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docker41 [41]
2 years ago
10

A major traffic problem in the Greater Cincinnati area involves traffic attempting to cross the Ohio River from Cincinnati to Ke

ntucky using Interstate 75. Let us assume that the probability of no traffic delay in one period, given no traffic delay in the preceding period, is 0.9 and that the probability of finding a traffic delay in one period, given a delay in the preceding period, is 0.6. Traffic is classified as having either a delay or a no-delay state, and the period considered is 30 minutes.Required:a. Assume that you are a motorist entering the traffic system and receive a radio report of a traffic delay. What is the probability that for the next 60 minutes (two time periods) the system will be in the delay state?b. What is the probability that in the long run the traffic will not be in the delay state?c. An important assumption of the Markov process model presented here has been the constant or stationary transition probabilities as the system operates in the future. Do you believe this assumption should be questioned for this traffic problem? Explain.
Mathematics
1 answer:
myrzilka [38]2 years ago
8 0

Answer:

a) 0.36

b) 0.3

c) Yes

Step-by-step explanation:

Given:

Probability of no traffic delay in one period, given no traffic delay in the preceding period = P(No_Delay) = 0.9

Probability of finding a traffic delay in one period, given a delay in the preceding period = P(Delay) = 0.6

Period considered = 30 minutes

a)

Let A be the probability that for the next 60 minutes (two time periods) the system will be in the delay state:

As the Probability of finding a traffic delay in one period, given a delay in the preceding period is 0.6 and one period is considered as 30 minutes.

So probability that for the next two time periods i.e. 30*2 = 60 minutes, the system in Delay is

P(A) = P(Delay) * P(Delay) =  0.6 * 0.6 = 0.36

b)

Let B be the probability that in the long run the traffic will not be in the delay state.

This statement means that the traffic will not be in Delay state but be in No_Delay state in long run.

Let C be the probability of one period in Delay state given that preceding period in No-delay state :

P(C) =  1 - P(No_Delay)

        = 1 - 0.9

P(C) = 0.1

Now using P(C) and P(Delay) we can compute P(B) as:

P(B) = 1 - (P(Delay) + P(C))

      = 1 - ( 0.6 + 0.10 )

      = 1 - 0.7

P(B) = 0.3

c)

Yes this assumption should be questioned for this traffic problem because it implies that  traffic will be in Delay state for the 30 minutes and just after 30 minutes, it will be in No_Delay state. However, traffic does not work like this in general and it makes this scenario unrealistic. Markov process model can be improved if probabilities are modeled as a function of time instead of being presented as constant (for 30 mins).

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Jonah was standing at an elevation somewhere between Negative 1 and one-half and Negative 2 and one-third meters with regards to
Agata [3.3K]

Options :

A number line going from negative 3 to positive 3 in increments of 1.

1 and StartFraction 5 Over 6 EndFraction meters

Negative 2 and StartFraction 3 Over 6 EndFraction meters

2 and StartFraction 3 Over 6 EndFraction meters

Negative 1 and StartFraction 5 Over 6 EndFraction meters

Answer:

Negative 1 and StartFraction 5 Over 6 EndFraction meters

Step-by-step explanation:

Jonah's position range :

Between - 1 1/2 meters and - 2 1/3 meters with respect to sea level

With regards to sea level and the range of position given ; Jonah's position will be below sea level, that is Jonah's position cannot be a positive location on the number line, thus all options with positive values are incorrect.

Hence Jonah's position will be any number on a number line located in between :

-2 1/3 and - 1 1/2

That is

-2 1/3 ≤ Jonah's location ≤ - 1 1/2

-2 1/2 ( this is less than - 2 1/3) (incorrect)

-1 5/6 lies in between the given range and is thus a possible position.

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2 years ago
Mai has 40 tens. write how many hundreds. write the number.
Olegator [25]
40 tens = how many hundreds.  40 x 10 = 400
4 0
2 years ago
Use the data below, showing a summary of highway gas mileage for several observations, to decide if the average highway gas mile
Murrr4er [49]

Answer:

Step-by-step explanation:

Hello!

You need to test at 1% if the average highway gas mileage is the same for three types of vehicles (midsize cars, SUV's and pickup trucks) to compare the average values of the three groups altogether, you have to apply an ANOVA.

                n  |  Mean |  Std. Dev.

Midsize  | 31 |  25.8   |  2.56

SUV’s     | 31 |  22.68 |  3.67

Pickups  | 14 |  21.29  |  2.76

Be the study variables :

X₁: highway gas mileage of a midsize car

X₂: highway gas mileage of an SUV

X₃: highway gas mileage of a pickup truck.

Assuming these variables have a normal distribution and are independent.

The hypotheses are:

H₀: μ₁ = μ₂ = μ₃

H₁: At least one of the population means is different.

α: 0.01

The statistic for this test is:

F= \frac{MS_{Treatment}}{MS_{Error}}~F_{k-1;n-k}

Attached you'll find an ANOVA table with all its components. As you see, to manually calculate the statistic you have to determine the Sum of Squares and the degrees of freedom for the treatments and the errors, next you calculate the means square for both and finally the test statistic.

<u>For the treatments:</u>

The degrees of freedom between treatments are k-1 (k represents the amount of treatments): Df_{Tr}= k - 1= 3 - 1 = 2

<u>The sum of squares is: </u>

SSTr: ∑ni(Ÿi - Ÿ..)²

Ÿi= sample mean of sample i ∀ i= 1,2,3

Ÿ..= grand mean, is the mean that results of all the groups together.

So the Sum of squares pf treatments SStr is the sum of the square of difference between the sample mean of each group and the grand mean.

To calculate the grand mean you can sum the means of each group and dive it by the number of groups:

Ÿ..= (Ÿ₁ + Ÿ₂ + Ÿ₃)/ 3 = (25.8+22.68+21.29)/3 = 23.256≅ 23.26

SS_{Tr}= (Ÿ₁ - Ÿ..)² + (Ÿ₂ - Ÿ..)² + (Ÿ₃ - Ÿ..)²= (25.8-23.26)² + (22.68-23.26)² + (21.29-23.26)²= 10.6689

MS_{Tr}= \frac{SS_{Tr}}{Df_{Tr}}= \frac{10.6689}{2}= 5.33

<u>For the errors:</u>

The degrees of freedom for the errors are: Df_{Errors}= N-k= (31+31+14)-3= 76-3= 73

The Mean square are equal to the estimation of the variance of errors, you can calculate them using the following formula:

MS_{Errors}= S^2_e= \frac{(n_1-1)S^2_1+(n_2-1)S^2_2+(n_3-1)S^2_3}{n_1+n_2+n_3-k}= \frac{(30*2.56^2)+(30*3.67^2)+(13*2.76^2)}{31+31+14-3} = \frac{695.3118}{73}= 9.52

<u>Now you can calculate the test statistic</u>

F_{H_0}= \frac{MS_{Tr}}{MS_{Error}} = \frac{5.33}{9.52}= 0.559= 0.56

The rejection region for this test is <em>always </em>one-tailed to the right, meaning that you'll reject the null hypothesis to big values of the statistic:

F_{k-1;N-k;1-\alpha }= F_{2; 73; 0.99}= 4.07

If F_{H_0} ≥ 4.07, reject the null hypothesis.

If F_{H_0} < 4.07, do not reject the null hypothesis.

Since the calculated value is less than the critical value, the decision is to not reject the null hypothesis.

Then at a 1% significance level you can conclude that the average highway mileage is the same for the three types of vehicles (mid size, SUV and pickup trucks)

I hope this helps!

4 0
2 years ago
Find the distance between the points L(7, -1) and M(-2, 4).
VLD [36.1K]

Answer:

(5-5)

Step-by-step explanation:

hope this helped

5 0
2 years ago
What is the y-intercept of line MN?<br><br> What is the equation of MN written in standard form?
bagirrra123 [75]

Answer:

y-intercept of the line MN = 2

Standard form of the equation ⇒ x + y = 2

Step-by-step explanation:

Coordinates of the ends of a line MN → M(-3, 5) and N(2, 0)

Slope of a line = \frac{y_2-y_1}{x_2-x_1}

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                        = -1

Equation of the line MN passing through (-3, 5) and slope = -1,

y - 5 = (-1)(x + 3)

y - 5 = -x - 3

y = -x + 2

This equation is in the y-intercept form,

y = mx + b

where m = slope of the line

b = y-intercept

Therefore, y-intercept of the line MN = 2

Equation in the standard form,

x + y = 2

8 0
2 years ago
Read 2 more answers
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