(y+6)^2-(y-2)^2 I got 16y+26 but it is wrong?

Mathematics

1 answer:

Gemiola [76]2 years ago

5 0

Answer:

16y + 32

Step-by-step explanation:

Expand each term.

(y+6)² - (y-2)²

= (y+6)(y+6) - (y-2)(y-2)

= y² + 12y + 36 - (y² - 4y + 4)

Subtract the second group by changing each term's signs

= y² + 12y + 36 - y² + 4y - 4

Collect like terms

= 16y + 32

You might be interested in

A fair coin is tossed repeatedly with results Y0, Y1, Y2, . . . that are 0 or 1 with probability 1/2 each. For n ≥ 1 let Xn = Yn

Gekata [30.6K]

Answer:

False. See te explanation an counter example below.

Step-by-step explanation:

For this case we need to find:

$P(X_{n+1} = | X_n =i, X_{n-1}=i') =P(X_{n+1}=j |X_n =i)$ for all $i,i',j$ and for $X_n$ in the Markov Chain assumed. If we proof this then we have a Markov Chain

For example if we assume that $j=2, i=1, i'=0$ then we have this:

$P(X_{n+1} = | X_n =i, X_{n-1}=i') =\frac{1}{2}$

Because we can only have $j=2, i=1, i'=0$ if we have this:

$Y_{n+1}=1 , Y_n= 1, Y_{n-1}=0, Y_{n-2}=0$ , from definition given $X_n = Y_n + Y_{n-1}$

With $i=1, i'=0$ we have that $Y_n =1 , Y_{n-1}=0, Y_{n-2}=0$

So based on these conditions $Y_{n+1}$ would be 1 with probability 1/2 from the definition.

If we find a counter example when the probability is not satisfied we can proof that we don't have a Markov Chain.

Let's assume that $j=2, i=1, i'=2$ for this case in order to satisfy the definition then $Y_n =0, Y_{n-1}=1, Y_{n-2}=1$

But on this case that means $X_{n+1}\neq 2$ and on this case the probability $P(X_{n+1}=j| X_n =i, X_{n-1}=i')= 0$ , so we have a counter example and we have that:

$P(X_{n+1} =j| X_n =i, X_{n-1}=i') \neq P(X_{n+1} =j | X_n =i)$ for all $i,i', j$ so then we can conclude that we don't have a Markov chain for this case.