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3 years ago

Can someone explain what's n+1-3n=-5+n+1-2

2 answers:

8 0

N+1-3n=5+n+1-2

Simplify it

-2n+1 = n+4

Add 2n to both sides

1 = 3n+4

Subtract 4 from both sides

-3 = 3n

Divide by 3

n = -1

trapecia [35]3 years ago

5 0

N + 1 - 3n = -5 + n + 1 - 2

Combine like terms:

n - 3n = -2n

-2n + 1 = -5 + n + 1 - 2

Combine like terms:

-5 + 1 - 2 = -6

-2n + 1 = n - 6

Subtract 'n' to both sides:

-3n + 1 = -6

Subtract 1 to both sides:

-3n = -7

Divide -3 to both sides:

n = 2 1/3 or 7/3

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4 years ago

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Ulleksa [173]

Step-by-step explanation:

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2 years ago

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

If y varies directly as the x^2 and y = 200 when x = 5, what is the value of y when x = -3?

gayaneshka [121]

As the variation is direct we have:
$y = kx ^ 2
$
We must find the value of k.
For this, we use the following data:
y = 200 when x = 5
Substituting values we have:
$200 = k5 ^ 2
$
Clearing k:
$k = 200/5 ^ 2

k = 200/25

k = 8$
Then, the function is:
$y = 8x ^ 2
$
We evaluate the function for x = -3
$y = 8 (-3) ^ 2

y = 8 (9)

y = 72$
Answer:
the value of y when x = -3 is:
c.
72

6 0

4 years ago

1. Solve and show your work for each question.

DedPeter [7]

Answer:

A) 0.36 with a line over both 36 means as a decimal the 36 repeats, so it would look like 0.363636

as a fraction it would be 36/99, divide both numbers by 9 to get 4/11

B) 0.36 with the line over the 6 means just the 6 repeats so it would look like 0.3666

as a fraction that would be 33/90 , divide both by 3 to get 11/30

C) 0.36 would be 36/100, divide both numbers by 4 to get 9/25

Stainep-by-step explain

8 0

3 years ago