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2 years ago

The probability distribution for a

Mathematics

1 answer:

Yanka [14]2 years ago

5 0

There are only two possible outcomes in the table that satisfy x ≤ -3; either x = -5 or x = -3. Then

P(x ≤ -3) = P(x = -5) + P(x = -3) = 0.17 + 0.13 = 0.3

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koban [17]

$\dfrac{\mathrm dy}{\mathrm dt}=-4x\implies x=-\dfrac14\dfrac{\mathrm dy}{\mathrm dt}\implies\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac14\dfrac{\mathrm d^2y}{\mathrm dt^2}$

Substituting this into the other ODE gives

$-\dfrac14\dfrac{\mathrm d^2y}{\mathrm dt^2}=-4y\implies y''-16y=0$

Since $x(t)=-\dfrac14y'(t)$ , it follows that $x(0)=-\dfrac14y'(0)=4\implies y'(0)=-16$ . The ODE in $y$ has characteristic equation

$r^2-16=0$

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$y_c=C_1e^{4t}+C_2e^{-4t}$

From the initial conditions we get

$y(0)=5\implies 5=C_1+C_2$

$y'(0)=16\implies-16=4C_1-4C_2$

$\implies C_1=\dfrac12,C_2=\dfrac92$

So we have

$\boxed{y(t)=\dfrac12e^{4t}+\dfrac92e^{-4t}}$

Take the derivative and multiply it by -1/4 to get the solution for $x(t)$ :

$-\dfrac14y'(t)=\boxed{x(t)=-\dfrac12e^{4t}+\dfrac92e^{-4t}}$

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