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bazaltina [42]
2 years ago
15

the formula for the volume of a cylinder is V=3.14r^2h. The volume of a cylinder is 3 times the volume of a cone with the same r

adius and height. If the volume of a cone with the same height as a cylinder equals the volume of the cylinder, the equation for the radius of cone R in terms of the radius of a cylinder is?
Mathematics
1 answer:
alukav5142 [94]2 years ago
6 0
Almost there sis SJSISJCNXND xndb dksjjbd jdjjd did staff Hibbert dub dub Dude
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12 – 7x ≤-2 <br> solve and show work
vova2212 [387]

Answer:

x≥2

Step-by-step explanation:

Let's solve your inequality step-by-step.

12−7x≤−2

Step 1: Simplify both sides of the inequality.

−7x+12≤−2

Step 2: Subtract 12 from both sides.

−7x+12−12≤−2−12

−7x≤−14

Step 3: Divide both sides by -7.

−7x/−7 ≤ −14/−7

x≥2

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Answer: 84 is your answer hope this helped

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Step-by-step explanation:

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If f(x)=√x+12 and g(x)=2√x, what is the value of (f – g)(144)?
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I need to know how to prove if an angle is congruent or not.
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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

8 0
3 years ago
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