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Alina [70]
3 years ago
8

The path of a breaching ocra whale can be modeled by the function k(x) = x^2 - 2x - 8, where x is the time elapsed in seconds un

der water, and k(x) is the height in ft below the water.
Determine HOW LONG is the ocra (killer) whale underwater to the nearest second. MUST BE SOLVED ALGEBRAICALLY!
Mathematics
1 answer:
____ [38]3 years ago
3 0

Answer:

The time duration of the ocra whale spends under water is 4 seconds

Step-by-step explanation:

The equation that model the path of the ocra whale is presented as follows;

k(x) = x² - 2·x - 8

Where;

x = The time elapsed in seconds under water

k(x) = The height in feet below water

Therefore, the point where the ocra whale enters the water, k(x) = 0

At the point of submerging below the water surface, we have;

k(x) = 0 = x² - 2·x - 8

Factorizing the quadratic equation gives;

0 = x² - 2·x - 8 = (x - 4)·(x + 2)

Therefore, when k(x) = 0, either;

(x - 4) = 0, which gives, x = 4

(x + 2) = 0, which gives, x = -2

Given that 'x' represent the time, the time elapsed while the whale is under water, x = 4 seconds

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Step-by-step explanation:

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The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.904 g and a standard deviation
Dovator [93]

Answer:

-0.674 = \frac{X -0.904}{0.295}

X= 0.904 -0.674*0.295 = 0.705

0.674 = \frac{X -0.904}{0.295}

X= 0.904 +0.674*0.295 = 1.103

So we expect about 50% of the middle values within 0.705 and 1.103 g

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".  

Solution to the problem

Let X the random variable that represent the amounts of nicotine in a certain brand of a population, and for this case we know the distribution for X is given by:

X \sim N(0.904,0.295)  

Where \mu=0.904 and \sigma=0.295

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}

We want to find a two values that accumulates the middel 50% of the data. So we need to find a quantile who accumulates (1-0.5)/2 = 0.25 of the area on each tail and we got:

z = \pm 0.674

Since P(Z<-0.674) = 0.25 and P(Z>0.674) =0.25. And using this value we can find the possible values of X

-0.674 = \frac{X -0.904}{0.295}

X= 0.904 -0.674*0.295 = 0.705

0.674 = \frac{X -0.904}{0.295}

X= 0.904 +0.674*0.295 = 1.103

So we expect about 50% of the middle values within 0.705 and 1.103 g

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