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2 years ago

2+y-8=19 what is the value of y

1 answer:

8 0

The value of y would be 25. What you would do is add your 8 on both sides of the equal sign. After doing this your equation should look like, 2+y=27. Next you would subtract 2 on both sides of your equal sign and get 25. Your finishing equation should look like y=25.
Hope this helps! (:

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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.

5 0

2 years ago

What is the value of X

xxMikexx [17]

Answer:

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

6 high school seniors choose from among 20 quotes for their yearbook. What is the probability that at least 2 of them choose the

shusha [124]

Using the binomial distribution, it is found that there is a 0.0328 = 3.28% probability that at least 2 of them choose the same quote.

<h3>What is the binomial distribution formula?</h3>

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem, we have that:

There are 6 students, hence n = 6.

There are 20 quotes, hence the probability of each being chosen is p = 1/20 = 0.05.

The probability of one quote being chosen at least two times is given by:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

P(X < 2) = P(X = 0) + P(X = 1).

Then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.(0.05)^{0}.(0.95)^{6} = 0.7351$

$P(X = 1) = C_{6,1}.(0.05)^{1}.(0.95)^{5} = 0.2321$

Then:

P(X < 2) = P(X = 0) + P(X = 1) = 0.7351 + 0.2321 = 0.9672.

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.9672 = 0.0328$

0.0328 = 3.28% probability that at least 2 of them choose the same quote.

More can be learned about the binomial distribution at brainly.com/question/24863377

6 0

1 year ago

Please helpp!!!!

fomenos

Answer:

Dom(gof)=Dom(f)={1,3,4}.

(gof)(1)=g{f(1)}=g(2)=3.

(gof)(3)=g{f(3)}=g(5)=1.

(gof)(4)=g{f(4)}=g(1)=3.

∴gof={(1,3),(3,1),(4,3)}.

7 0

2 years ago

Please help

Airida [17]

no, it is constant for a while then changes at 90gallons

6 0

2 years ago