Melody

3 divided by 4       3/4   times

It seems that 2 in 8 are red

so

\(\frac{x}{12}=\frac{2}{8}\\ \frac{x}{12}=\frac{1}{4}\\ x=\frac{12}{4}\\ x=3\)

Based on experiemntal probability, 3 are likely to be red.

Yes, not big deal.

I liked your approach  :)

Geno, your answer would be great if it was 199! but it isn't.  All the even numbers are missing.

odd multiples of 3,    6n-3  from n=1 to   n=round down[(199+3)/6] = 33          33 of them

odd multiples of 9      9,27,.....189                 9*1 to 9*21                                  11  of them

odd multiples of 27     27  ,...  179                 27*1 to 27*7                                        4 of them

odd multiples of 81     81  that is the only one                                                          1 of them

33+11+4+1 = 49

So guest and I agree.  But i used Genos method  :)

g(x) = -.03x^2+x+6=0

0= -.03x^2+x+6

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-1 \pm \sqrt{1^2-4*-0.03*6} \over 2*-0.03}\\ x = {-1 \pm \sqrt{1+0.72} \over -0.06}\\ \)

you can simplify it,

The one that is not choosy is going to be a Setter.  That way there are 5 setters  and 5 spikers 

Take a setter, makes no difference which one.  She can be matched with any of 5 spikers

Take another setter.  She can be matched with any of the remaining 4 spikers.

etc

So the answer will be   5! = 120 ways

Try proofreading your question.   

Your wasting your own and everybody elses time.

Nah, I do not think that is true :)

Let a and b real numbers such that   \(x^4+2x^3-4x^2+ax+b = (Q(x))^2\) for some polynomial Q(x).

What is the value of a + b?

\(x^4+2x^3-4x^2+ax+b = (Q(x))^2\\ (Q(x))^2=(x^2+mx+t)(x^2+mx+t) \\ (Q(x))^2=x^4+mx^3+tx^2\;\;+mx^3+m^2x^2+mtx\;+\;tx^2+mtx+t^2\\ (Q(x))^2=x^4+2mx^3+2tx^2\;\;+m^2x^2+2mtx+t^2\\ (Q(x))^2=x^4+2mx^3+(2t+m^2)x^2+2mtx+t^2\\\)
\(2m=2\;  \;so\;\; m=1\\ 2t+m^2=-4\\ 2t+1=-4\\ t=-2.5\\~\\ 2mt=a\\ 2*1*-2.5=a\\ a=-5\\~\\ t^2=b\\ b=(-2.5)^2\\ b=6.25\\~\\ a+b=6.25-5=1.25 \)

Latex

x^4+2x^3-4x^2+ax+b = (Q(x))^2\\
(Q(x))^2=(x^2+mx+t)(x^2+mx+t) \\
(Q(x))^2=x^4+mx^3+tx^2\;\;+mx^3+m^2x^2+mtx\;+\;tx^2+mtx+t^2\\
(Q(x))^2=x^4+2mx^3+2tx^2\;\;+m^2x^2+2mtx+t^2\\
(Q(x))^2=x^4+2mx^3+(2t+m^2)x^2+2mtx+t^2\\

Can you please write right that without the unrendered LaTex!