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.
\($5.00 - $0.05 = $4.95\)
A stem and leaf diagram has the units digit in a separate column, so we have C. 55 as the answer.
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.
There are \(9\) members who are not Fred, so we can simply do 9C3 which equals \(84\).
https://web2.0calc.com/questions/math-question_122
\(400 = 20^2\), so we have \((2^4 + 2^2)^2 = (2^4)^2 + 2(2^4)(2^2)+(2^2)^2 = 2^8 + 2^7 + 2^4\). \(8+7+4=19\).
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 \(\frac{81}{90} = \frac{9}{10}\).
We have 10C8 \(* (\frac{1}{2})^2 ≈ 0.044\)
+399 