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Vlad [161]
3 years ago
13

The freefall function to calculate velocity, v, of an object that begins at rest and falls for distance d is v(d) = 2gd , where

g is the acceleration due to gravity. The Storm Runner roller coaster at Hershey Park drops 55 m. Use the free-fall function to calculate to the nearest tenth its velocity at the end of the freefall. Use g = 9.8 m/s2 Enter your answer, rounded to the nearest tenth, in the box. v(55) = m/s​
Mathematics
1 answer:
andrey2020 [161]3 years ago
3 0

9514 1404 393

Answer:

  32.8 m/s

Step-by-step explanation:

The function is actually ...

  v(d) = \sqrt{2gd}

Filling in the given values, the velocity is about ...

  v(55) = \sqrt{2\cdot9.8\cdot55}=\sqrt{1078}\approx32.8\text{ m/s}

The velocity at the end of the free-fall is about 32.8 meters per second.

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Help me solve the binary operations please ​
son4ous [18]

Answer:

  • 3*2=\sf{\pink{\dfrac{4}{5}}}

Step-by-step explanation:

<u>Given</u><u> </u><u>that</u><u>,</u><u> </u>

x*y=\dfrac{(2x-y)}{(x+y}

<h3><u>SO</u><u>,</u><u> </u></h3>

3*2=

\dfrac{2(3)-2}{3+2}\\\\\dfrac{6-2}{5}\\\\\dfrac{4}{5}

<u>Therefore</u><u>:</u><u>-</u>

\therefore  {3*2}=\dfrac{4}{5}

6 0
2 years ago
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
If rain is falling at the rate of 2 inches per hour how many inches of rain will fall in x minutes
katen-ka-za [31]
2/60= .0333
So your answer would be .0333x
5 0
3 years ago
Easy "mean" question :)
Gnesinka [82]

Answer:

Answer choice C

Step-by-step explanation:

Adding zero doesn't changed the sum, therefore doesn't change the median

7 0
3 years ago
Read 2 more answers
Help for 15 points:<br> What does x equal?: -5(x - 4) = -30
Sergeu [11.5K]
-5(x-4) =30

-5x+20=-30
      -20  -20

-5x=-50
 
-50 divided by -5 = 10
6 0
3 years ago
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