20 points to help with this

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

2 years ago

What is the reactance of a 25 mF capacitor when the applied frequency is 400 Hz?

LuckyWell [14K]

The Reactance of a Capacitor is t<u>he resistance it has when alternating current (A/C) with a specific frequency (f) passes through it. </u>This is measured in ohms (Ω) and its formula is:

$X_{C}=\frac{1}{2\pi fC}$ (1)

Where:

$X_{C}$ is the Reactance in ohms (Ω)

$f$ is the frequency of the alternating current in Hertz ( $Hz$ )

$C$ is the Capacitance of the Capacitor in Farad ( $F$ )

In this case we have a 25 mF capacitor with an applied frequency of 400 Hz.

Note $1mF=1*10^{-3}F=0.001F$

Well, with the given data we have to solve (1):

$X_{C}=\frac{1}{2\pi(400Hz)(25*10^{-3}F)}$

Then:

XC=0.0159 Ω This is the Reactance of the Capacitor

4 0

2 years ago

Pleaseee helppp mmeeee!

nexus9112 [7]

Answer:

They are both 45 degrees

Step-by-step explanation:

The square in the middle means a right angle (90) and 5 and 6 are equal too so they both equal 45

6 0

2 years ago

Can someone explain to me how to remove all the perfect squares ? I’m kind of confused and my teacher is unavailable to help and

mixer [17]

Answer:

b^2√30b

Step-by-step explanation:

√30b^5 - Original

Break down the radical with its factors (factor tree)

√3*2*5*b*b*b*b*b

See what perfect squares you can take out (Can take out four b terms) You cant take out any perfect squares involving 30 since there are no perfect squares

b^2√3*2*5*b

b^2√30b

4 0

1 year ago

We probably couldn't go to all the stars because they're too far away. It would take a long time to get there.

Dvinal [7]

Answer:

The correct answer is B. Contain generalizations and contractions.

Step-by-step explanation:

A generalization occurs when a specific idea is applied to a general category, in simple terms it occurs if a person states something always occur because this person saw it once. In this case, generalization occurs as this person affirms all stars are too far away and thus it will take a long time to get to those stars, however not all stars are as far away as this person believes and this does not mean it will take a long time to get there as new technology can be created, thus, this statement is not true and need revision. On the other hand, contractions refers to the process of shortening words by putting them together, this can be seen in "couldn't" ( could + not) and "they're" (they + are), contradictions are not correct as they should not be used in formal language and be avoided, thus, contradictions also need to be revised.

4 0

2 years ago