1.5 / 2 = 1 / x # cups / weight = # cups / weight

cross multiply 
1.5x = 2 

divide both sides by 1.5 

x = 1.3(the 3 keeps going) or 1 and 1/3 pounds
So the answer is 1/3
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8 0

3 years ago

The amount of kerosene, in thousands of liters, in a tank at the beginning of any day is a random amount Y from which a random a

Tems11 [23]

Find the marginal density for $Y$ by integrating the joint PDF over all possible values of $x$ :

$f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\int_0^y2\,\mathrm dx$

$\implies f_Y(y)=\begin{cases}2y&\text{for }0$

Then the density of $X$ conditioned on $Y$ is

$f_{X|Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\frac1y$

for $0$ and undefined elsewhere.

Thus

$P(0.25$

= = =

I'm not seeing the TeX rendering on any of my browsers. Maybe it's a site-wide issue? In any case, I'm attaching a picture of the text above in case it's incomprehensible.

5 0

3 years ago