Cssa mx1 (1 Viewer)

Hermes1

Banned
Joined
Oct 4, 2010
Messages
1,282
Gender
Male
HSC
2011
i mean because they are roots of the questions. if you substitute back to the original P(x), it should be P(x)=0 right? but apparently it's not.
i just typed them both into my calculator in that equation and they are working for P(x) = 0.
just try it again yourself.
 

Hermes1

Banned
Joined
Oct 4, 2010
Messages
1,282
Gender
Male
HSC
2011
i got -6 x 10^(-12) and -1.2 x 10^(-12)
i think you should post this question in MX1 forum for drongoski to take a look. bcuz im pretty sure the question is fine but u should get a second opinion.

did u at least leave the answers u got? bcuz testing they equalled zero wasnt a part of the question. so u wont lose any marks.
 

muzeikchun852

Active Member
Joined
Mar 29, 2009
Messages
1,590
Gender
Undisclosed
HSC
2011
i think you should post this question in MX1 forum for drongoski to take a look. bcuz im pretty sure the question is fine but u should get a second opinion.

did u at least leave the answers u got? bcuz testing they equalled zero wasnt a part of the question. so u wont lose any marks.
yeah i did.
 
Last edited:

Dylanamali

Active Member
Joined
Jul 7, 2009
Messages
1,248
Gender
Male
HSC
2011
I forgot the specifics of the q, but I'm preety sure p(red light) = 3/10, p(amber) = 1/10, p(green) = 6/10

Therefore p(redlight) = 0.3, q(non redlight) = 0.7

(p+q)^5 =......

Answer: 5C4(0.3)(0.7)^4
 

Dylanamali

Active Member
Joined
Jul 7, 2009
Messages
1,248
Gender
Male
HSC
2011
btw can anyone remember the last 2 questions and the solution, why alpha is the greatest when x^2 = hH or wateva..
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top