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zheka24 [161]
3 years ago
5

A dice is rolled 50 times.

Mathematics
2 answers:
Deffense [45]3 years ago
6 0

Answer:

b.) 3000 ÷ 5 = 600

c.) 5 x 12 = 60

Step-by-step explanation:

I tried hope it helps

NemiM [27]3 years ago
3 0

Answer:

(a) 9/50=0.18

(b) 12/3000=4×10^-3

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A cargo ship is carrying nine shipping crates. Each rate is equal in , and the
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The answer is 523 because 4,707 divide 9 to be equal numbers for each one I hope it helps
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Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a
Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

P(X

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}

The cumulative distribution for this function is given by:

F(X) = 1- e^{-\lambda x}, x\ geq 0

We know the value for the mean on this case we have that :

mean = \frac{1}{\lambda}

\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: Var(X) =\frac{1}{\lambda^2}

And the deviation would be:

Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725

And the mean is given by Mean = 2.725

Two deviations correspond to 5.540, so we want this probability:

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

P(X

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3 years ago
I just completed 12 questions on a test the progress bar shows 20% is complete how many total questions are on the test?
Mademuasel [1]
If 12 is 20%, then 24 is 40%, then 36 is 60%, then 48 is 80%, then 60 is 100%

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7 0
3 years ago
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<span>4 hundreds 13 tens 5 ones = 400+13*10+5
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Solve f(x)=0 and Compute f(0) given f(x)=2x-7.
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Step-by-step explanation:

f(x)=2x-7.

by substituting value 0 for x,

f (0)=2×0-7

f (0)=-7

7 0
3 years ago
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