Hello: <span>
<span> an equation is : y =ax+b ( a is a slope )

parallel to the line (q) : y=6/5x-1 a =6/5 ( same slope )
the equation of (r) : is : y = 6/5x+b</span></span>

the line through point (5,5) : 5= (6/5)(5)+b ...b=-1

an equation of (r) : is : y = 6/5x-1

or : 6/5x-y = 1

6 0

3 years ago

A quantity x has cumulative distribution function p(x) = x − x2/4 for 0 ≤ x ≤ 2 and p(x) = 0 for x < 0 and p(x) = 1 for x &gt

34kurt

$F_X(x)=\begin{cases}0&\text{for }x2\end{cases}$

Recall that the PDF is given by the derivative of the CDF:

$f_X(x)=\dfrac{\mathrm dF_X(x)}{\mathrm dx}=\begin{cases}1-\dfrac x2\\\\0&\text{otherwise}\end{cases}$

The mean is given by

$\mathbb E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\int_0^2\left(x-\dfrac{x^2}2\right)\,\mathrm dx=\frac23$

The median is the number $M$ such that $F_X(x)=\mathbb P(X\le M)=\dfrac12$ . We have

$F_X(M)=M-\dfrac{M^2}4=\dfrac12\implies M=2\pm\sqrt2$

but both roots can't be medians. As a matter of fact, the median must satisfy $0\le M\le2$ , so we take the solution with the negative root. So $M=2-\sqrt2$ is the median.

3 0

3 years ago