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vladimir1956 [14]
3 years ago
7

In a tornado, the wind velocity in meters per second can be described by the function v(p)-12.3 1134 -3p where p is the air pres

sure in millibars. What is the air pressure of a tornado in which the wind velocity is 49.3 meters per second? Round your answer to the nearest whole millibar.
Mathematics
2 answers:
anastassius [24]3 years ago
8 0

The answer is 373 millibars.

natali 33 [55]3 years ago
5 0

In a tornado, the wind velocity in meters per second can be described by the function

 v(p)-12.3=1134 -3p

Whee,p=Pressure in Millibars

When, Wind Velocity =49.3 meters per second

We have to Calculate air Pressure.

49.3-12.3=1134-3p

37=1134-3p

3p=1134-37

3p=1097

\rightarrow p=\frac{1097}{3}\\\\p=365.67\\\\p=366\text{approx}

So, Pressure=366 Millibar

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Select the correct product.<br><br> (y + z)3
Gre4nikov [31]

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Step-by-step explanation:

(y + z)3

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3y + 3z

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Hope this helps :)

7 0
3 years ago
Laura says that the distance between (4,-2) and (4,-7) in the coordinate plane is equal to |-2-(-7)|units Ryan say that the dist
son4ous [18]

Answer:

Both are correct.

Step-by-step explanation:

In the point (4,-2) and (4,-7), the x-coordinates are same. So, the distance between these points is the absolute difference between the y-coordinates of these points.

Distance=|-2-(-7)|=5\text{ units}

So, Laura is correct.

In the point (2,3) and (-9,3), the y-coordinates are same. So, the distance between these points is the absolute difference between the x-coordinates of these points.

Distance=|2-(-9)|=|2+9|=11\text{ units}

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3 0
3 years ago
At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population pro
Gnom [1K]

Answer:

A sample of 1068 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

95% confidence level

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At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?

We need a sample of n.

n is found when M = 0.03.

We have no prior estimate of \pi, so we use the worst case scenario, which is \pi = 0.5

Then

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.03\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.03}

(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}

n = 1067.11

Rounding up

A sample of 1068 is needed.

8 0
3 years ago
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