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sveticcg [70]
3 years ago
9

Can someone please answer. There is one problem. There's a picture. Thank you!

Mathematics
1 answer:
AveGali [126]3 years ago
3 0
Using SOH, the answer is O/H therefore is b/a.
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4. Dean Pelton wants to perform calculations to impress the accreditation consultants, but upon asking for information about GPA
Leya [2.2K]

Answer:

the minimum records to be retrieved by using Chebysher - one sided inequality is 17.

Step-by-step explanation:

Let assume that n should represent the number of the students

SO, \bar x can now be the sample mean of number of students  in GPA's

To obtain n such that P( \bar x \leq 2.3 ) \leq .04

⇒ P( \bar x \geq 2.3 ) \geq .96

However ;

E(x) = \int\limits^4_2 Dx (2+e^{-x} ) 4x = D  \\ \\ = D(e^{-x} (e^xx^2 - x-1 ) ) ^D_2 = 12.314 D

E(x^2) = D\int\limits^4_2 (2+e^{-x})dx \\ \\ = \dfrac{D}{3}[e^{-4} (2e^x x^3 -3x^2 -6x -6)]^4__2}}= 38.21 \ D

Similarly;

D\int\limits^4_2(2+ e^{-x}) dx = 1

⇒ D*(2x-e^{-x} ) |^4_2 = 1

⇒ D*4.117 = 1

⇒ D= \dfrac{1}{4.117}

\mu = E(x) = 2.991013 ; \\ \\ E(x^2) = 9.28103

∴  Var (x)  = E(x^2) - E^2(x) \\ \\  = .3348711

Now; P(\bar \geq 2.3) = P( \bar x - 2.991013 \geq 2.3 - 2.991013) \\ \\ = P( \omega  \geq .691013)  \ \ \ \  \ \ \ \ \ \ (x = E(\bar x ) - \mu)

Using Chebysher one sided inequality ; we have:

P(\omega \geq -.691013) \geq \dfrac{(.691013)^2}{Var ( \omega) +(.691013)^2}

So; (\omega = \bar x - \mu)

⇒ E(\omega ) = 0 \\ \\ Var (\omega ) = \dfrac{Var (x_i)}{n}

∴ P(\omega \geq .691013) \geq \dfrac{(.691013)^2}{\frac{.3348711}{n}+(691013)^2}

To determine n; such that ;

\dfrac{(.691013)^2}{\frac{.3348711}{n}+(691013)^2} \geq 0.96 \\ \\ \\ (.691013)^2(1-.96) \geq \dfrac{-3348711*.96}{n}

⇒ n \geq \dfrac{.3348711*.96}{.04*(.691013)^2}

n \geq 16.83125

Thus; we can conclude that; the minimum records to be retrieved by using Chebysher - one sided inequality is 17.

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Step-by-step explanation:

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Step-by-step explanation:

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