What is c - b for six grade math

Mathematics

1 answer:

JulsSmile [24]3 years ago

4 0

Answer:

specify?

Step-by-step explanation:

give us a image

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What is the solution is to this 3/5(2m-10)=2/3m+10

aivan3 [116]

To find the solution to the given expression, we simply isolate the unknown variable m and perform the necessary operations. The solution is as follows.

(3/5)*(2m-10) = (2/3)*m + 10
15*[(3/5)*(2m-10) = (2/3)*m + 10]
9(2m-10) = 10m + 150
18m-90 =10m + 150
18m - 10m = 150 + 90
8m = 240
m = 240/8
m = 30

First, we multiply the whole expression to the least common denominator (LCD) which is 15. Multiplication and division were then performed accordingly. The terms containing the variable m were transposed to the left side of the equation. Mathematical operations were then applied to get the final value of m which is equal to 30.

8 0

3 years ago

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.