KnockOut

I'd love to answer these except for the fact that they're under moderation right now...It says that moderation is at 74%, then it goes back down to 8%, and so on. I've been seeing this on almost every graph I see here for the past two days...weird.

$\mathrm{Range\:of\:}-3^x+4:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)<4\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:4\right)\end{bmatrix}$

The range of an exponential function of the form $-c \cdot n^{ax+b} + k$ is $f(x) < k$

Therefore, $k = 4$.

So, the solution is $f(x) < 4$ which would give us answer choice $\boxed B$

There are two ways that we can end up with an even number by adding two numbers.

Either both numbers have to be even, or both have to be odd.

Starting off, let's find out the probability that we will roll two cubes and they will both show an even number.

For cube A, we have $6$ possible outcomes. Out of those six, three numbers are even ($2 , 4, 6$).

Therefore, we have $\frac{3}{6} = \frac{1}{2}$ ways to roll one cube and have an even number show up.

This is the same for cube B. In total, we have $\frac{1}{2} * \frac{1}{2} = \frac{1}{4}$ ways to roll two cubes and have both of them show even numbers.

If we do the same thing for the odd numbers, we would get the same result of $\frac{1}{4}$.

Adding these two possibilities together, we would get $\frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

Therefore, our answer is $\boxed{\frac{1}{2}}$

I'm just earning hints as we go. After I asked about what you said, it seems as if you are not supposed to consider the seperate teams as much, but instead the gymnasts by themselves.

The answer is $\boxed 5$ but I need to type up an explanation now if someone can help. 

By the way. I tried this answer and it came out as incorrect. 

Okay, so let me explain this problem better so that someone can explain the answer to me.

Alrighty.

So yes, the teams are supposed to be the same size. There are $x$ number of teams, and $y$ number of gymnasts on each team.

EVERY gymnast must shake hands with EVERYONE else except for herself.

After this handshaking process finished, a single coach on a certain team came down and only shook the hands of the gymnasts on that certain team with $y$ gymnasts.

In total, there were 281 handshakes. We are trying to solve for the number of handshakes that the single coach on that certain team made, and since the coach only shook the hands of the gymnasts on her own team, I guess we are solving for $y$.

So rewriting the problem, we get: Solve for the lowest possible value of $y$.

@Guest, so what this problem is trying to ask, is that there is one certain team, and the gymnasts from that certain team is doing handshakes with gymnasts from other teams. The "coach" is the coach from that certain team.

I honestly do not know. This is all the problem provided me with. So I'm guessing that team size does not necessarily matter, but I'm not sure.

Off Topic.....Thanks!

CPhill, didn't you forget about the V?