VCE Methods Question Thread!

[Tau]

What is the notation for defining normal and binomial distributions with known parameters n and p (or mean and sd)?

Normal distribution: \(X\sim \mathcal N(\mu, \sigma^2)\)
Binomial distribution: : \(X\sim \text{Bi}(n, p)\)

[^^^111^^^]

Ahh thanks!!
Totally forgot about polynomials !!!
And for the other question I'm asking about the stationary point when the gradient is 0 of the derived graph. because for instance 3x(x+3)^2, the x-value  of the stationary point (m=0) of the cubic will be -3  I've not sure if this will work for quartic, x^5, x^6 graphs

hope that makes sense >>.<<

For quartics I think that it is different, I am not too sure, (You might want to ask like someone else). And DW about x^5 or x^6 graphs, (They are not in the study design ). From what I know, the roots of a quartic are solvable by radicals but then that becomes very very complicated. So yeah soz I can't help too much out there.

[DBA-144]

Normal distribution: \(X\sim \mathcal N(\mu, \sigma^2)\)
Binomial distribution: : \(X\sim \text{Bi}(n, p)\)

Does the ~ not define approximately? and are we allowed to write 0.025^2 if sd=0.025 or do we need to find its square?

[Sine]

Does the ~ not define approximately? and are we allowed to write 0.025^2 if sd=0.025 or do we need to find its square?

The post Tau made is proper probability notation. In this context, ~ does not mean approximately

[Tau]

Does the ~ not define approximately? and are we allowed to write 0.025^2 if sd=0.025 or do we need to find its square?

The tilde indicates that the random variable has a specific distribution, not that it is approximately equal to something. I would say that is better to write 0.025^2 then to explicitly give the squared value. That’s makes it explicit that the sd is the unsquared value.

[aspiringantelope]

Ok thanks ^^^111^^^^

I've also got another question =[
How do you make
\(3\left(x+2\right)^2\cdot \left(x^2+1\right)-2x\left(x+2\right)^3\) equal \(\left(x+2\right)^2\left(x-3\right)\left(x-1\right)\) ?
I'm having some problems >.<

[^^^111^^^]

Ok thanks ^^^111^^^^

I've also got another question =[
How do you make
\(3\left(x+2\right)^2\cdot \left(x^2+1\right)-2x\left(x+2\right)^3\) equal \(\left(x+2\right)^2\left(x-3\right)\left(x-1\right)\) ?
I'm having some problems >.<

Hey,
The first step would be to simplify both sides of the equation by dividing them both by (x+2)^2. Then we will have 3(x²+1) - 2x(x+2)³ = (x-3)(x-1). Now by expanding both sides and simplifying we get:
-22x² -2x⁴ - 12x -12x³. (Let me know if you want clarification how i ended up with this).
After that we can solve for x by using the factor theorem and long division. So using the factor theorem of polynomials we can say that (x+1) is a factor of the  quartic polynomial. Then, by using long division and repeating the process again for the cubic polynomial derived, we can safely say that x = -1 , -2, 0,  (I'll let u find the last one )
Please tell me if I had made any mistakes.
Hope that helps

[DrDusk]

Hey,
The first step would be to simplify both sides of the equation by dividing them both by (x+2)^2. Then we will have 3(x²+1) - 2x(x+2)³ = (x-3)(x-1). Now by expanding both sides and simplifying we get:
-22x² -2x⁴ - 12x -12x³. (Let me know if you want clarification how i ended up with this).
After that we can solve for x by using the factor theorem and long division. So using the factor theorem of polynomials we can say that (x+1) is a factor of the  quartic polynomial. Then, by using long division and repeating the process again for the cubic polynomial derived, we can safely say that x = -1 , -2, 0,  (I'll let u find the last one )
Please tell me if I had made any mistakes.
Hope that helps

This works but it's a bit of a roundabout way to do it.

This method would be better as you don't need to solve a 3rd degree polynomial which can be tedious.
Firstly observe that both the equations already have (x+2)^2 in them. So first thing we can do is factor it out of the first one.

[^^^111^^^]

This works but it's a bit of a roundabout way to do it.

This method would be better as you don't need to solve a 3rd degree polynomial which can be tedious.
Firstly observe that both the equations already have (x+2)^2 in them. So first thing we can do is factor it out of the first one.

Fair enough! 

[breathe123]

Hey,
The first step would be to simplify both sides of the equation by dividing them both by (x+2)^2. Then we will have 3(x²+1) - 2x(x+2)³ = (x-3)(x-1). Now by expanding both sides and simplifying we get:
-22x² -2x⁴ - 12x -12x³. (Let me know if you want clarification how i ended up with this).
After that, we can solve for x by using the factor theorem and long division. So using the factor theorem of polynomials we can say that (x+1) is a factor of the quartic polynomial. Then, by using long division and repeating the process again for the cubic polynomial derived, we can safely say that x = -1, -2, 0,  (I'll let u find the last one )
Please tell me if I had made any mistakes.
Hope that helps

Why are you in a Methods thread when you are in year 9? Followed you from the JMSS entrance 2019 thread.

[^^^111^^^]

Why are you in a Methods thread when you are in year 9? Followed you from the JMSS entrance 2019 thread.

Idk what do you mean, I just want to contribute my knowledge that's all.

P.S. Please send me a PM  instead.

[SynX]

Ok thanks ^^^111^^^^

I've also got another question =[
How do you make
\(3\left(x+2\right)^2\cdot \left(x^2+1\right)-2x\left(x+2\right)^3\) equal \(\left(x+2\right)^2\left(x-3\right)\left(x-1\right)\) ?
I'm having some problems >.<

\(3\left(x+2\right)^2\cdot \left(x^2+1\right)-2x\left(x+2\right)^3\)
In this case the factor (X+2)² is obvious on the both side of the plus sign. So you should divide that first. So it becomes a product of (X+2)²and other stuff .
This factorization is not hard as it only contains a single pronumeral and the (x+2)².  So you basically factorize a binomial instead of trinomial... where you have all kinds of crazy stuff like a³-b³=(a-b)(a²+b²+ab), a³+b³=(a+b)(a²+b²-ab), (a+b)³=a³+a²b+ab²+b. Imagine going ^5 ,^7 and more (odd number powers are usually harder)
It becomes much harder and almost impossible as you go ^3 and more without an obvious common factor.

[^^^111^^^]

\(3\left(x+2\right)^2\cdot \left(x^2+1\right)-2x\left(x+2\right)^3\)
In this case the factor (X+2)² is obvious on the both side of the plus sign. So you should divide that first. So it becomes a product of (X+2)²and other stuff .
This factorization is not hard as it only contains a single pronumeral and the (x+2)².  So you basically factorize a binomial instead of trinomial... where you have all kinds of crazy stuff like a³-b³=(a-b)(a²+b²+ab), a³+b³=(a+b)(a²+b²-ab), (a+b)³=a³+a²b+ab²+b. Imagine going ^5 ,^7 and more (odd number powers are usually harder)
It becomes much harder and almost impossible as you go ^3 and more without an obvious common factor.

Remember trinomial is an expression with three terms and binomial is two terms.
So a trinomial is like 2x + 5y² -3, and binomial is like 5x-1.

This works but it's a bit of a roundabout way to do it.

This method would be better as you don't need to solve a 3rd degree polynomial which can be tedious.
Firstly observe that both the equations already have (x+2)^2 in them. So first thing we can do is factor it out of the first one.

I merely used another way as a continuation from what I told aspiringantelope previously (about the factor theorem and so on).

[SynX]

Remember trinomial is an expression with three terms and binomial is two terms.
So a trinomial is like 2x + 5y² -3, and binomial is like 5x-1.I merely used another way as a continuation from what I told aspiringantelope previously (about the factor theorem and so on).

Oh, i mess up with those terms again. I mean...cubic equations...maybe?

[^^^111^^^]

Oh, i mess up with those terms again. I mean...cubic equations...maybe?

DW , it really doesn't matter for now . Yeah you can say cubic equations,but usually for cubics and quartics, they are collectively called as polynomials.