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xxTIMURxx [149]

2 years ago

13

Four runners start at the same point: Lin, Elena, Diego, Andre. For each runner write a multiplication equation that describes t

heir journey.
I NEED A REAL ANSWER!! NOT "I DON'T KNOW!!" IF YOU DON"T KNOW, DON'T ANSWER!!!

Lin runs for 25 seconds at 8.2 meters per second. What is her finish point?
Elena runs for 28 seconds and finishes at 250 meters. What is her velocity?
Diego runs for 32 seconds at -8.1 meters per second. What is his finish point?
Andre runs for 35 seconds and finishes at -285 meters. What is his velocity?

2 answers:

kirill [66]2 years ago

6 0

Mark the formulas

Velocity=Displacement/Time

#Lin

We need distance

$\\ \tt\hookrightarrow 25(8.2)=205m$

#Elena

We need velocity

$\\ \tt\hookrightarrow 250/28\approx 9m/s$

#Diego

We nend distance

$\\ \tt\hookrightarrow 32(-8.1)=-259.2m$

#Andre

We need velocity

$\\ \tt\hookrightarrow -285/35\approx -8m/s$

Semenov [28]2 years ago

3 0

Answer:

Lin = 8.2 m/s x 25 s =<u> 205 m </u> (<em>Finish Point)</em>

Elena = V = 250 m / 28 s <u>V = 8.93/s </u> (<em>Velocity</em>)

Diego = -8.1 m/s x 32 s =<u> -259.2 meters</u> (<em>Finish Point</em>)

Andre = V = -285 m /35 s V= -8.14ms (<em>Velocity</em>)

Step-by-step explanation:

I hope this helps!

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Answer:

a) $P(X$

$P(z$

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$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

d) $z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the life of a particular auto transmission of a population, and for this case we know the distribution for X is given by:

$X \sim N(72000,12000)$

Where $\mu=72000$ and $\sigma=12000$

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using excel or the normal standard table and we got:

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Part b

$P(X>65000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>-0.583)=1-P(Z$

Part c

$P(X>100000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

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Sicne this probability just represent 1% of the data we can consider this value as unusual.

Part d

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

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