This is the question Below.

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.

5 0

3 years ago

Round 222,977 to the nearest hundred

erastovalidia [21]

The place the value of 9 is in is the hundreds place. So, any number greater than 5 is rounded up. 7 is greater than 5, so the 9 becomes a 0 and the next number is increased by 1. All the values after are marked as a 0. The answer is 223,000.

4 0

3 years ago

Read 2 more answers

Peterkin park has a square fountain with a walkway aroind it. The fountain measures 12 feet on each side. The walkway is 3 1/2 f

Black_prince [1.1K]

Answer: 829.44 ft²

Step-by-step explanation:

Correction - The walkway is 31.2 feet wide.

To solve this, first find the area of the fountain:

Area of a square = Length * Width

= 12 * 12

= 144 ft²

The find the area of the walkway including the fountain:

= 31.2 * 31.2

= ‭973.44‬ ft²

To find the area of the walkway alone, subtract the area of the square from the area of the rectangle:

= 973.44 - 144

= 829.44 ft²

3 0

3 years ago

A shoreline is eroding at a rate of 1/3 foot per 2 months how many feet are eroding this year

mylen [45]

Answer:

2 feet

Step-by-step explanation:

The shoreline is eroding at 1/3 foot per 2 months. There are 12 months in a year so you multiply 1/3 x 6 and you get 2.

8 0

2 years ago

Answer all for brainliest

sergey [27]

Answer: 1/10 is greater; niether its equal; 2 1/2, .25, -.2, -1/2, -4/5

Step-by-step explanation:

1/10 is a tenth and .09 is a hunderedth

7 0

3 years ago