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3 years ago

The percent of area associated with z = 1.4 is ____%

Mathematics

2 answers:

Solnce55 [7]3 years ago

7 0

The answer is A. 38.49

GarryVolchara [31]3 years ago

4 0

Answer with explanation:

To find area associated with, z=1.4, we will take help of z table.

On the vertical axis, we will look at value at 1.4.

$Z_{(1.4)}=0.9192$

At,Z=1.4,Area Occupied = 91.92%

In the normal curve,at

Z=50%,that is at the point,where

Mean =Median =Mode, in the middle of curve

Area Occupied =50%

Z-value at 50% = 0.5000=50%

So,percent of area associated with z = 1.4 is

= Area on the right of mean

= 91.92 % - 50%

=41.92%

Option B

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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

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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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2 5 -18 0

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