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3 years ago

7

Find the solution of the given initial value problem. y(4) + 2y''' + y'' + 8y' − 12y = 12 sin t − e−t; y(0) = 7, y'(0) = 0, y''(

0) = −1, y'''(0) = 4 y(t) =

1 answer:

Allisa [31]3 years ago

4 0

Consider, please, this option.
Note, 1. check was not performed, 2. make design according the local requirements.

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12. Kim bought a poster that cost $8.95 and

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3 years ago

suppose x is a non-negative integer valued random variable which follows a probability distribution such that p x=0 = 06 and for

Andre45 [30]

Not sure if the given probability that $X=0$ is 0.06 or 0.6, or something else altogether. I'll just refer to it by the number $p$ .

$\begin{cases}P(X=0)=p\\P(X=i+1)=\frac15P(X=i)&\text{for }i\ge1\end{cases}$

As is, the probability that $X=1$ is indeterminate, so I think you intended to write

$\begin{cases}P(X=0)=p\\P(X=i+1)=\frac15P(X=i)&\text{for }i\ge\boxed{0}\end{cases}$

Then by this definition,

$i=0\implies P(X=1)=\dfrac15P(X=0)=\dfrac p5$

$i=1\implies P(X=2)=\dfrac15P(X=1)=\dfrac p{5^2}$

and so on.

Then

$\boxed{P(X\le2)=P(X=0)+P(X=1)+P(X=2)=p+\dfrac p5+\dfrac p{5^2}=\dfrac{31p}{25}}$

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4 years ago