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alexandr1967 [171]
2 years ago
9

Thousands of people are going to a Grateful dead concert in Pauley Pavillion at UCLA. They park their the foot cars on several o

f the long streets near the arena. There are no lines to tell the drivers where to park, so they park at random locations, and end up leaving spacings between the cars that are independent and uniform on .0; 10/. In the long run, what fraction of the street is covered with cars
Mathematics
1 answer:
BabaBlast [244]2 years ago
7 0

This question is incomplete, the complete question is;

Thousands of people are going to a Grateful dead concert in Pauley Pavilion at UCLA. They park their 10 foot cars on several of the long streets near the arena. There are no lines to tell the drivers where to park, so they park at random locations, and end up leaving spacings between the cars that are independent and uniform on ( 0, 10 ). In the long run, what fraction of the street is covered with cars?.

Answer:

in the long run, the fraction of street that will be covered with cars is 0.6667

Step-by-step explanation:

 given the data in the question;

the gaps between two cars follows uniformly ) 0, 10 )

therefore in the long run, average gap will be;

⇒ (0 + 10) / 2 = 10 / 2 = 5 feet

meaning that, in the long run a single car will occupy;

⇒ length of car + gap

⇒ 10 + 5 = 15 feet

∴ the fraction of road covered with cars will be;

⇒ length of car / space or length of road occupied by one car

⇒ 10 / 15

⇒ 0.6667

Therefore, in the long run, the fraction of street that will be covered with cars is 0.6667

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Jeannine runs a mobile dog bathing service, and the amount of money she earns varies directly with the number of dogs she bathes
Ilia_Sergeevich [38]
The cost of bathing service per dog :
$324/27 = $12


x = <span>number of dogs she washes
</span>y = revenue
so the equation:

y = 12x

1 dog = $12
2 dogs = $12 x 2  = $24
3 dogs = $12 x 3 = $36
4 dogs = $12 x 4 = $48

use these coordinate points to graph it.

x - coordinate = <span>number of dogs she washes
</span>y - coordinate = total cost ($)
8 0
2 years ago
Read 2 more answers
There are 50 students in an auditorium, of which 2x are boys and y are girls. After (y - 6) boys leave the auditorium and (2x -
Dmitrij [34]
Total number of students=50
Number of boys=2x
Number of girls=y
total will be:
2x+y=50
⇒y=50-2x

when (y-6) boys left the auditorium the new number of boys was:
2x-(y-6)
=2x-y+6
but y=50-2x
thus the new number will be:
2x-(50-2x)+6
=4x-44

when (2x-5) girls left the auditorium the remaining number will be:
y-(2x-5)
=y-2x+5
but 
y=50-2x
thus the new number of girls will be:
50-2x-2x+5
=55-4x
new total number of students:
(55-4x)+(4x-44)
=11

probability of selecting a girl at random will be:
(55-4x)/11=9/13
13(55-4x)=9*11
715-52x=99
616=52x
x=12
thus
y=50-12=38
thus
x=12 and y=38


7 0
2 years ago
In a small town, 50% of single family homes have a front porch. 48 single family homes are randomly selected. Let X represent th
Reil [10]

Answer:

N(24, 3.46)

Step-by-step explanation:

We are given the following in the question:

Percentage of family homes having front poach = 50%

p = 50\% = 0.50

Sample size, n = 48

Normal approximation to the given distribution:

\mu = np = 48(0.50) = 24

\sigma = \sqrt{np(1-p)} = \sqrt{48(0.50)(1-0.50)} = 3.46

Thus, the distribution of single family homes is best approximated by the normal distribution N(24, 3.46) where mean is 24 and standard deviation is 3.46

6 0
2 years ago
A rectangular prism is shown below 15mm 10mm 9mm
Verdich [7]
Area of cross section x height
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6 0
2 years ago
There are two misshapen coins in a box; their probabilities for landing on heads when they are flipped are, respectively, .4 and
Leokris [45]

Answer:

E(X) = 6.0706

Step-by-step explanation:

1) Define notation

X = random variable who represents the number of heads in the 10 first tosses

Y = random variable who represents the number of heads in range within toss number 4 to toss number 10

And we can define the following events

a= The first coin has been selected

b= The second coin has been selected

c= represent that we have 2 Heads within the first two tosses

2) Formulas to apply

We need to find E(X|c) = ?

If we use the total law of probability we can find E(Y)

E(Y) = E(Y|a) P(a|c) + E(Y|b)P(b|c) ....(1)

Finding P(a|c) and using the Bayes rule we have:

P(a|c) = P(c|a) P(a) / P(c) ...(2)

Replacing P(c) using the total law of probability:

P(a|c) = [P(c|a) P(a)] /[P(c|a) P(a) + P(c|b) P(b)] ... (3)

We can find the probabilities required

P(a) = P(b) = 0.5

P(c|a) = (3C2) (0.4^2) (0.6) = 0.288

P(c|b) = (3C2)(0.7^2) (0.3) = 0.441

Replacing the values into P(a|c) we got

P(a|c) = (0.288 x 0.5) /(0.288x 0.5 + 0.441x0.5) = 0.144/ 0.3645 = 0.39506

Since P(a|c) + P(b|c) = 1. With this we can find P(b|c) = 1 - P(a|c) = 1-0.39506 = 0.60494

After this we can find the expected values

E(Y|a) = 7x 0.4 = 2.8

E(Y|b) = 7x 0.7 = 4.9

Finally replacing the values into equation (1) we got

E(Y|c) = 2.8x 0.39506 + 4.9x0.60494 = 4.0706

And finally :

E(X|c) = 2+ E(Y|c) = 2+ 4.0706 = 6.0706

6 0
2 years ago
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