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3 years ago

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Ben is making math manipulatives to sell. He wants to make at least $450. Each manipulative costs $18 to make. He is selling the

m for $30 each. What is the minimum number he can sell to track his goal?

1 answer:

lys-0071 [83]3 years ago

6 0

7 is the answer because 7 is great and is always the answer

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Margarita is 15 years older than Silvia the sum of their ages is 41

Nimfa-mama [501]

One of them is 13 and the other is 28

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3 years ago

Read 2 more answers

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:

$P(X$

8 0

3 years ago

hat is the difference between the areas of a circular table with diameter 6 ft and a circular table with diameter 8 ft?

prohojiy [21]

I believe it’s pie times radius squared so 6ft = 3.14•3 squares and 8ft = 3.14•4 squares and then subtract

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3 years ago

Grayson goes to BestBuy to buy a computer for $499, a mouse for $19.99, and a cable cord for $8.99. Grayson buys this on credit

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Answer:

$44

Step-by-step explanation:

499 + 19.99 + 8.99 = 527.98

527.98/12 = 43.9983333 or $44.00

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HHHHHEEEEELLLLLPPPPP!!!!!!!!!!!!!!!!!!!!!!!

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<span>x/y where x and y are integers and y≠0 (rational)

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