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Harrizon [31]
3 years ago
7

Find the equation of the sphere if one of its diameters has endpoints (4, 2, -9) and (6, 6, -3) which has been normalized so tha

t the coefficient of x2 is 1.
Mathematics
1 answer:
Pavel [41]3 years ago
8 0

Answer:

(x - 5)^2 + (y - 4)^2 + (z - 6)^2 = 14.

(Expand to obtain an equivalent expression for the sphere: x^2 - 10\,x + y^2 - 8\, y + z^2 - 12\, z + 63 = 0)

Step-by-step explanation:

Apply the Pythagorean Theorem to find the distance between these two endpoints:

\begin{aligned}&\text{Distance}\cr &= \sqrt{\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2 + \left(z_2 - z_1\right)^2} \cr &= \sqrt{(6 - 4)^2 + (6 - 2)^2 + ((-3) - (-9))^2 \cr &= \sqrt{56}}\end{aligned}.

Since the two endpoints form a diameter of the sphere, the distance between them would be equal to the diameter of the sphere. The radius of a sphere is one-half of its diameter. In this case, that would be equal to:

\begin{aligned} r &= \frac{1}{2} \, \sqrt{56} \cr &= \sqrt{\left(\frac{1}{2}\right)^2 \times 56} \cr &= \sqrt{\frac{1}{4} \times 56} \cr &= \sqrt{14} \end{aligned}.

In a sphere, the midpoint of every diameter would be the center of the sphere. Each component of the midpoint of a segment (such as the diameter in this question) is equal to the arithmetic mean of that component of the two endpoints. In other words, the midpoint of a segment between \left(x_1, \, y_1, \, z_1\right) and \left(x_2, \, y_2, \, z_2\right) would be:

\displaystyle \left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right).

In this case, the midpoint of the diameter, which is the same as the center of the sphere, would be at:

\begin{aligned}&\left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right) \cr &= \left(\frac{4 + 6}{2},\, \frac{2 + 6}{2}, \, \frac{(-9) + (-3)}{2}\right) \cr &= (5,\, 4\, -6)\end{aligned}.

The equation for a sphere of radius r and center \left(x_0,\, y_0,\, z_0\right) would be:

\left(x - x_0\right)^2 + \left(y - y_0\right)^2 + \left(z - z_0\right)^2 = r^2.

In this case, the equation would be:

\left(x - 5\right)^2 + \left(y - 4\right)^2 + \left(z - (-6)\right)^2 = \left(\sqrt{56}\right)^2.

Simplify to obtain:

\left(x - 5\right)^2 + \left(y - 4\right)^2 + \left(z + 6\right)^2 = 56.

Expand the squares and simplify to obtain:

x^2 - 10\,x + y^2 - 8\, y + z^2 - 12\, z + 63 = 0.

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We can use 2 point form, or point-slope form.

Let's use </span>point-slope form.

the slope m is \frac{-5-2}{2-(-3)}= \frac{-7}{5}, then use any of the points to write the equation. (ex, pick (2, -5))

y-(-5)=(-7/5)(x-2)

y+5=(-7/5)x+14/5

y= (-7/5)x+14/5 - 5 =(-7/5)x+14/5 - 25/5 =(-7/5)x-11/5


Thus, the lines are 

i) y=-ax+4      and  ii) y=(-7/5)x-11/5

the slopes are the coefficients of x: -a and (-7/5),

the product of the slopes of 2 perpendicular lines is -1, 

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(-a)(-7/5)=-1

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Answer:

a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

Problems of normally distributed samples can be solved using 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.

Mildly obese

Normally distributed with mean 373 minutes and standard deviation 67 minutes. So \mu = 373, \sigma = 67

A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?

So n = 5, s = \frac{67}{\sqrt{5}} = 29.96

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

Z = \frac{410 - 373}{29.96}

Z = 1.23

Z = 1.23 has a pvalue of 0.8907.

So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

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Normally distributed with mean 526 minutes and standard deviation 107 minutes. So \mu = 526, \sigma = 107

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So n = 5, s = \frac{107}{\sqrt{5}} = 47.86

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

Z = \frac{410 - 526}{47.86}

Z = -2.42

Z = -2.42 has a pvalue of 0.0078.

So there is a 1-0.0078 = 0.9922 = 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

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