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3 years ago

What is the midpoint of line segment EF? O (2,2) O (-3,-2) O (-2,1) • (-4,2)

Mathematics

2 answers:

natima [27]3 years ago

6 0

Answer:

(-2, 1)

Step-by-step explanation:

In image

Hope this helps!

Love you.

- doomdabomb

dimaraw [331]3 years ago

3 0

E which is (-5,3) and F is (1,-1)
You add the X’s and Y’s and diced them by 2
-5+1 3+(-1)
——— , ——- = (-2,1)
2 2

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Answer:

Step-by-step explanation:

From the given information:

The uniform distribution can be represented by:

$f_x(x) = \dfrac{1}{1500} ; o \le x \le \ 1500$

The function of the insurance is:

$I(x) = \left \{ {{0, \ \ \ x \le 250} \atop {x -20 , \ \ \ \ \ 250 \le x \le 1500}} \right.$

Hence, the variance of the insurance can also be an account forum.

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here;

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$E[I(x)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250) \ dx$

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Similarly;

$E[I^2(x)] = \int f_x(x) I^2 (x) \ sx$

$E[I(x)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250)^2 \ dx$

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∴

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Finally, the standard deviation of the insurance payment is:

$= \sqrt{Var(I(x))}$

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