VCE Methods Question Thread!

[BubbleWrapMan]

They should always tell you what you need to label so there's no chance of confusion

[Homer]

i defined a function on my casio classpad 330 cas using "x" as the function name and "t" as the variable. Even after clearing all the variables, evaluating simple equation now with "x" gives an "Incorrect agrument". Im not sure what to do now? How do I get it back to normal? Thanks

[Homer]

Also- X is a normally distributed variable with standard deviation Z. If Pr(X<a)=0.6 then Pr(X<a+Z) is equal to? ANS: 0.895

[Alwin]

i defined a function on my casio classpad 330 cas using "x" as the function name and "t" as the variable. Even after clearing all the variables, evaluating simple equation now with "x" gives an "Incorrect agrument". Im not sure what to do now? How do I get it back to normal? Thanks

Since my school might be one of the few others that uses the ClassPad, I may be able to help you. Go to:

          Action > Command > Clear_a_z > execute

Usually that, combined with Clear All Variables usually does the trick

[darklight]

I got 1003 ml, answers state 1004 ml for question 8.

[Nato]

could someone please help me with the graph of


thanks

[Zealous]

I got 1003 ml, answers state 1004 ml for question 8.

I just did the working myself and got 1003ml. To be exact I got 1003.13mL.
Dunno if we're both wrong or if the answer is a little off.

could someone please help me with the graph of

thanks

Whenever I sketch trig functions that have translations, I like to start with the graph without translations, then translate important points and connect them.

So look at the graph of , which doesn't have a horizontal translation, and mark in the main points, like the x intercepts, and maximum and minimum points onto your graph.

What I do then is take the x value of these important points and move them all units to the left, mark in the new points, rub out the original points and connect the new, translated points.

That's what I do when I sketch translated graphs, but there are many other methods that others will do.

[darklight]

It's saying that it's false, because it is false

You're typing in which is equal to , not 0.25. It is equal to 0.25 for a certain value of a, not a in general. You should be solving the expression for a.

From the beginning, assuming you have given the correct equations in the original part of the question, your expression should be something along the lines of solve(int(2*x*e^(-x^2),x,0,a)=1/4,a)

You could quite easily do the integration by hand and end having to solve , which gives:



This is approximately 0.536, the value you were looking for.

Hey,

Hm, that's the part I'm not understanding. If the antiderivative is e^(-x^2), and the graph is viable for values of x greater than 0 then a must be greater than 0. Therefore, when you do the calculations shouldn't it be:

[e^(-x^2)] with a being the upper limit, 0 being the lower limit.
This would produce e^(-a^2) - 1 which is not equal to 1 - e^(-a^2).

I just did the working myself and got 1003ml. To be exact I got 1003.13mL.
Dunno if we're both wrong or if the answer is a little off.

I got 1003.13 mL too. Maybe the book is wrong...

[Phy124]

Hey,

Hm, that's the part I'm not understanding. If the antiderivative is e^(-x^2)...

is the antiderivative of but the function was equal to so the antiderivative is .

[darklight]

is the antiderivative of but the function was equal to so the antiderivative is .

I see the light!! Thank you

[Smiley_]

Find the derivative and then use this to anti diff

(2x-3x^2)^6

I can diff it then I get confused

[Daenerys Targaryen]

actually is there more information to this question?

[pi]

actually is there more information to this question?

Forgot the "+c".

Regards.

[Nato]

Hey guys, need help with a question


i am trying to graph and finding the intercepts by letting y=0.
i got stuck at . how can i find values for x (can't find any that equal )

thanks

[clueless123]

There is no x-intercepts for that graph