+122 Not sure about this... But here's my rational
We know that we have \(15\) scores of \(10\).
Since the mode is \(7\) we need to have \(16\) scores of \(7\) (and no more)
The median is \(9\), so the \(20\) and \(21st\) scores must be \(9...\) to keep the average as small as possible.
And the other \(3\) scores need to be as small as possible so let them \(=1\)
We have
\(1, 1, 1, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7 ,7, 7, 9 ,9, 9, 9, 9, 9,10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10\)
\(\text{Median} = 9\)
\(\text{Mode} = 7\)
\(\text{Mean} = 7.975 = 8\)
Credit to CPhill; https://web2.0calc.com/questions/basic-stats_2#r1
P.S. I changed up CPhill's answer a little, so please give me a little credit also!
+122 I think this is supposed to be :
\(( 10x^3 - 1 / (2x^2) ) ^5\)
The constant term is :
\(- C(5, 3) \cdot (10x^3)^2 \cdot (\frac{1}{2x^2})^3 =\)
\(-10 \cdot (100 x^6) \cdot (\frac{1}{8x^6} ) =\)\(- [10 \cdot \frac{100}{8}] \cdot (\frac{x^6}{x^6}) =\)
\(- 125\)
Credit to CPhill; https://web2.0calc.com/questions/find-the-constant-term-in-the-expansion-of-10x-3-1-2x-2-5#r1
P.S. I changed up CPhill's answer a little, so please give me a little credit also!
+122 John has a \(100\)% chance of being there if he arrives at \(2:20\) or earlier. Add \(\frac13\).
At \(3:00\), he has a \(\frac13\) chance. We can take this and take the average from \(2:20\) and \(3:00\) to get \(\frac23\) for this \(40\) minute period. Multiply this by the portion of time. Add \(\frac49\) to the probability.
Result: He has a \(\frac79\) chance of catching the train.
Credit to helperid1839321; https://web2.0calc.com/questions/john-needs-to-catch-a-train#r1
P.S. I changed up helperid1839321's answer a little, so please give me a little credit also!
!
