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3 years ago

Cho f(t)=(t-4)u(t-2)phep bien doi laplace la F(s)=e^-2s/s^2-2e^-2s/s62

Mathematics

1 answer:

FinnZ [79.3K]3 years ago

4 0

Step-by-step explanation:

cho f(t)=(t-4)u(t-2)phep bien doi laplace la

F(s)=e^-2s/s^2-2e^-2s/s62

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A player chooses one of the numbers 1 through 4. After the choice has been made, two regular four-sided (tetrahedral) dice are r

d1i1m1o1n [39]

Answer: $=\frac{-1}{16}$

Step-by-step explanation:

given data:

number to be chosen from = 1,2,3,4.

<u><em>Solution:</em></u>

ways the number can show

$2*3*1=6$ ways for your number to appear once.

$1*1=1$ ways for your number to appear twice.

$3*3=9$ ways for your number not to appear at all

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$= \frac{6+2-9}{16}$

$=\frac{-1}{16}$

expected return of the player is $=\frac{-1}{16}$

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3 years ago

Find the generating function for the sequence 1,-2,4,-8, 16, ...

kirill [66]

Answer:

$a(x)=\dfrac{1}{1+2x}$

Step-by-step explanation:

The generating function a(x) produces a power series ...

$a(x)=a_0+a_1x+a_2x^2+a_3x^3+\dots$

where the coefficients are the elements of the given sequence.

We observe that the given sequence has the recurrence relation ...

$a_0=1;a_n=-2a_{n-1} \quad\text{for n $>$ 0}$

This can be rearranged to ...

$a_n+2a_{n-1}=0$

We can formulate this in terms of a(x) as follows, then solve for a(x).

$\sum\limits^{\infty}_{n=1} {a_{n}x^n} =a(x)-a_0 \quad\text{and}\\\\\sum\limits^{\infty}_{n=1} {2a_{n-1}x^n} =(2x)a(x) \quad\text{so}\\\\\sum\limits^{\infty}_{n=1} {(a_n+2a_{n-1})x^n}=0=a(x)-a_0+2xa(x)\\\\a(x)=\dfrac{a_0}{1+2x}=\dfrac{1}{1+2x}$

The generating function is ...

a(x) = 1/(1+2x)

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