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Charra [1.4K]
3 years ago
10

Please go answer my questions posted I will be happy if you do plz

Mathematics
1 answer:
Taya2010 [7]3 years ago
4 0

Answer:

yes

Step-by-step explanation:

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I WILL GIVE BRAINLIEST IF YOU HELP THIS IS REALLY HARD FOR ME I HAVE THE FIRST TWO JUST THE LAST ONE PLEASE ALSO EXPLAIN HOW YOU
asambeis [7]

Answer:

n ≥ -1 or n ≤ 3

Combined:

-1 ≤ n ≤ 3

Since the inequalities are ≤, it'll be a closed circle

The closed circle will start AT -1 and another closed circle will end at 3

5 0
3 years ago
In the spinner below, what is true about the probability of landing on 1 and the probability of landing on 6?​
Tanzania [10]

Answer:

1/6 for 1 and 1/6 for 6

Step-by-step explanation:

4 0
3 years ago
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Because sample means were being tested the ____ is used to calculate the z-score
SSSSS [86.1K]

Answer: standard deviation

Step-by-step explanation:

7 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
5)
Paladinen [302]

Answer:183.4

Step-by-step explanation:

90/11=8.181818  1500/8.1818 = ≈183.4 time you would roll six  

6 0
3 years ago
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