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J= obtuse
K=right
L=acute
M=acute
Answer:
Step-by-step explanation:
Assume that f(x) = 0 for x outside the interval [4,9]. We will use the following
![Var(X) = E[X^2}- (E[X])^2](https://tex.z-dn.net/?f=Var%28X%29%20%3D%20E%5BX%5E2%7D-%20%28E%5BX%5D%29%5E2)
Standard deviation = ![\sqrt[]{Var(X)}](https://tex.z-dn.net/?f=%20%5Csqrt%5B%5D%7BVar%28X%29%7D)
Mean =
Then,
![E[X] = \int_{4}^{9}\frac{1}{5}dx = \frac{9^2-4^2}{2\cdot 5} = \frac{13}{2}](https://tex.z-dn.net/?f=E%5BX%5D%20%3D%20%5Cint_%7B4%7D%5E%7B9%7D%5Cfrac%7B1%7D%7B5%7Ddx%20%3D%20%5Cfrac%7B9%5E2-4%5E2%7D%7B2%5Ccdot%205%7D%20%3D%20%5Cfrac%7B13%7D%7B2%7D)
![E[X^2] = \int_{4}^{9}\frac{x^2}{5}dx = \frac{9^3-4^3}{3\cdot 5} = \frac{133}{3}](https://tex.z-dn.net/?f=E%5BX%5E2%5D%20%3D%20%5Cint_%7B4%7D%5E%7B9%7D%5Cfrac%7Bx%5E2%7D%7B5%7Ddx%20%3D%20%5Cfrac%7B9%5E3-4%5E3%7D%7B3%5Ccdot%205%7D%20%3D%20%5Cfrac%7B133%7D%7B3%7D)
Then, 
Then the standard deviation is ![\frac{5}{2\sqrt[]{3}}](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B2%5Csqrt%5B%5D%7B3%7D%7D)
Answer:
1/4
Step-by-step explanation:
Hope this . helped :) can i have brainliest