Recall that the PDF is given by the derivative of the CDF:
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](https://tex.z-dn.net/?f=%5Cmathbb%20E%5BX%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20x%5C%2Cf_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cint_0%5E2%5Cleft%28x-%5Cdfrac%7Bx%5E2%7D2%5Cright%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac23)
The median is the number
such that
. We have
but both roots can't be medians. As a matter of fact, the median must satisfy
, so we take the solution with the negative root. So
is the median.
Answer:
1/5
Step-by-step explanation:
Up 1, over 5
Answer:
that is the solution to the question
i believe u r confused about the 5 tanks because 18 can not fit into 5. so here is what it would accurately be.
to find out how many tanks you need your going to need to find the least common multiple. the least common multiple for both are below
18: 1, 2, 3, 6, 9, 18
24:1, 2, 3, 4, 6, 8, 12, 24
the least common multiple is 3. you will need three tanks. now for part b of this equation, to find out how many of each need to be put into the tank. to find this out you need to divide 18 plant and 24 fish by 3 tanks.
18/3 = 6
24/3 = 8
6 plants will go in each tank and 8 fish will go in each tank
the ratio is 6:8 per tank
Your equations would be
c+d=10
5c+6d=56 (c= # of cats, d= # of dogs) Then solve
I am using elimination and multiply the first equation by -5
-5c-5d=-50
5c+6d=56
Then eliminate
-d=-6 Divide by -1 d=6
Then solve for c
c+6=10
c=4
Hope this helps!
