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kramer
3 years ago
13

The American Community Survey showed that residents of New York City have the longest travel times to get to work compared to re

sidents of other cities in the United States. According to the latest statistics available, the average travel time to work for residents of New York City is 38.3 minutes. What is the probability it will take a resident of this city between 17.24 and 42.13 minutes to travel to work?
Mathematics
1 answer:
LiRa [457]3 years ago
6 0

Answer:

The probability it will take a resident of the city between 17.24 and 42.13 minutes to travel to work is 0.3046.

Step-by-step explanation:

The random variable <em>X</em> can be defined as the travel times to get to work.

The expected travel time is, <em>μ</em> = 38.3 minutes.

The distribution of random variable <em>X</em> can be defined as the distribution of time interval between which a person reaches their work place at a constant average rate.

This implies that <em>X</em> follows an Exponential distribution with parameter, \lambda=\frac{1}{\mu}=\frac{1}{38.3}.

The probability density function of <em>X</em> is:

f_{X}(x)=\lambda e^{-\lambda x};\ x\geq 0

Compute the probability it will take a resident of the city between 17.24 and 42.13 minutes to travel to work as follows:

P(17.24\leq X\leq 42.13)=\int\limits^{42.13}_{17.24}{\frac{1}{38.3} e^{-x/38.3}}}\, dx

                                   =\frac{1}{38.3}\times \int\limits^{42.13}_{17.24}{ e^{-x/38.3}}}\, dx

                                   =\frac{1}{38.3}\times| \frac{e^{-x/38.3}}{-1/38.3}}|^{42.13}_{17.24}\\

                                   =-e^{42.13/38.3}+e^{17.34/38.3}\\=-0.3329+0.6375\\=0.3046

Thus, the probability it will take a resident of the city between 17.24 and 42.13 minutes to travel to work is 0.3046.

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LenKa [72]

Answer:

The answer is D

Step-by-step explanation:

I just used an algebra calculator

5 0
3 years ago
The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,
Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

If the mean number of arrivals is 10

This means that \mu = 10

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is P(X \leq 10)

P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045

P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045

P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023

P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076

P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189

P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378

P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631

P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901

P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126

P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251

P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251

P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831

58.31% probability that there are no more than 10 arrivals.

8 0
3 years ago
Triangle ABC is similar to triangle FED, find the measure of x. A. X = 9 B. X = 11 C. X = 110 D. X = 231
just olya [345]

Answer:

A+B+C=130

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Step-by-step explanation:

From the question we are told that:

Dimension of ABC

A=9

B=11

C=110

Dimension of FED

D=231

Generally the equation for the similar triangles is mathematically given by

\frac{A}{F}=\frac{B}{E}=\frac{C}{D}

Therefore solving for F

\frac{9}{F}=\frac{110}{231}

F=\frac{9*231}{110}

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Therefore solving for E

\frac{11}{E}=\frac{110}{231}

E=\frac{11*231}{110}

E=23.1

Measure of \triangle ABC

A+B+C=130

Measure of \triangle FED

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Answer is 144 my dude
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