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IceJOKER [234]
2 years ago
5

At time t is greater than or equal to zero, a cube has volume V(t) and edges of length x(t). If the volume of the cube decreases

at a rate proportional to its surface area, which of the following differential equations could describe the rate at which the volume of the cube decreases?
A) dV/dt=-1.2x^2
B) dV/dt=-1.2x^3
C) dV/dt=-1.2x^2(t)
D) dV/dt=-1.2t^2
E) fav/dt=-1.2V^2
Mathematics
2 answers:
algol [13]2 years ago
8 0

Answer:

  A)  dV/dt=-1.2x^2

Step-by-step explanation:

The rate of change of volume is given by dV/dt. Surface area is proportional to x^2. Since the volume is decreasing, the constant of proportionality between surface area and rate of volume change will be negative. Hence a possible equation might be ...

  dV/dt = -1.2x^2

VARVARA [1.3K]2 years ago
6 0

Answer:

C

Step-by-step explanation:

V(t) = [x(t)]³

A(t) = 6[x(t)]²

dV/dt = k × 6[x(t)]²

Where k < 0

From the options,

taking k = -0.2

dV/dt = -1.2[x(t)]²

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Answer:

a) Mean blood pressure for people in China.

b) 38.21% probability that a person in China has blood pressure of 135 mmHg or more.

c) 71.30% probability that a person in China has blood pressure of 141 mmHg or less.

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Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If X is two standard deviations from the mean or more, it is considered unusual.

In this question:

\mu = 128, \sigma = 23

a.) State the random variable.

Mean blood pressure for people in China.

b.) Find the probability that a person in China has blood pressure of 135 mmHg or more.

This is 1 subtracted by the pvalue of Z when X = 135.

Z = \frac{X - \mu}{\sigma}

Z = \frac{135 - 128}{23}

Z = 0.3

Z = 0.3 has a pvalue of 0.6179

1 - 0.6179 = 0.3821

38.21% probability that a person in China has blood pressure of 135 mmHg or more.

c.) Find the probability that a person in China has blood pressure of 141 mmHg or less.

This is the pvalue of Z when X = 141.

Z = \frac{X - \mu}{\sigma}

Z = \frac{141 - 128}{23}

Z = 0.565

Z = 0.565 has a pvalue of 0.7140

71.30% probability that a person in China has blood pressure of 141 mmHg or less.

d.)Find the probability that a person in China has blood pressure between 120 and 125 mmHg.

This is the pvalue of Z when X = 125 subtracted by the pvalue of Z when X = 120. So

X = 125

Z = \frac{X - \mu}{\sigma}

Z = \frac{125 - 128}{23}

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X = 120

Z = \frac{X - \mu}{\sigma}

Z = \frac{120 - 128}{23}

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Z = -0.35 has a pvalue of 0.3632

0.4483 - 0.3632 = 0.0851

8.51% probability that a person in China has blood pressure between 120 and 125 mmHg.

e.) Is it unusual for a person in China to have a blood pressure of 135 mmHg? Why or why not?

From b), when X = 135, Z = 0.3

Since Z when X = 135 is less than two standard deviations from the mean, it is not unusual for a person in China to have a blood pressure of 135 mmHg.

f.) What blood pressure do 90% of all people in China have less than?

This is the 90th percentile, which is X when Z has a pvalue of 0.28. So X when Z = 1.28. Then

X = 120

Z = \frac{X - \mu}{\sigma}

1.28 = \frac{X - 128}{23}

X - 128 = 1.28*23

X = 157.44

So

157.44mmHg

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