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2 years ago

Solve for x if 523x = 523. please

Mathematics

2 answers:

Olenka [21]2 years ago

8 0

Answer:

x = 1

Step-by-step explanation:

523x = 523

Divide each side by 523 to get the x alone

523/523 = 1

vfiekz [6]2 years ago

7 0

Answer:

x=1

Step-by-step explanation:

To get x by itself you need to divide 523 by the other side of the equal sign so divide 523 by 523 which is 1.

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Lelu [443]

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Step-by-step explanation:

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Simora [160]

Using it's formula, it is found that the mean of the discrete random variable is given by:

B. 30.47.

<h3>What is the mean of a discrete distribution?</h3>

The expected value of a discrete distribution is given by the <u>sum of each outcome multiplied by it's respective probability</u>.

Hence, considering the table, the mean of the discrete distribution is:

E(X) = 23 x 0.16 + 25 x 0.09 + 26 x 0.18 + 31 x 0.12 + 34 x 0.24 + 38 x 0.21 = 30.47.

Hence option B is correct.

More can be learned about the mean of a discrete random variable at brainly.com/question/26660401

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2 years ago

(b) dy/dx = (x - y+ 1)^2

Elanso [62]

Substitute $v(x)=x-y(x)+1$ , so that

$\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}$

Then the resulting ODE in $v(x)$ is separable, with

$1-\dfrac{\mathrm dv}{\mathrm dx}=v^2\implies\dfrac{\mathrm dv}{1-v^2}=\mathrm dx$

On the left, we can split into partial fractions:

$\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx$

Integrating both sides gives

$\dfrac{\ln|1-v|+\ln|1+v|}2=x+C$

$\dfrac12\ln|1-v^2|=x+C$

$1-v^2=e^{2x+C}$

$v=\pm\sqrt{1-Ce^{2x}}$

Now solve for $y(x)$ :

$x-y+1=\pm\sqrt{1-Ce^{2x}}$

$\boxed{y=x+1\pm\sqrt{1-Ce^{2x}}}$

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