<span>I think you meant this: (3m/m-6)*(5m^3-30m^2/5m^2)
If that's the case, the simplified form:
=3m</span>
Answer:
Pov:شما به ترجمه گوگل رفت برای دیدن آنچه من گفتم و دیدم این xD
Step-by-step explanation:
Pov:شما به ترجمه گوگل رفت برای دیدن آنچه من گفتم و دیدم این xD
Answer:
8
Step-by-step explanation:
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From the graph of the given function , the value of f(1) = -1.
As given in the question,
From the graph of the given function,
Two coordinates from the graph are as follow:
( x₁ , y₁) = (1, -1)
( x₂ , y₂ ) = ( 0, -3 )
Equation of the line representing the function is given by:
(y - y₁) /(x-x₁) = ( y₂ -y₁)/ (x₂ -x₁)
⇒(y +1)/ (x-1) = (-3 +1)/ (0-1)
⇒ (y +1)/ (x-1) = 2
⇒y +1 = 2x -2
⇒ y = 2x -3
To get the value of x we have,
y = f(x)
⇒f(x) = 2x -3
⇒f(1) = 2(1) -3
⇒f(1) = -1
Therefore, from the graph of the given function , the value of f(1) = -1.
Learn more about graph here
brainly.com/question/17267403
#SPJ1
Answer:
In order to find the variance we need to calculate first the second moment given by:
And the variance is given by:
![Var(X) = E(X^2) +[E(X)]^2 = 23.36 -[4.74]^2 = 0.8924](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20%2B%5BE%28X%29%5D%5E2%20%3D%2023.36%20-%5B4.74%5D%5E2%20%3D%200.8924)
And the deviation would be:
Step-by-step explanation:
Previous concepts
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
Solution to the problem
For this case we have the following distribution given:
X 3 4 5 6
P(X) 0.07 0.4 0.25 0.28
We can calculate the mean with the following formula:
In order to find the variance we need to calculate first the second moment given by:
And the variance is given by:
And the deviation would be:
