RainbowSquirrel53

Suppose Ben had \(b\) coins originally. Then Skylar has \(3b/5\) coins (also originally).

Since after Ben gives Skylar ten coins they each have the same amount, we can get the equation

\(b-10=3b/5+10\)

Solving, we have \(b=50\), so then Skylar had \(30\) coins. Therefore, the two friends had \(\boxed{80}\) coins altogether.

The question isn't complete.

Let's say that there are \(6x\) women, and \(7y\) men. So then, \(5x\) women volunteered and \(3y\) men volunteered.

From this, we have the system of equations

\(6x+7y=1143\)

\(5x+3y=672\)

Solving, we get \(x=75\) and \(y=99\).

(a) Since \(y=99\), we know that there were \(3\cdot99=\boxed{297}\) male volunteers.

(b) Since \(x=75\), there were \(6\cdot75=450\) women, so then \(\frac{450}{1143}=\boxed{\frac{50}{127}}\) of the members were women.

Let the common ratio of the first three numbers is \(r\) and the numbers be \(x_1,x_2,x_3,\) and \(x_4,\) respectively. Note that we can replace the value of the first number anytime with the fourth.

From the fact that the last three numbers are in an arithmetic sequence, we can rewrite the set as 

\(x_2+12,x_2,x_2+6,x_2+12\)

We can then get the pair of equations

\(r(x_2+12)r=x_2\)

\(rx_2=x_2+6\)

Solving, we find that \(x_2=-4\) and \(r=-1/2\). Therefore, the set of four numbers is

\(\boxed{8,-4,2,8}\)

The y-intercept is \((-10,0)\).

Something went wrong with my Latex on line one, it's supposed to read \(3t-5 .

\(3t-5

Adding \(4t\) gives \(7t-5<5t-2\leq15\).

Solving \(7t-5<5t-2\) and \(5t-2\leq15\) gives \(t<1.5\) and \(t\leq3.4\), respectively.

Combining these then shows that \(t<1.5\), so therefore in interval notation the answer is \((-\infty,1.5)\).

I believe it's \((-\infty,1)\cup(4,\infty)\).

There are 29 pairs.

For the second method, you (correct me if I'm wrong) overlooked the fact that the probability of each of the possibilities may not be the same.