VCE Methods Question Thread!

[Shadowxo]

I am so sorry! i didnt mean to sound rude, i totally understand this is a website for all to use and share - and i completely shouldnt be taking it granted for! i was just panicking because my teacher sent this to us for practise before our actual methods SAC today, and i didnt know how to do it, so i got a little scared!

I had a look at the explanations you did, and it 100% makes sense, and i have completed them, i dont know how to upload a file for you to check, onto here because i did it on paper, but we can assume i (hopefully) did it right!

but, thank you so much!

I am extremely sorry, as i didnt seem to sound rude! I feel really bad!

Thank you for taking your time in helping me, means alot!
Have a good day

Good luck
Glad you understand it now, just stay calm and remember what you've been taught

[Mattjbr2]

Let's talk ourselves through it. An intersection is an event in common. For example, what do dogs and cats have in common? Both have four legs, tails, fur. Now, when we talk about the intersection of number sets, we instead want to know what numbers are shared between them. Consider on a die, the event of rolling an even number or a prime number - in this case, the intersection is only the number 2.

Now, let's look at the cases given. We have two events - one in which the random variable X takes on the value of 15 (that is, you score exactly 15), and one in which the random variable X takes on the value 12, or higher (that is, you score 12 or more). For what values of X can you score exactly 15 but also more than 12? This sentence might sound weird because it is an odd thing to say, because obviously the only time you can score exactly 15 and more than 12 is when you score 15, that is when X=15. That's why your intersection is equal to X=15. Same logic for your second one.

Let me know if this hasn't properly answered your question.

Can't believe i didn't think of it that way initially. Thanks!

[Lavar Big BBB Balls]

Hi,

For this: http://imgur.com/a/5zBYc

For part b) 2nd column line 1, I'm not sure what a * e^(-a/lambda) is equal to after I evaluate the limit. I don't think it's 1 like I wrote..

How else am I supposed to do this?

Thanks

[VanillaRice]

Hi,

For this: http://imgur.com/a/5zBYc

For part b) 2nd column line 1, I'm not sure what a * e^(-a/lambda) is equal to after I evaluate the limit. I don't think it's 1 like I wrote..

How else am I supposed to do this?

Thanks

Apologies in advance if I've misread your writing (correct me if I have). Let's think about the 'a' which is multiplied by the exponent. That's essentially a dilation factor, and we'll ignore that for now. When you compute the limit, you'll get something like looks like exp(-infinity), and we'll remove the lambda since infinity divided by anything is still infinity. If you think about a graph of the form y=exp(-x), as x approaches infinity, what's the value of y?
It's the same concept that you used for your first term.

Spoiler

The graph would approach zero, so the limit can be written as zero.

Also, for the last term, it's a multiple of zero, which is zero, not one 

Hope this helps 

[lovelyperson]

Hello! Can I please get some help with the following questions? Thanks!

1) The function f is defined by f: (-∞, -2) → R, f(x) = log_e((x-1)² - 8 ). Show that there are no solutions to the equation f(x) = 0.
2) For what value of k, where k is a real constant, does x² - 10 = k have 4 solutions?

[VanillaRice]

Hello! Can I please get some help with the following questions? Thanks!

1) The function f is defined by f: (-∞, -2) → R, f(x) = log_e((x-1)² - 8 ). Show that there are no solutions to the equation f(x) = 0.
2) For what value of k, where k is a real constant, does x² - 10 = k have 4 solutions?

1) It might help to rewrite the equation as

Have a go at solving this for x, and then try to answer the question, keeping in mind the domain.

2) I don't think there's any value of k that will give you 4 solutions for a quadratic. The question is essentially asking for a y-value that will intersect the graph of x² - 10 four times. If you sketch a graph you'll see the maximum number of solutions is two (for k > -10).

Good luck, and let us know if you run into any trouble

[captkirk]

Hello, can someone tell me how to do part a by using the Casio classpad. Thanks! http://imgur.com/a/s2B3Y

[gnaf]

Hi

How do you do this question which asks to find the sample size using binomial distribution? Do I have to use guess and check?

[VanillaRice]

Hi

How do you do this question which asks to find the sample size using binomial distribution? Do I have to use guess and check?

I think 50 shots is too big of a number to be able to use the method for finding a sample size for a binomial distribution. I don't really see any way to do it otherwise, so you most likely would have to do a guess and check method on your calculator.

[A TART]

I think 50 shots is too big of a number to be able to use the method for finding a sample size for a binomial distribution. I don't really see any way to do it otherwise, so you most likely would have to do a guess and check method on your calculator.

After spamming my CAS, the closest thing i got is 85.
http://imgur.com/2sAkgCA

[gnaf]

Thanks so much VanillaRice, I'll give it a go

A TART- that's some real spam there, so impressed 

[VanillaRice]

After spamming my CAS, the closest thing i got is 85.
http://imgur.com/2sAkgCA

Since 85 is the closest you can get to 0.99, and we want Pr(X >= 50) > 0.99, the minimum number of shots would be 86, since this is the lowest number that gives a probability of > 0.99 

[Shadowxo]

If I were you I'd probably check with your teacher. I can't remember probability but it seems as though there should be a calculator ability to solve it (tried to find it and couldn't). No harm in checking though!

[VanillaRice]

If I were you I'd probably check with your teacher. I can't remember probability but it seems as though there should be a calculator ability to solve it (tried to find it and couldn't). No harm in checking though!

This particular question is the last one in an exercise where they try to get you to use a particular method. Usually the question would ask for 1 or 2 successful shots (i.e. Pr(X >= 1) > 0.99) rather than 50. If this were the case (let's use 'at least 1 successful shot'), we could calculate this easily by using:


And we can write an expression for Pr(X=0)

Thereby allowing us to solve for n. But 50 is way too high to solve using this method. I guess one could try Pr(X>=50) = 1 - Pr(X<50), but then we'd have to write expressions for Pr(X=0) all the way up to Pr(X=49), and even if we did, trying to solve it would probably crash the calculator ¯\_(ツ)_/¯

[Shadowxo]

This particular question is the last one in an exercise where they try to get you to use a particular method. Usually the question would ask for 1 or 2 successful shots (i.e. Pr(X >= 1) > 0.99) rather than 50. If this were the case (let's use 'at least 1 successful shot'), we could calculate this easily by using:


And we can write an expression for Pr(X=0)

Thereby allowing us to solve for n. But 50 is way too high to solve using this method. I guess one could try Pr(X>=50) = 1 - Pr(X<50), but then we'd have to write expressions for Pr(X=0) all the way up to Pr(X=49), and even if we did, trying to solve it would probably crash the calculator ¯\_(ツ)_/¯

Sounds like you have it nailed I was looking on my old calculator seeing if I could find some inverse binomial function but it doesn't exist so guess-and-check looks like the most appropriate method