Andrew estimates the attendance at the game to be 550 people. There ar

The route used by a certain motorist in commuting to workcontains two intersections with traffic signals. The probabilitythat he

zavuch27 [327]

Answer:

a) 0.2

b) 0.2

c) 0.5

Step-by-step explanation:

Let $S$ be the event "the car stops at the signal.

In the attached figure you can see a tree describing all possible scenarios.

For the first question the red path describes stopping at the first light but not stopping at the second. We can determine the probability of this path happening by multiplying the probabilities on the branches of the tree, thus

$P(a)=0.4\times0.5=0.2$

For the second one the blue path describes the situation

$P(b)=0.4\times 0.5=0.2$

For the las situation the sum of the two green path will give us the answer

$P(c)=0.6\times 0.5 + 0.4\times 0.5=0.3+0.2=0.5$

7 0

3 years ago

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Steps to find the answer of this picture pretty please

Kazeer [188]

add the sides together

3 0

3 years ago

Need the answer for number 7 ASAP!!!!

cupoosta [38]

B is the answer because 60/ 12.50 equals 12/2.5

3 0

3 years ago

The city has an average of 13 days of rainfall for April.

zhenek [66]

Using the Poisson distribution, we have that:

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

<h3>What is the Poisson distribution?</h3>

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

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

The parameters are:

x is the number of successes

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

For this problem, the mean is given as follows:

$\mu = 13$

The probability of having exactly 10 days of precipitation in the month of April is P(X = 10), hence:

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

$P(X = 10) = \frac{e^{-13}13^{10}}{(10)!} = 0.0859$

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

The probability of having less than three days of precipitation in the month of April is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

In which:

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

$P(X = 0) = \frac{e^{-13}13^{0}}{(0)!} \ approx 0$

$P(X = 1) = \frac{e^{-13}13^{1}}{(1)!} = 0.00003$

$P(X = 2) = \frac{e^{-13}13^{2}}{(2)!} = 0.00019$

Then:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0.00003 + 0.00019 = 0.00022

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

For more than 15 days, the probability is:

P(X > 15) = P(X = 16) + P(X = 17) + ... + P(X = 20)

Applying the formula for each of these values and adding them, we have that P(X > 15) = 0.2364, hence:

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

More can be learned about the Poisson distribution at brainly.com/question/13971530

#SPJ1

6 0

2 years ago

VvLabel Slope and Y-intercept in each Equation

Kamila [148]

1- slope is 4 y intercept 0,2

2- slope is 6 y intercept 0,5

3- slope is 2 y intercept is 0,5

4 -slope is 1 y intercept is 0, 3/4

5- slope is 10 y intercept is 0,3

6- slope is 4 y intercept is 0,1/2

7- slope is 4 y intercept is 0, 3/2

8- slope is 1 y intercept is 0,2

4 0

3 years ago

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