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victus00 [196]
2 years ago
10

Emi computes the mean and variance for the population data set 87, 46, 90, 78, and 89. She finds the mean is 78. Her steps for f

inding the variance are shown below.
sigma squared = StartFraction (87 minus 78) squared + (46 minus 78) squared + (90 minus 78) squared + (78 minus 78) squared + (89 minus 78) squared Over 5 EndFraction. = StartFraction 9 squared minus 32 squared + 12 squared + 0 squared + 11 squared Over 5 EndFraction. = StartFraction 81 minus 1,024 + 144 + 0 + 121 Over 5 EndFraction. = StartFraction negative 678 Over 5 EndFraction. = negative 135.6

What is the first error she made in computing the variance?
Emi failed to find the difference of 89 - 78 correctly.
Emi divided by N - 1 instead of N.
Emi evaluated (46 - 78)2 as -(32)2.
Emi forgot to take the square root of -135.6.
Mathematics
2 answers:
Alex787 [66]2 years ago
7 0

Answer:

its d

Step-by-step explanation:

edg 2021

stiks02 [169]2 years ago
4 0

Answer:

135.6

i think

Step-by-step explanation:

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2 years ago
Where does the helix r(t) = cos(πt), sin(πt), t intersect the paraboloid z = x2 + y2? (x, y, z) = What is the angle of intersect
Colt1911 [192]

Answer:

Intersection at (-1, 0, 1).

Angle 0.6 radians

Step-by-step explanation:

The helix r(t) = (cos(πt), sin(πt), t) intersects the paraboloid  

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\bf (cos(\pi t), sin(\pi t), t)

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\bf cos^2(\pi t)+sin^2(\pi t)=1

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(cos(π), sin(π), 1) = (-1, 0, 1)

The angle of intersection between the helix and the paraboloid is the angle between the tangent vector to the curve and the tangent plane to the paraboloid.

The <em>tangent vector</em> to the helix in t=1 is

r'(t) when t=1

r'(t) = (-πsin(πt), πcos(πt), 1), hence

r'(1) = (0, -π, 1)

A normal vector to the tangent plane of the surface  

\bf z=x^2+y^2

at the point (-1, 0, 1) is given by

\bf (\frac{\partial f}{\partial x}(-1,0),\frac{\partial f}{\partial y}(-1,0),-1)

where

\bf f(x,y)=x^2+y^2

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\bf \frac{\partial f}{\partial x}=2x,\;\frac{\partial f}{\partial y}=2y

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(-2,0,-1)

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\bf (0,-\pi,1)-((0,-\pi,1)\bullet(-2,0,-1))(-2,0,1)=(0,-\pi,1)-(-2,0,1)=(2,-\pi,0)

The angle between the tangent vector to the curve and the tangent plane to the paraboloid equals the angle between the tangent vector to the curve and the vector we just found.  

But we now

\bf (2,-\pi,0)\bullet(0,-\pi,1)=\parallel(2,-\pi,0)\parallel\parallel(0,-\pi,1)\parallel cos\theta

where  

\bf \theta= angle between the tangent vector and its projection onto the tangent plane. So

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and

\bf \theta=arccos(0.8038)=0.6371\;radians

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