to be continuous (as the question states), the value of m has to be 8000.

[Homer]

Thankyou Hutchoo! I got mixed up between continuous graph and continuous probability distribution, and tried finding the area under the graph. haha

[soccerboi]

Can someone help me with this: A discrete random variable X has a s.d of 2. Then the s.d of Y where Y=4X-3 is what?

[b^3]

Hint: and

...should get .

[barydos]

Sorry to barge, but I've got a probability related question (binomial)


It's part c) which has got me stumped! D:

[Hancock]

Sorry to barge, but I've got a probability related question (binomial)
(Image removed from quote.)

It's part c) which has got me stumped! D:

So, we know that;
p = 0.05
n = 10

So, presuming that you are getting this A and B ok, I'll move to C.

 So, the probability that two batches are tested is if the first one fails. This occurs when Pr(X=2).

Pr(X=2) = nCr(10,2)(0.05)^2(0.95)^8 = 0.07463
Now, we know that if there are any faulty articles in the 2nd batch, it fails. So there will never be a 3rd batch. There is also always 1 batch tested.

So, we know that the Pr(one batch) + Pr(2 batches) = 1, so Pr(one batch) = 0.92537. Drawing up a table, and working out the mean of a discrete distribution, you get;

Now, E(X) = (0.92537)(1) + (0.07463)(2) = 1.07463 batches.

SHIT, articles, not batches. Damn it. Well, we know that one batch is equal to 10 articles, so

E(X)articles = 10.7463

May be wrong, but you get the idea. You can make a discrete table and work out the mean from there. A similar question is on the Q2 of 2010 VCAA Methods.

[barydos]

Sorry to barge, but I've got a probability related question (binomial)
(Image removed from quote.)

It's part c) which has got me stumped! D:

So, we know that;
p = 0.05
n = 10

So, presuming that you are getting this A and B ok, I'll move to C.

 So, the probability that two batches are tested is if the first one fails. This occurs when Pr(X=2).

Pr(X=2) = nCr(10,2)(0.05)^2(0.95)^8 = 0.07463
Now, we know that if there are any faulty articles in the 2nd batch, it fails. So there will never be a 3rd batch. There is also always 1 batch tested.

So, we know that the Pr(one batch) + Pr(2 batches) = 1, so Pr(one batch) = 0.92537. Drawing up a table, and working out the mean of a discrete distribution, you get;

Now, E(X) = (0.92537)(1) + (0.07463)(2) = 1.07463 batches.

SHIT, articles, not batches. Damn it. Well, we know that one batch is equal to 10 articles, so

E(X)articles = 10.7463

May be wrong, but you get the idea. You can make a discrete table and work out the mean from there. A similar question is on the Q2 of 2010 VCAA Methods.

Wow, you're amazing, thank you so much!

[dinosaur93]

How do you find the the answer for question b?

Im getting 0.0961 which is not the right answer?