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Dafna11 [192]
4 years ago
9

andy mailed two packages. the first package weighed 2.48 pounds, and the second package weighed 2.6 pounds. andy said the first

package will cost more to send because it weighs more
Mathematics
2 answers:
Alex787 [66]4 years ago
8 0

The mistake is that 2.48 does not weigh more. The second weighs more because 2.6 weighs more then 2.48, which means, 2.6 is greater than 2.48

rodikova [14]4 years ago
3 0
The second package weighs more, 2.6 or 2.60 weighs more than 2.48
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Answer:

4,5,6

Step-by-step explanation:

square root of 4²+5²= approximately 6.4

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The nutrition facts label on a container of dry roasted cashews indicates there are 161 calories in 28 grams. You eat 9 cashews
o-na [289]
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5 0
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Read 2 more answers
At the city museum child admission is $5.70 and the adult admission is $9.60 on Thursday,186 ticket were sold for a total sales
romanna [79]
A + c = 186.....a = 186 - c
9.60a + 5.70c = 1399.50

9.60(186 - c) + 5.70c = 1399.50
1785.60 - 9.60c + 5.70c = 1399.50
-9.60c + 5.70c = 1399.50 - 1785.60
-3.90c = - 386.10
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7 0
3 years ago
slader The magnitude of earthquakes recorded in a region of North America can be modeled ashaving an exponential distribution wi
Akimi4 [234]

Answer:

a) P(X>3.0)=1-P(X \leq 3.0)= 1- [1- e^{-\frac{1}{2.4} 3.0}]=e^{-\frac{1}{2.4} 3.0}=0.287

b) P(2

P(2

Step-by-step explanation:

Definitions and concepts

The Poisson process is useful when we want to analyze the probability of ocurrence of an event in a time specified. The probability distribution for a random variable X following the Poisson distribution is given by:

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

And the parameter \lambda represent the average ocurrence rate per unit of time.

The exponential distribution is useful when we want to describ the waiting time between Poisson occurrences. If we assume that the random variable T represent the waiting time btween two consecutive event, we can define the probability that 0 events occurs between the start and a time t, like this:

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

We can express in terms of the mean \mu =\frac{1}{\lambda}

P(X>x)= e^{-\frac{1}{\mu} x}

And the cumulative function would be given by the complement rule like this:

P(X\leq x)=1- e^{-\frac{1}{\mu} x}

Solution for the problem

For this case we have that X the random variable that represent the magnitude of earthquakes recorded in a region of North America, we know that the distribution is given by:

X\sim Expon(\mu =2.4)

a) Find theprobability that an earthquake striking this region will a exceed 3.0 on the Richter scale

So for this case we want this probability:

P(X>3.0)=1-P(X \leq 3.0)= 1- [1- e^{-\frac{1}{2.4} 3.0}]=e^{-\frac{1}{2.4} 3.0}=0.287

b) fall between 2.0 and 3.0 on the Richter scale.

For this case we want this probability:

P(2 < X

And replacing we have this:

P(2

P(2

7 0
3 years ago
Does anyone know the answer ?
alexira [117]

Answer:

b = √a² + c² is incorrect and different from other 3

Step-by-step explanation:

4 0
3 years ago
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