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2 years ago

8

Kyle spent more than 37 minutes swimming last night. Which of the following amounts of time could Kyle have been swimming? Selec

t two that apply.
A
2050 seconds

B
2150 seconds

C
2250 seconds

D
2350 seconds

1 answer:

Komok [63]2 years ago

7 0

Answer:

2,350 seconds.

Step-by-step explanation:

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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.

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2 years ago

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Answer:

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Step-by-step explanation:

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KatRina [158]

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Find a round to the nearest tenth a 150 12 10cm c a=?cm

Kryger [21]

Answer:

<h3> $Using~ the ~ law~ of ~sinse:-$ </h3><h3> $\frac{a}{SinA} =\frac{b}{SinB}$ </h3>

$\frac{a}{Sin150}$ $=\frac{b}{SinB}$

$a=\frac{10~sin~150}{Sin~12}$

$a=[24] ~cm$

<h3><u>-----------------------</u></h3><h3><u>hope it helps..</u></h3><h3><u>have a great day!!</u></h3>

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2 years ago

Could you please help me!?

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Answer:

Answer: 42.5

Step-by-step explanation:

It says 1 inch equals 5 feet. So, every inch is on the map is an extra 5 feet in real space. So if you think about it, all this is is a simple multiplication problem.

Here is the math:

8.5 * 5 = 42.5

Think about it, for every inch there is on the map there is an extra 5 feet in real space. There is 8 inches, so we need 5 extra feet for each of those 8 inches. So what we get for 1 inch is 5 feet, for 2 inches 10 feet, 3 inches 15 feet, etc etc until we reach 8 inches. For the 0.5 inches alone that would equal 2.5 feet in real life since 1 inch is 5 feet.

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2 years ago