The equality used is Subtraction property of equality
Option A is correct
Step-by-step explanation:
We need to identify which option best explains or justifies Step 1.
Step 1 is: 
The given equation is: 
To get the equation for step 1 we need to subtract c on both sides of equation i.e using subtraction property of equality


So, The equality used is Subtraction property of equality
Option A is correct.
Keywords: Solving Quadratic Equations
Learn more about Solving Quadratic Equations at:
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2x+y=9
3x+5y=19
I will do this problem in 2 ways. I.)Substitution II.)Elimination
Solution I.) Substitution
We can subtract 2x from both sides in the first equation.
y=9-2x
Now we can substitute the y in the second equation with 9-2x
3x+5(9-2x)=19
-7x+45=19
-7x=-26
x=26/7
y=9-2(26/7)=11/7
Solution II.)Elimination
We can multiply both side of first equation by 5 to get a 5y in both equations.
10x+5y=45
Now because both are positive 5y we just need to do simple subtraction of the 2 equation, each side respectively.
(10x+5y)-(3x+5y)=45-19
7x=26
x=26/7
2*26/7+y=9
y=11/7
Ultimately you get the same answer, both are viable methods, some problems are faster with one method but I recommend mastering both since they are very useful.
Answer:
Simplified: 4X^3
Step-by-step explanation:
Simplify the expression.
Answer:
10
Step-by-step explanation:
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:
![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:
