Guide to Using the Ti-nspire for METHODS - The simple and the overcomplicated

[paulsterio]

1)
2)

cosh(x), sinh(x) and tanh(x) are known as "hyperbolic trigonometric purposes"
apart from being a model real life situations (such as hanging wire), they are also used to evaluate "imaginary hyperbolic angles" which have to do with the hyperbola

in order to evaluate things such as sin(x), cos(x) and tan(x) where x is imaginary, we developed the relations of sinh, cosh and tanh
they come from the relationship that cis(x) = e^(ix)

1) sinh(x) = -i sin(ix)
2) cosh(x) = cos(ix)

Hence, in order to evaluate things such as cos(i)
We will say that cos(i) = cosh(pi) =

So it becomes interesting that the cosine of an imaginary number is, in fact, a real number

That's essentially a little bit about the hyperbolic trigonometric functions

[enwiabe]

Wow, just happened upon this and I have to say great work b^3!

[b^3]

Ok guys I've found a small blunder on my part. Under normal distributions, I originally incorrectly stated what the pdf function does. It was stated that it did the values from negative infinity to the value you specified but what it really does is give the height of the probability function at the point you specify.

Sorry about that mistake guys & girls. I'll make a note at the top of the post as I can't update the printable version.

Again I'm sorry guys, lucky I caught it before the exam though.

[vgardiy]

Hey, there is an easier say to find the gradient at a point.

type

d/dx(f(x))|x=1 and it will find f'(1)

the d/dx () is the proper template for the derivative

its easier than defining the function then  defining the derivative etc.

[b^3]

Hey, there is an easier say to find the gradient at a point.

type

d/dx(f(x))|x=1 and it will find f'(1)

the d/dx () is the proper template for the derivative

its easier than defining the function then  defining the derivative etc.

I'm guessing that is the [menu] [4] [2] method? It's easier if you only have to work out the derivative once. If you have to work out multiple derivatives (say in an exercise, not the exam) then the other one is quicker. I prefer to use the definitions cause they are quicker for me, but work with what you feel better with and are quicker with.

[vcehuge]

Posted this already but here seems more appropriate > Any idea how i would go about solving this on the calculator?

The probability of winning a single game of change is 0.15, and whether or not the game is won is independant of any other game, suppose jodie plays a sequence of n games.

If the probability of jodie winning at least one game is more than 0.95, the smallest value n can take is closest to >

Any help would be amazing

[b^3]

Let X=number of games won
Pr(X>=1)=>0.95
Pr(X<1)<0.05
Pr(X=0)<0.05
You know that Pr(X=0)=pⁿ (as ⁿC₀=1 and (1-p)₀=1)
Then solve on calc 0.15ⁿ=<0.05 for n.
Then round up the number.

EDIT: I have a feeling that is partly incorrect (in what I've written, it will still work, can someone just check it for me please)

[vcehuge]

Looks good tyvm

[diligent18]

My calculate isn't co-operating with me for the "Getting Exact Values On the Graph Screen" method.
I enter "1/2", but the points immediately convert back to non-exact forms.
Help?

[SteliosV]

Appreciated