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3 years ago

2(x-1) + 3x = 3 Can you help me

Mathematics

2 answers:

Tcecarenko [31]3 years ago

5 0

Answer:

Step-by-step explanation:

2x-2+3x=3

5x-2=3

5x=5

x=1

abruzzese [7]3 years ago

3 0

The answer is 20x23=45 hope this helps byeeee

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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

3 years ago

5.44nL to mL convert

aniked [119]

$5.44nl \times \frac{1ml}{1000000nl} = 0.00000544ml$

6 0

4 years ago

Please help i attached it

lapo4ka [179]

Answer:

C. plumber A has a greater initial value

4 0

2 years ago

For the pair of points find the distance between them and the midpoint of the line segment joining them.(720,50), (125,18)The di

Lana71 [14]

Given two points

$(x_1,y_1)$

and

$(x_2,y_2)$

The distance between them is >>>

$D=\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2}$

The points given are (Sqrt(20), Sqrt(50)) and (Sqrt(125), Sqrt(8)), so their distance is >>>

$\begin{gathered} D=\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2} \\ D=\sqrt[]{(\sqrt[]{8}-\sqrt[]{50})^2+(\sqrt[]{125}-\sqrt[]{20})^2} \\ D=\sqrt[]{(\sqrt8)^2-2(\sqrt[]{8})(\sqrt[]{50})+(\sqrt[]{50})^2^{}+(\sqrt[]{125})^2-2(\sqrt[]{125})(\sqrt[]{20})+(\sqrt[]{20})^2} \\ D=\sqrt[]{8-2(2\sqrt[]{2})(5\sqrt[]{2})+50+125-2(5\sqrt[]{5})(2\sqrt[]{5})+20} \\ D=\sqrt[]{8-40+50+125-100+20} \\ D=\sqrt[]{63} \\ D=3\sqrt[]{7} \end{gathered}$

----------------------------------------------------------------------------------------------------------

The midpoint formula is >>>

$M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Plugging in the points, we have >>>

$\begin{gathered} M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \\ M=(\frac{\sqrt[]{20}+\sqrt[]{125}}{2},\frac{\sqrt[]{50}+\sqrt[]{8}}{2}) \\ M=(\frac{2\sqrt5+5\sqrt[]{5}}{2},\frac{5\sqrt[]{2}+2\sqrt[]{2}}{2}) \\ M=(\frac{7\sqrt[]{5}}{2},\frac{7\sqrt[]{2}}{2}) \end{gathered}$

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1 year ago

An elevator descends 1,000 feet in 8 seconds what is the change in height per second

Svetlanka [38]

You multiple 1000 by 8 and have tour answer which is 8000 per second

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3 years ago