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Gemiola [76]
3 years ago
7

Please help and explain

Mathematics
1 answer:
Alja [10]3 years ago
7 0

corn = 15%

angle = (15/100)×360

= 54°

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What is the area of the parallelogram?
stira [4]

Answer:

a=bh

Step-by-step explanation:

the answer is 70

im sorry if thats not right

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3 years ago
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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.

6 0
3 years ago
Help me out<br>........1​
Fiesta28 [93]

Answer: -13

Step-by-step explanation:

2-(3^2)+4-3-7 = -13

6 0
2 years ago
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For what value of k does the equation 6(x + 1) + 2 = 3(k5x + 1) + 3 have no solution?
NARA [144]

Answer:

k = (6/15)

Step-by-step explanation:

The equation is:

6*(x + 1) + 2 = 3*(k*5*x + 1) + 3

To have no solutions, we need to have something like:

x + 7 = x + 4

where we can remove x in both sides and end with

7 = 4

So this equation is false, meaning that there is no value of x such that this equation is true, then the equation has no solutions.

First, let's try to simplify our equation:

6*(x + 1) + 2 = 3*(k*5*x + 1) + 3

6*x + 6 + 2 = 3*k*5*x + 3*1 + 3

6*x + 8 = 15*k*x + 6

if 15*k = 6, then the system clerly has no solution.

then:

k = 6/15

then we get:

6*x + 8 = (6/15)*15*x + 6

6*x + 8 = 6*x + 6

8 = 6

The system has no solutions.

8 0
2 years ago
What is the factor of z^3+64
Leto [7]
I hope this helps you

8 0
4 years ago
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