dierdurst

Let's use a Stars and Bars approach on this. There are 30 votes(stars) and 4 choices, but to get 4 catergories we only need 3 bars. In order to find where these bars would go, we can do 33C3 which is 5456.

- Daisy

You can draw triangle \(PAS\) with \(DS\) perpendicular(the altitude) to \(AP\). You know \(PS = 10\). Let's name the midpoint of \(PS\) the letter \(F\). \(PF = 5, PA = 13, \) so \(AF = 12\). Now, cos \(PAS = \frac{AD}{AS}\). We already know \(AS = 13,\) so we just need to solve for \(AD\). To find \(AD\), we first need to solve for \(DS\), which is one of the altitudes of triangle \(PAS\). We know that the area of a triangle is \(\frac{1}{2}bh\), so we can make an equation. \(\frac{1}{2}\times10\times12=\frac{1}{2}\times13\times DS\). From this, we get \(DS = \frac{120}{13}\). Now, we can use Pythagorean Theorum to find \(AD\). \(12^2-(\frac{120}{13})^2=AD^2\). After calculating this, we can get \(AD = \frac{12\sqrt{69}}{13}\). Now, lets plug this into our fraction. We can get \(\frac{\frac{12\sqrt{69}}{13}}{13}\) which simplifies to \(\frac{12\sqrt{69}}{169}\).

- Daisy

Hi Guest, 

You should not delete your question, because other people may have that question as well.

- Daisy

That method works, but if you are looking for a much easier method here is one:

Reflect \(A\) and \(B\) \(180°\)down, and mark those new points as \(A'\) and \(B'\). You can see that running from \(A\) to \(B'\) is the same distance as a route that would fit in the requirments of the race. Now, we have to find the shortest route from \(A\) to \(B'\). The shortest route is a straight line. This straight line's length is \(1442\) m. 

- Daisy

\($5.00 - $0.05 = $4.95\)

- Daisy

A stem and leaf diagram has the units digit in a separate column, so we have C. 55 as the answer.

- Daisy

We cannot see a 0 in the 1's row, so A is not correct. We cannot see a 1 in the 2's row, so B is not correct. We do see a 4 in the 4's column, so C. is the answer.

- Daisy

There are \(9\) members who are not Fred, so we can simply do ₉C₃ which equals \(84\).

- Daisy

\(400 = 20^2\), so we have \((2^4 + 2^2)^2 = (2^4)^2 + 2(2^4)(2^2)+(2^2)^2 = 2^8 + 2^7 + 2^4\). \(8+7+4=19\).

- Daisy

From the \(10\) to \(99\), there are \(90\) numbers. Out of these numbers, \(9\) have the same digits(\(11, 22, 33, 44, 55, 66, 77, 88, 99\)). So your probability would be \(\frac{81}{90} = \frac{9}{10}\). 

- Daisy

We have ₁₀C₈ \(* (\frac{1}{2})^2 ≈ 0.044\)

- Daisy