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MA_775_DIABLO [31]

3 years ago

9

A set of data with a mean of 45 and a standard deviation of 8.3 is normally distributed. Find the value that is +1 standard devi

ation away from the mean.

2 answers:

Anuta_ua [19.1K]3 years ago

7 0

Answer:

53.3

Step-by-step explanation:

Add one standard deviation to the mean:

45 +1(8.3) = 53.3

Sliva [168]3 years ago

4 0

Answer:

53.3

Step-by-step explanation:

+1 standard deviation away from the mean is 45+1(8.3)

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Use separation of variables to solve dy dx − tan x = y2 tan x with y(0) = √3. Find the value of c in radians, not degrees

a_sh-v [17]

Answer:

$y(x)=tan(-log(cos(x))+\frac{\pi }{3} )$

Step-by-step explanation:

Rewrite the equation as:

$\frac{dy(x)}{dx}-tan(x)=y(x)^{2} *tan(x)$

Isolating $\frac{dy}{dx}$

$\frac{dy}{dx} =tan(x)+tan(x)*y^{2}$

Factor:

$\frac{dy}{dx} =tan(x)*(1+y^{2} )$

Dividing both sides by $(1+y^{2} )$ and multiplying them by $dx$

$\frac{dy}{1+y^{2} } =tan(x)dx$

Integrate both sides:

$\int\ \frac{dy}{1+y^{2} } = \int\ tan(x) dx$

Evaluate the integrals:

$arctan(y)=-log(cos(x))+C_1$

Solving for y:

$y(x)=tan(-log(cos(x))+C_1)$

Evaluating the initial condition:

$y(0)=\sqrt{3} =tan(-log(cos(0))+C_1)=tan(-log(1)+C_1)=tan(0+C_1)$

$\sqrt{3} =tan(C_1)\\arctan(\sqrt{3} )=C_1\\60=C_1$

Converting 60 degrees to radians:

$60degrees*\frac{\pi }{180degrees} =\frac{\pi }{3}$

Replacing $C_1$ in the diferential equation solution:

$y(x)=tan(-log(cos(x))+\frac{\pi }{3} )$

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dmitriy555 [2]

Answer:

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Step-by-step explanation:

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