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[i2amroy]

Number of points is the number of hits. Turns is hits plus misses.

I got that, but my question was just double-checking if I was correct in assuming that the percentage given as "percentage of points per turn" could simply be taken from 1 to get the "percentage of misses per turn," which is what I'm actually looking for. (As I noted earlier, the important number I'm really looking for here is how good the guy is at dodging when the other guy has the anger hits compared to when the other guy doesn't have anger hits). I mean I'm pretty sure that that's what I do at this point, but if the small amount of statistics I've taken has taught me anything it's that you always need to be careful because statistic calculations are often extremely unintuitive.

[RedWarrior0]

Now for my own question. For a recent assignment, I was give the following function and told to find the residue at pi. Mostly I'm just wondering if I completely missed some easy method, or if the only way a senior undergrad would reasonably know is the excruciating way. Mathematica was no help here.

[Reelya]

-snip-

I think I'm understanding you properly, but now is changing the points per turn into a "dodge percentage chance" (which is what I'm looking for as a final answer) just as simple as doing 1-ans%?

If you know the # of hits per unit of time, then yes you can derive the dodge% by subtracting that percentage from 1, since we're talking probability, there are two outcomes, and they have to add up to 1.

My method for solving the problem was based on realizing that there are two independent sequences (normal hit, vs dodge-hit-hit-hit), we know the likelihood of each sequence occurring, and we know the amount of time that each sequence takes. from that we can work out the proportion of time spent in each sequence, and therefore average number of hits (or dodges) per unit time.

[3man75]

Quick question: Graph wise what makes a even function even and a Odd function odd?

Again this is without any equations/numbers being in place.

[frostshotgg]

An even function looks the same if you mirror it across the middle line without a horizontal shift. An odd equation looks the same if you rotate it 180 degrees around without either a horizontal or vertical shift. For even, think y = x^2. It looks the same as you go either positive or negative from x = 0. For odd, think y = x. It goes up when it goes right the same speed as it goes down when it goes left. If you flipped the graph upside down it'd look the same.

[TheBiggerFish]

Even:Symmetric about y=0

Odd:Symmetric about y=x (or) You can rotate it 180 degrees and have the same function.

Dang it ninja.

[3man75]

Thank you everyone!

[3man75]

If i'm suppose to list the potential rational zeros of the following polynomial: 5x^3 - x^2 + 3

Then how do I do it step by step?

I know the answers are 1/5, 3/5, 1, and 3 (plus or minus of course) but i'm not sure how to get that. Why is 5 not included as a potential zero?

Note: this is finding potential zeros NOT finding zeros.

[Dozebôm Lolumzalìs]

...using what process?

I can limit all zeros of a polynomial to nontranscendental real numbers - that doesn't tell me much.

It really depends on the method. How, roughly, did you get those numbers?

[Karlito]

Using the rational root test, presumably. Any rational number (that isn't 0) that is a root of a polynomial is of the form r/s, where r is a factor of the last coefficient and s is a factor of the first coefficient (and r and s are relatively prime i.e. the fraction is in lowest terms.)

No matter what class this is, I'd expect the rational root test to just be something that's given in your textbook.

[Mostali]

If i'm suppose to list the potential rational zeros of the following polynomial: 5x^3 - x^2 + 3

Then how do I do it step by step?

Why is 5 not included as a potential zero?

List all factors of the constant term - {1,3} and list all factors of the leading coefficient - {1,5}.  Then list all combinations of ratios with an element of the first set in the numerator and an element of the second set in the denominator - {1/1,1/5,3/1,3/5}.  If the polynomial has a rational root then it will be the positive or negative of one of these ratios.

The specific reason that 5 is not a possible zero is a modular argument.  Consider x=5, then 5(5)^3-(5)^2+3=0 => 5(5^3-5)=-3 => 5^3-5=-3/5.  Notice that the LHS of the last equation is an integer while the RHS is not, and therefore an impossibility.  You could repeat this with any polynomial with a constant term that isn't divisible by 5.

[TheBiggerFish]

...How do you apply L'Hôpital's Rule to things that aren't 0/0 or infinity/infinity?

[frostshotgg]

Find a way to rewrite the term until it's one of those two.

[MagmaMcFry]

...How do you apply L'Hôpital's Rule to things that aren't 0/0 or infinity/infinity?

You don't, because you don't need to, because you're (usually) done already.

Example: If f(x) goes to 7 and g(x) goes to 4, then f(x)/g(x) goes to 7/4.

Example: If f(x) goes to infinity and g(x) goes to -3, then f(x)/g(x) goes to -infinity.

[TheBiggerFish]

I meant other indeterminate forms.