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3 years ago

Point O is between points Mand N on line segment MN. O is the midpoint of line segment MN if

1 answer:

s2008m [1.1K]3 years ago

8 0

Diagram:
M------------O-------------N
................. ^ midpoint
Obviously if O is the *mid*point, then it is exactly halfway between M and N. That also means it has bisected the segment into two equal parts.
On one side you have MO (or OM, if you like)
On the other side you have ON.
In order for O to be the midpoint:
ON = OM

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The average number of annual trips per family to amusement parks in the UnitedStates is Poisson distributed, with a mean of 0.6

IrinaK [193]

Answer:

a) 0.5488 = 54.88% probability that the family did not make a trip to an amusement park last year.

b) 0.3293 = 32.93% probability that the family took exactly one trip to an amusement park last year.

c) 0.1219 = 12.19% probability that the family took two or more trips to amusement parks last year.

d) 0.8913 = 89.13% probability that the family took three or fewer trips to amusement parks over a three-year period.

e) 0.1912 = 19.12% probability that the family took exactly four trips to amusement parks during a six-year period.

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 interval.

Poisson distributed, with a mean of 0.6 trips per year

This means that $\mu = 0.6n$ , in which n is the number of years.

a.The family did not make a trip to an amusement park last year.

This is P(X = 0) when n = 1, so $\mu = 0.6$ .

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

$P(X = 0) = \frac{e^{-0.6}*(0.6)^{0}}{(0)!} = 0.5488$

0.5488 = 54.88% probability that the family did not make a trip to an amusement park last year.

b.The family took exactly one trip to an amusement park last year.

This is P(X = 1) when n = 1, so $\mu = 0.6$ .

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

$P(X = 1) = \frac{e^{-0.6}*(0.6)^{1}}{(1)!} = 0.3293$

0.3293 = 32.93% probability that the family took exactly one trip to an amusement park last year.

c.The family took two or more trips to amusement parks last year.

Either the family took less than two trips, or it took two or more trips. So

$P(X < 2) + P(X \geq 2) = 1$

We want

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1) = 0.5488 + 0.3293 = 0.8781$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.8781 = 0.1219$

0.1219 = 12.19% probability that the family took two or more trips to amusement parks last year.

d.The family took three or fewer trips to amusement parks over a three-year period.

Three years, so $\mu = 0.6(3) = 1.8$ .

This is

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = \frac{e^{-1.8}*(1.8)^{0}}{(0)!} = 0.1653$

$P(X = 1) = \frac{e^{-1.8}*(1.8)^{1}}{(1)!} = 0.2975$

$P(X = 2) = \frac{e^{-1.8}*(1.8)^{2}}{(2)!} = 0.2678$

$P(X = 3) = \frac{e^{-1.8}*(1.8)^{3}}{(3)!} = 0.1607$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1653 + 0.2975 + 0.2678 + 0.1607 = 0.8913$

0.8913 = 89.13% probability that the family took three or fewer trips to amusement parks over a three-year period.

e.The family took exactly four trips to amusement parks during a six-year period.

Six years, so $\mu = 0.6(6) = 3.6$ .

This is P(X = 4). So

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

$P(X = 4) = \frac{e^{-3.6}*(3.6)^{4}}{(4)!} = 0.1912$

0.1912 = 19.12% probability that the family took exactly four trips to amusement parks during a six-year period.

4 0

3 years ago

Add [1/5x-4+2y]+[2/5x+5-4y]=

alisha [4.7K]

<h3>I'll teach you how to solve (1/5x-4+2y)+(2/5x+5-4y)</h3>

-----------------------------------------------------------------

(1/5x-4+2y)+(2/5x+5-4y)

Remove parentheses:

1/5x-4+2y + 2/5x+5-4y

Group like terms:

1/5x+2/5x+2y-4y-4+5

Add similar elements:

3/5x+2y-4y-4+5

Add similar elements:

3/5x-2y-4+5

Multiply:

3x/5-2y-4+5

Add subtract the numbers:

3x/5+1-2y

Your Answer Is 3x/5+1-2y

plz mark me as brainliest :)

6 0

3 years ago

I NEED CORRECT ANSWER

RUDIKE [14]

Answer:

Step-by-step explanation:

Activity 3

Q1) consistent, independent

Q2) inconsistent

Q3) consistent, dependent

Q4) consistent, independent

4 0

3 years ago

Adding unlike fractions<br><br> 1/15+38/100

Daniel [21]

Answer:

134/300 or 67/150

Step-by-step explanation:

the LCM of 15 and 100 is 300. 15 *20 is 300 so we multiply both the numerator and denominator of 1/15 by 20 to get 20/300, with 38/100 we multiply by 3 to get 114/300. Now we add 20/300 and 114/300 to get 134/300 which has a common factor of 2 so we can simply it to 67/150

4 0

3 years ago

Someone help me with #51

Orlov [11]

850 x 0.05 = 42.5
850 + 42.5 = 892.5
She earned 42.5
And 892.5 she wardens at the end of four years

8 0

3 years ago

Read 2 more answers