Answer:
One answer because it is not a repeating fraction/decimal.
Step-by-step explanation:
Answer:
Step-by-step explanation:
Assume that f(x) = 0 for x outside the interval [4,9]. We will use the following
![E[X^k] = \int_{4}^{9}x^k f(x) dx](https://tex.z-dn.net/?f=E%5BX%5Ek%5D%20%3D%20%5Cint_%7B4%7D%5E%7B9%7Dx%5Ek%20f%28x%29%20dx)
![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 = ![E[X]](https://tex.z-dn.net/?f=E%5BX%5D)
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:
-3/2
Step-by-step explanation:
the rise (change in y) is -3 and the run (change in x) is 2
rise over run
-3/2
Step-by-step explanation:
6x-12=12+5x
6x - 5x = 12 + 12
x = 24
Answer:
y=-3/4x+17/2
Step-by-step explanation:
the equation for slope intercept form is y=mx+b, where m is the slope and b is the y intercept.
since we know the slope of the line, we can substitute it into the equation
y=-3/4x+b
we need to find b though.
since the equation will pass through the point (6,4) that means that if we plug it into the equation, it'll be a true statement; in this case, we can use it to solve for b.
substitute (6,4) into the equation
4=-3/4(6)+b
multiply
4=-9/2+b
17/2=b
the y intercept is 17/2
therefore the equation of the line is y=-3/4x+17/2
hope this helps :)
