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3 years ago

I have 18 ones,15 hundreds,15 tens,and 8 thousands.what number am i?

Mathematics

1 answer:

antoniya [11.8K]3 years ago

3 0

15818 ?? This is making me get a headache

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Builders and architects use scale models to help them design and build new buildings. An architecture

laiz [17]

Answer:

sss,not,not

Step-by-step explanation:

input those answers in the correct order. brainliest plz

7 0

2 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

StartLayout 1st row StartRoot StartFraction 250 c cubed Over 9 d Superscript 6 Baseline EndFraction EndRoot space c greater-than

MAVERICK [17]

The simplest form of the given algebraic expression √(250c³/9d⁶) is; 5c * (√10c)]/3d³

<h3>How to simplify algebra problems?</h3>

We are given the algebra problems as;

√(250c³/9d⁶)

Now 250c³ can be broken down into 250 = 25c² × 10c

Thus, we now have;

√(250c³/9d⁶) = [√(25c²) * (√10c)]/√9d⁶

When we breakdown the above expression, we have;

5c * (√10c)]/3d³

Read more about Algebra Problems at; brainly.com/question/723406

#SPJ1

8 0

1 year ago

Number 18.

Rashid [163]

f(x) = -5/2x + 4

f(3) = (-5/2)(3) + 4

f(3) = -15/2 + 4

f(3) = -7.5 + 4

f(3) = -3.5

Answer: B. –3.5

4 0

3 years ago

Question 7 of 10

Varvara68 [4.7K]

Answer:

Step-by-step explanation:

3 0

3 years ago