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Hermes1

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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

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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

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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.
 
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Dylanamali

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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

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btw can anyone remember the last 2 questions and the solution, why alpha is the greatest when x^2 = hH or wateva..
 

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