Melody

f(x)=4^x5+8x+2

if it is even then

f(x)=g(-x)

and if that happens then it means the graph has the x axis as the axis of symmetry.

Does f(1)=f(-1)   if it doesn't then it is NOT even. If it does then it might be even you will need to think a little.

If it is odd then

f(-x)=-f(x)  and it will have 180 degrees rotational symetry about the origin.

does f(-1)= -f(1)  ??

You can cheek you answer by graphing it here

10!= 1*2*3*4*5*6*7*8*9*10

= 2*3*2*2*5*2*3*7*2*2*2*9*2*5

= 2*2*2*2*2*2*2*2*3*5*3*7*9*5

=   4 *  4  *  4  *  4  *3*5*3*7*9*5

=        4^4              *3*5*3*7*9*5

So the highest value of x is 4

in a clinical study , volunteers are tested for a gene that has been found to increase the risk for a disease the probability that a person carries the gene is 0.2 

p=0.2     q=0.8

1) what is the probablilty that 8 or more people will have to be tested before 3 with the gene are detected 

   P(8 or more need to be tested before 3 are detected)

=P(2 or less will have it if 7 people are tested)

=P(0 out 7 have it) + P(1 out 7 have it) + P(2 out 7 have it)

\(=  7C0 * 0.8^7 + 7C1 * 0.2*0.8^6 +7C2 * 0.2^2 * 0.8^5\\ = 0.8^7 + 7* 0.2*0.8^6 +21 * 0.04 * 0.8^5\\\)

0.8^7 + 7* 0.2*0.8^6 +21 * 0.04 * 0.8^5 = 0.851968

approx 85.2%

2) ??

Thanks Heureka, that is great    

Here is a part of the tree, you just have to get rid of the 0.5 probablities and repalce them with the correct ones.

I have already started it.

Multiply the probabilities on each of the branches to get the prob of each event happening.

That is simple for a computer but I do not think it is how the question was intended to be done.

I'd really like to see someone tackle this question.  :/

Problem:

A club with five men and six women wish to form a committee. The number of men must be between 1 and 3 (inclusive), and the number of women must be between 2 and 4 (inclusive). How many different committees can be formed?

1 man and 2 women    5     *   6C2

1 man and 3 women    5     *   6C3

1 man and 4 women    5      *  6C4

2 man and 2 women    5C2  * 6C2

2 man and 3 women    5C2  * 6C3

2 man and 4 women    5C2  * 6C4

3 man and 2 women    5C3  * 6C2

3 man and 3 women   5C3  * 6C3

3 man and 4 women   5C3  * 6C4

Work all those out and add them together.

Thank you. 

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