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Answer:
<h2>x = 5 or x = -2</h2>Step-by-step explanation:
Look at the photo below for the details.
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1 answer:
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Answer:
Step-by-step explanation:
Previous concepts
In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".
The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).
And the standard deviation of a random variable X is just the square root of the variance.
Solution to the problem
For this case we can calculate the properties required with the following table:
Interval Mid point (x) f x*f x^2 *f
_________________________________________
1-10 5.5 40 220 1210
11-20 15.5 15 232.5 3603.75
21-30 25.5 23 586.5 14955.75
>31 35.5 10 355 12602.5
________________________________________
Total 88 1394 32372
We assume that the mid point for the class >31 is 35.5 using the problem information.
For this case the expected value would be given by:
The variance owuld be given by this formula"
And if we replace we got:
The standard deviation would be just the square root of the variance:
Answer:
(x + 4)² + (y - 2)² = 8
Step-by-step explanation:
The equation of a circle in standard form is
(x - h)² + (y - k)² = r²
where (h, k) are the coordinates of the centre and r is the radius
Here (h, k ) = (- 4, 2 ) , then
(x - (- 4))² + (y - 2)² = r²
(x + 4)² + (y - 2)² = r²
CA is the radius of the circle
To find r use the distance formula
r =
with (x₁, y₁ ) = C(- 4, 2) and (x₂, y₂ ) = A(- 2, 4)
r =
=
= , then r² = (
)² = 8
Then
(x + 4)² + (y - 2)² = 8 ← equation of circle
Answer:
(3, 0)
Step-by-step explanation:
Given the inequality y > -2x + 3 and y ≤ x - 2
The graph of the inequalities are plotted using the geogebra graphing online calculator.
The portion of the graph that is shaded with dark blue, represents the portion that supports the equation.
All the ordered pair points given in the question are also labelled in the graph.
From the graph we can see that only point (3, 0) falls in the area that supports the equation. Hence (3, 0) makes both inequalities true.
