The expected value is
![E[X]=\displaystyle\sum_xx\,P(X=x)=\frac1{11}\sum_{x=0}^{10}x=\dfrac{0+1+\cdots+9+10}{11}=\dfrac{55}{11}=\boxed{5}](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Csum_xx%5C%2CP%28X%3Dx%29%3D%5Cfrac1%7B11%7D%5Csum_%7Bx%3D0%7D%5E%7B10%7Dx%3D%5Cdfrac%7B0%2B1%2B%5Ccdots%2B9%2B10%7D%7B11%7D%3D%5Cdfrac%7B55%7D%7B11%7D%3D%5Cboxed%7B5%7D)
The standard deviation is the square root of the variance, which is
![V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2](https://tex.z-dn.net/?f=V%5BX%5D%3DE%5B%28X-E%5BX%5D%29%5E2%5D%3DE%5BX%5E2%5D-E%5BX%5D%5E2)
where
![E[X^2]=\displaystyle\sum_xx^2\,P(X=x)=\frac1{11}\sum_{x=0}^{10}x^2=\dfrac{0^2+1^2+\cdots+9^2+10^2}{11}=\dfrac{385}{11}=35](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%5Cdisplaystyle%5Csum_xx%5E2%5C%2CP%28X%3Dx%29%3D%5Cfrac1%7B11%7D%5Csum_%7Bx%3D0%7D%5E%7B10%7Dx%5E2%3D%5Cdfrac%7B0%5E2%2B1%5E2%2B%5Ccdots%2B9%5E2%2B10%5E2%7D%7B11%7D%3D%5Cdfrac%7B385%7D%7B11%7D%3D35)
so that
![V[X]=35-5^2=10](https://tex.z-dn.net/?f=V%5BX%5D%3D35-5%5E2%3D10)
making the standard deviation
![\sqrt{V[X]}=\sqrt{10}\approx\boxed{3.16}](https://tex.z-dn.net/?f=%5Csqrt%7BV%5BX%5D%7D%3D%5Csqrt%7B10%7D%5Capprox%5Cboxed%7B3.16%7D)
Answer:
it is x=27.if im wrong im sorry im 10 and in collage. im smart but i have a lot of work to do.
Step-by-step explanation:I have a lot of thinges on my mind.
Answer:
9
Step-by-step explanation:
just plugged in 4 for x
y=1.5(4)+3
hope this helps and is right haha <3
Millimeter. 1mm= 1mL, 100mm= 100mL, 1000mm = 1000mL / 1L