Rom

34

{
 {{9, 91}},
 {{39, 91}},
 {{52, 91}},
 {{1, 8, 91}},
 {{1, 9, 91}},
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 {{1, 52, 91}},
 {{8, 9, 91}},
 {{8, 39, 91}},
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 {{9, 39, 52}},
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 {{1, 8, 9, 91}},
 {{1, 8, 39, 52}},
 {{1, 8, 39, 91}},
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 {{1, 9, 39, 52}},
 {{1, 9, 39, 91}},
 {{1, 9, 52, 91}},
 {{1, 39, 52, 91}},
 {{8, 9, 39, 52}},
 {{8, 9, 39, 91}},
 {{8, 9, 52, 91}},
 {{8, 39, 52, 91}},
 {{9, 39, 52, 91}},
 {{1, 8, 9, 39, 52}},
 {{1, 8, 9, 39, 91}},
 {{1, 8, 9, 52, 91}},
 {{1, 8, 39, 52, 91}},
 {{1, 9, 39, 52, 91}},
 {{8, 9, 39, 52, 91}},
 {{1, 8, 9, 39, 52, 91}}
}

what are you going on about partialmathematician?

\(\text{for (d) you've already calculated }\mu=\dfrac{13}{12} \text{ and }\eta=1\\ \text{so you are asked to find }P\left[X < 3 \cdot \dfrac{13}{12} - 1\right] =\\ P\left[X <\dfrac 9 4\right]\)

\(\text{this will be}\\ \displaystyle \int \limits_0^1~\dfrac 1 2~dx + \int_1^{\frac 9 4}\dfrac{3-x}{4}~dx = \dfrac{119}{128}\\ \)

deleted

the two are identical

\(\dfrac 4 9 + \dfrac 1 6 + \dfrac{5}{18} = \\ \dfrac{8}{18}+\dfrac{3}{18}+\dfrac{5}{18} = \\ \dfrac{8+3+5}{18} = \\ \dfrac{16}{18} = \\ \dfrac 8 9\)

I don't really understand what you are saying but for example take k=3

\(\dbinom{4}{3}2^3 \left(\dfrac{3}{x^2}\right)^{4-3} = \\ 4\cdot 8 \cdot \dfrac{3}{x^2} = \\ 32\dfrac{3}{x^2} = \dfrac{96}{x^2}\\ b=96\)

I think this is supposed to be a demonstration of the binomial theorem.

\((a+b)^n = \sum \limits_{k=0}^n~\dbinom{n}{k}a^k b^{n-k}\)

In this case

\(a = 2\\ b= \dfrac{3}{x^2}\)

\(\left(2 + \dfrac{3}{x^2}\right)^4 = \sum \limits_{k=0}^4~\dbinom{4}{k}2^k\left(\dfrac{3}{x^2}\right)^{4-k}\)

you can plug in k=4, 0 to get a-e

none of the numbers in your post showed up.

You can see two points on this line (0,1) and (2,0).  From this we can find the line.

\(\text{the slope }m \text{ is given by}\\ m=\dfrac{y_2-y_1}{x_2-x_1} = \dfrac{0-1}{2-0} = -\dfrac 1 2\\ \text{using the first point and the slope we write the point slope form of the line}\\ (y-1) = -\dfrac 1 2 (x-0)\\ \text{and with a bit of algebra we get }\\ \dfrac 1 2 x +y= 1 \text{ or}\\ x+2y = 2\)

\(\text{Given two points }p_1,~p_2, \text{ the midpoint of the segment joining them is given by}\\ m=\dfrac{p_1+p_2}{2} \\ \text{in this case }\\ m =\left( \dfrac{-4+28}{2},~\dfrac{9-29}{2}\right)=?\)

I leave that bit of arithmetic to you.