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andre [41]
3 years ago
5

Find the absolute maximum and minimum values of the following functions on the given curves functions:

Mathematics
1 answer:
bagirrra123 [75]3 years ago
6 0

Answer:

Solution has explained below:

Step-by-step explanation:

(a) f(x,y) = x+y

To find out the maximum and minimum values, we need to find first and second derivatives, we have

fx= 1, fx₁=0 and

fy= 1 and fyy=0

For stationary points fx=fy=0, which gives,

1=1=0, so that there is just one stationary point, (x,y)=(0,0)

If  fx₁  < 0 and fyy < 0, function is maximum

If  fx₁ > 0 and fyy > 0, function is minimum.

(b) g(x,y) = xy

Sol: To find out the maximum and minimum values, we need to find first and second derivatives, we have

fx= 1 and fy  = 1

fx₁ = 0 and fyy =0

for stationary points fx = fy =0, which gives 1=1=0, so that there is just one stationary points, (x,y) =(0,0)

If fx₁ < 0 and fyy < 0, function is maximum.

If  fx₁  > 0 and fyy> 0, function is minimum.

c) h(x,y) = 2x2 + y2

sol: To find out the maximum and minimum values, we need to find first and second derivatives, we have

fx = 4x  and fy = 2y

fx₁= 4  and fyy = 2

Now, taking fx = 0 and fy = 0

Gives x=0 and y=0,

Stationary points are (x,y) = (0,0)

If fx₁ < 0 and fyy < 0, solution is maximum,

If fx₁ > 0 and fyy > 0, solution is minimum,

Here, fx₁ = 4 > 0 and fyy = 2 > 0.

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The resting heart rate for an adult horse should average about µ = 47 beats per minute with a (95% of data) range from 19 to 75
KatRina [158]

Answer:

a. 0.0582 = 5.82% probability that the heart rate is less than 25 beats per minute.

b. 0.1762 = 17.62% probability that the heart rate is greater than 60 beats per minute.

c. 0.7656 = 76.56% probability that the heart rate is between 25 and 60 beats per minute

Step-by-step explanation:

Empirical Rule:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean:

\mu = 47

(95% of data) range from 19 to 75 beats per minute.

This means that between 19 and 75, by the Empirical Rule, there are 4 standard deviations. So

4\sigma = 75 - 19

4\sigma = 56

\sigma = \frac{56}{4} = 14

a. What is the probability that the heart rate is less than 25 beats per minute?

This is the p-value of Z when X = 25. So

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

Z = \frac{25 - 47}{14}

Z = -1.57

Z = -1.57 has a p-value of 0.0582.

0.0582 = 5.82% probability that the heart rate is less than 25 beats per minute.

b. What is the probability that the heart rate is greater than 60 beats per minute?

This is 1 subtracted by the p-value of Z when X = 60. So

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

Z = \frac{60 - 47}{14}

Z = 0.93

Z = 0.93 has a p-value of 0.8238.

1 - 0.8238 = 0.1762

0.1762 = 17.62% probability that the heart rate is greater than 60 beats per minute.

c. What is the probability that the heart rate is between 25 and 60 beats per minute?

This is the p-value of Z when X = 60 subtracted by the p-value of Z when X = 25. From the previous two items, we have these two p-values. So

0.8238 - 0.0582 = 0.7656

0.7656 = 76.56% probability that the heart rate is between 25 and 60 beats per minute

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