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vfiekz [6]
3 years ago
7

A driver and a passenger are in a car accident. Each of them independently has probability 0.3 of being hospitalized. When a hos

pitalization occurs, the loss is uniformly distributed on [0, 1]. When two hospitalizations occur, the losses are independent. Calculate the expected number of people in the car who are hospitalized, given that the total loss due to hospitalizations from the accident is less than 1.
Mathematics
1 answer:
kompoz [17]3 years ago
5 0

Answer:

0.534

Step-by-step explanation:

p(0 losses) = 0.7² = 0.49

p(1 loss) = 2 x 0.3 x 0.7 = 0.42

p(2 losses) = 0.09

This is a conditional probability problem. If the number of people hospitalized is 0 or 1, then the  total loss will be less than 1. However, if two people are hospitalized, the probability that the  total loss will be less than 1 is 0.5. we need to exclude the 50% x 0.09 chance of a double loss costing more than 1. So

P(Cost < 1)

= 0.49 + 0.42 +0.045

= 0.955

P(0 losses | Cost < 1)

= P(0 losses and Cost < 1) / P(Cost < 1)

= 0.49 / 0.955 = 0.513

P(1 loss | Cost < 1)

= 0.42 / 0.955 = 0.440

P(2 losses | Cost < 1) = 0.045 / 0.955 = 0.047

Now take the expectation:

E[X] = (0)(0.513) + (1)(0.440) + (2)(0.047)

= 0.440 + 0.094

= 0.534

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Answer:

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2 years ago
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ivanzaharov [21]

Answer: 3*z^\frac{2}{9}

Step-by-step explanation:

The exponent 1/3 is the same thing as cube rooting the expression.

\sqrt[3]{27\sqrt[3]{z^2} }

3*z^{2/9}

Hope it helps <3

4 0
3 years ago
This problem uses the teengamb data set in the faraway package. Fit a model with gamble as the response and the other variables
hichkok12 [17]

Answer:

A. 95% confidence interval of gamble amount is (18.78277, 37.70227)

B. The 95% confidence interval of gamble amount is (42.23237, 100.3835)

C. 95% confidence interval of sqrt(gamble) is (3.180676, 4.918371)

D. The predicted bet value for a woman with status = 20, income = 1, verbal = 10, which shows a negative result and does not fit with the data, so it is inferred that model (c) does not fit with this information

Step-by-step explanation:

to)

We will see a code with which it can be predicted that an average man with income and verbal score maintains an appropriate 95% CI.

attach (teengamb)

model = lm (bet ~ sex + status + income + verbal)

newdata = data.frame (sex = 0, state = mean (state), income = mean (income), verbal = mean (verbal))

predict (model, new data, interval = "predict")

lwr upr setting

28.24252 -18.51536 75.00039

we can deduce that an average man, with income and verbal score can play 28.24252 times

using the following formula you can obtain the confidence interval for the bet amount of 95%

predict (model, new data, range = "confidence")

lwr upr setting

28.24252 18.78277 37.70227

as a result, the confidence interval of 95% of the bet amount is (18.78277, 37.70227)

b)

Run the following command to predict a man with maximum values ​​for status, income, and verbal score.

newdata1 = data.frame (sex = 0, state = max (state), income = max (income), verbal = max (verbal))

predict (model, new data1, interval = "confidence")

lwr upr setting

71.30794 42.23237 100.3835

we can deduce that a man with the maximum state, income and verbal punctuation is going to bet 71.30794

The 95% confidence interval of the bet amount is (42.23237, 100.3835)

it is observed that the confidence interval is wider for a man in maximum state than for an average man, it is an expected data because the bet value will be higher than the person with maximum state that the average what you carried s that simultaneously The, the standard error and the width of the confidence interval is wider for maximum data values.

(C)

Run the following code for the new model and predict the answer.

model1 = lm (sqrt (bet) ~ sex + status + income + verbal)

we replace:

predict (model1, new data, range = "confidence")

lwr upr setting

4,049523 3,180676 4.918371

The predicted sqrt (bet) is 4.049523. which is equal to the bet amount is 16.39864.

The 95% confidence interval of sqrt (wager) is (3.180676, 4.918371)

(d)

We will see the code to predict women with status = 20, income = 1, verbal = 10.

newdata2 = data.frame (sex = 1, state = 20, income = 1, verbal = 10)

predict (model1, new data2, interval = "confidence")

lwr upr setting

-2.08648 -4.445937 0.272978

The predicted bet value for a woman with status = 20, income = 1, verbal = 10, which shows a negative result and does not fit with the data, so it is inferred that model (c) does not fit with this information

4 0
3 years ago
5 + x / 4 = 1 what is x?
Ratling [72]
X/4= -4
x= -16.......
4 0
3 years ago
Read 2 more answers
When solving a system of equations of A and B, we get no solution. When solving B and C, the solution is (0,3). When solving A a
Aliun [14]

Step-by-step explanation:

How do we set about finding the points in which two graphs y = f(x) and y = g(x) intersect?

We already know how to find where the graph of f(x) cuts the x−axis. That’s where y = 0. We calculate it by solving the equation  f(x) = 0 .

When the graphs of y = f(x) and  y = g(x)  intersect , both graphs have exactly the same x and y values. So we can find the point or points of intersection by solving the equation  f(x) = g(x). The solution of this equation will give us the x value(s) of the point(s) of intersection. We can then find the y value by putting the value for x that we have found into one of the original equations. That is by calculating either f(x) or g(x).

Example 1 

Calculate the point of intersection of the two lines f(x) = 2x − 1 and g(x) = x + 1.  First let’s look at a graph of the two functions. We can see the point of intersection is (2, 3).



We calculate the point of intersection by solving the equation f(x) = g(x). That is:

    2x − 1 = x + 1

    2x − x = 1 + 1

            x = 2

The y coordinate can now be found by calculating f(2):

    f(2) = 2×2 − 1 = 3

The point of intersection is (2, 3).

The example shows that we can find the point of intersection in two ways.

Either graphically, by drawing the two graphs in the same coordinate system, or algebraically by solving the equation such as the one in the above example.

Solving an equation graphically is easy with a graphical calculator or a computer program such as Excel.

Some equations cannot be solved algebraically but we can find solutions that are correct to as many significant figures as we want by using computers and calculators

4 0
2 years ago
Read 2 more answers
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