Silly Questions Thread

[brightsky]

Or by substitution:



Let , then .

The integrand becomes:

[cameron_15]

A cubic graph has a turning point at (3,0), y-intercept at (0,18) and passes through (1,16)

Find the equation of the graph f(x)

At first I though y=ax^3+bx^2+etc...

Then I thought y=(x-3)^2(x-a)... or it could also be (3-x)^2(x-a)...

Now I am confused... I'm usually pretty good with all the simultaneous equation stuff.

[brightsky]

The cubic has turning point at (3,0), so:

...(1)

...(2)

Passes through (0,18) and (1,16) so:

...(3)

...(4)

Use these system of equations to solve for a, b, c, d.

EDIT: Previous post was rubbish. :p

[cameron_15]

Use turning point form, , where is the turning point and are constants.

I was thinking that, how does it handle situations in which there are 2 turning points?

[brightsky]

Use turning point form, , where is the turning point and are constants.

I was thinking that, how does it handle situations in which there are 2 turning points?

Ignore that, it was rubbish. xD The form is used for stationary points of inflexion. Edited post.

[cameron_15]

I had 3 of the equations, just not the derivative one 

Thanks for your help!

[brightsky]

I had 3 of the equations, just not the derivative one 

Thanks for your help!

No problem.

[Aqualim]

I need help sketching a multiple absolute value equation (I think that is the correct terminology)



so essentially its the abs(3*(abs(x))-abs(x^3))

[7132]

The heights of boys in year 12 are normally distributed with a mean of 175cm and a standard deviation of 10cm. it is known that pr(Z>0.5)=0.3

a) Using this info find probability that a boy selected at random is greater than 170cm

b) find probability that a randomly selected boy is more than 180cm given that his height is above mean

[Whatlol]

Pr (X > 170) = Pr ( Z > -0.5 ) = 1- PR (Z>0.5) = 0.7

Thats because Pr(Z < -1/2 ) = Pr(Z > 1/2)

Pr(X>180| X >175) = Pr(Z>0.5) / 0.5  = 0.3/0.5 = 3/10 x 2 = 3/5

[Juddinator]

In the 2007 Insight Exam 2, one of the answers to a question is . However I used my CAS to graph the function and found the answer to be 0.314, converting it to . Would I still get the answer correct in a VCAA exam as it is still in exact answer form?

[Whatlol]

In the 2007 Insight Exam 2, one of the answers to a question is . However I used my CAS to graph the function and found the answer to be 0.314, converting it to . Would I still get the answer correct in a VCAA exam as it is still in exact answer form?

no you would not, your answer is not an exact answer it is an approximate answer.

[m@tty]

In the 2007 Insight Exam 2, one of the answers to a question is . However I used my CAS to graph the function and found the answer to be 0.314, converting it to . Would I still get the answer correct in a VCAA exam as it is still in exact answer form?

Can't you get the exact answer with your calc?

Anyway, no you wouldn't get the mark, because, as Whatlol said, your fraction is just the same number as you had earlier, containing only 3 dp. Anything involving pi must have pi in the answer, unless you are specifically asked to approximate the value.

[Juddinator]

In the 2007 Insight Exam 2, one of the answers to a question is . However I used my CAS to graph the function and found the answer to be 0.314, converting it to . Would I still get the answer correct in a VCAA exam as it is still in exact answer form?

Can't you get the exact answer with your calc?

Anyway, no you wouldn't get the mark, because, as Whatlol said, your fraction is just the same number as you had earlier, containing only 3 dp. Anything involving pi must have pi in the answer, unless you are specifically asked to approximate the value.

Well I tried to get an exact value on my calc but it wouldn't spit one out for me.... I have a TI CAS so if anyone knows if it's possible could you explain it? I have already tried 'converting decimal to fraction'.

Thanks

[Aqualim]

Pr (X > 170) = Pr ( Z > -0.5 ) = 1- PR (Z>0.5) = 0.7

Thats because Pr(Z < -1/2 ) = Pr(Z > 1/2)

Pr(X>180| X >175) = Pr(Z>0.5) / 0.5  = 0.3/0.5 = 3/10 x 2 = 3/5

Shouldn't Pr(Z > 0.5) correlate to the mean, since it's the middle number? How do you know that 170 is the middle number, when it says the mean is 175? or do you just make that assumption when the question asks that?