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zimovet [89]
3 years ago
11

I need to find what is equal to m

Mathematics
1 answer:
USPshnik [31]3 years ago
5 0

Answer: 0

Step-by-step explanation:

Point A = (3 , 4)

Point B = (7 , 4)

\frac{yB - yA}{xB - xA}  = \frac{0}{5}

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A company is about to launch a new cell phone model. In the past, 40% of its cell phones have launched successfully. Before any
Eva8 [605]
There is a 70% chance that te phone will receive a favorable report.
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3 years ago
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Prove that this question:.​
Vikki [24]

This follows directly from the double angle identity for cosine and the Pythagorean identity:

\cos2\theta=\cos^2\theta-\sin^2\theta=\begin{cases}2\cos^2\theta-1\\1-2\sin^2\theta\end{cases}

So we have

\dfrac{1-2\sin^2\theta}{2\cos^2\theta-1}=\dfrac{\cos2\theta}{\cos2\theta}=1

8 0
3 years ago
Please help right away
Vilka [71]

Answer:

5 11/12

Step-by-step explanation:

Convert each fraction into like denominators: 2/3=8/12, 1/4=3/12

Add: 2 8/12 + 3 3/12 = 5 11/12

7 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.

6 0
3 years ago
Simplify the expression.
OlgaM077 [116]

Answer:

5.1

Step-by-step explanation:

-44.288-31.6 / -3.1(6-1.2)

Add -44.288 and -31.6

= -75.888 / -3.1(6-1.2)

Subtract 1.2 from 6

= -75.888 / -3.1(4.8)

Multiply -3.1 and 4.8

= -75.888 / -14.88

Divide -75.888 by -14.88

-75.888 / -14.88

=5.1

6 0
3 years ago
Read 2 more answers
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