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guapka [62]
3 years ago
9

A penguin can dive 48 meters below the surface of the water to reach a fish. It dives 2% of the total depth every 1 second. What

is the total depth of the penguin after 4 seconds?
A. −3.84 meters
B. −0.96 meters
C. −1.92 meters
D. −2.88 meters
Mathematics
1 answer:
Brut [27]3 years ago
7 0

Answer:

A. -3.84 meters

Step-by-step explanation:

Given:

A penguin can dive 2% of the total depth every 1 second.

Total depth = 48 meters.

Now we have to find 2% of 48.

= \frac{2}{100} .48

= \frac{96}{100}

= 0.96 meter

So, the penguin dives -0.96 meters per seconds down the water.

The negative sign represents below the surface of the water.

The total depth of the penguin after 4 seconds = 4 times -0.96

= -3.84 meters

Therefore, the answer is A. -3.84 meters

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Answer:

a) We are interested on this probability

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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>13.17)=P(\frac{X-\mu}{\sigma}>\frac{13.17-\mu}{\sigma})=P(Z>\frac{13.17-13}{0.5})=P(z>0.34)=0.367

b) P(\bar X >13.17)

The new z score is given by:

z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

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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 weights of a population, and for this case we know the distribution for X is given by:

X \sim N(13,0.5)  

Where \mu=13 and \sigma=0.5

We are interested on this probability

P(X>13.17)

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>13.17)=P(\frac{X-\mu}{\sigma}>\frac{13.17-\mu}{\sigma})=P(Z>\frac{13.17-13}{0.5})=P(z>0.34)=0.367

Part b

Since the distribution for X is normal then we can conclude that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we want this probability:

P(\bar X >13.17)

The new z score is given by:

z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

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