How many square feet of outdoor carpet will we need for this hole??? ANSWER QUICKLY

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

2 years ago

A hardcover book sells for $24 at the bookmark been pays a total of $25.02 for the book what is the sales tax rate

lys-0071 [83]

To find the tax rate you need to divide

24/25.02=<span>0.959232613909

0.95 would be your answer</span>

3 0

3 years ago

There were 80 runners to start a race. In the first half of the race, 1/4 of them dropped out. In the second half of the race, 2

IgorLugansk [536]

Answer:

36 runners finished the race.

Step-by-step explanation:

In the first half of the race, 1/4 of them dropped, so we need to calculate 1/4 of 80:

(1/4) * 80 = 80/4 = 20

If 20 runners dropped out, now we have 80 - 20 = 60 runners.

Then, in the second half, 2/5 of the remaining runners dropped out, so:

(2/5) * 60 = 2*60/5 = 120/5 = 24

If 24 runners dropped out, now we have 60 - 24 = 36 runners.

So, 36 runners finished the race.

7 0

2 years ago

Use the given information to prove that RSU=TSV

RoseWind [281]

Answer:

Statement, Reason

Given segment RV = segment TU, Given

m∠R = m∠T, Given

Segment RS ≅ segment ST, Base angles theorem

ΔRSV ≅ ΔTSU, SAS rule of congruency

Segment SV ≅ segment SU, CPCTC

VU ≅ UV, Reflexive property

RU = RV + VU, TV = TU + UV, Addition of segments

RV + VU ≅ TU + UV, Addition property of equality

RU ≅ TV, Transitive property of addition

ΔRSU ≅ ΔTSV, SSS rule of congruency

Step-by-step explanation:

Where:

SAS- Side Angle Side

CPCTC -Congruent Parts of Congruent Triangles are Congruent

SSS -Side Side Side

3 0

2 years ago

find the area and perimeter using the lengthy and with step by step I will give 40 points please answer right number 6,7,9,10,11

N76 [4]

Answer:

6. 16 meters^2, 16 meters

7. 100 milimieters^2, 40 milimeters

9. 56 centimeters^2, 30 centimeters

10. 49 inches^2, 28 inches

11. 9 meters^2, 12 meters

12. 72 yards^2, 36 yards

Step-by-step explanation:

s is for a square

A=s^2

P=4s

l+w is for a rectangle

A=lw

P=2l+2w

6. A=4^2

A= 16 meters^2

P=4*4

P= 16 meters

7. A=10^2

A=100 milimieters^2

P=4*10

P= 40 milimeters

9. A=8*7

A= 56 centimeters^2

P=2(8)+2(7)

P=30 centimeters

10. A=7^2

A=49 inches^2

P=4*7

P=28 inches

11. A= 3^2

A=9 meters^2

P=4*3

P= 12 meters

12.

A=12*6

A=72 yards^2

P= 2(12)+2(6)

P= 36 yards

Let me know if you have any questions. Also, don't forget units!

4 0

2 years ago