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Naddik [55]
3 years ago
6

A bulb after certain testing had a life of 25 months. The standard deviation based on this sample of size one is

Mathematics
1 answer:
lapo4ka [179]3 years ago
8 0

Answer:

<h2>25 months</h2>

Step-by-step explanation:

Using the formula for calculating the standard error of the mean to get the standard deviation. The standard error of the mean is expressed as;

SE = S/√n where;

S is the standard deviation

n is the sample size

Given SE = 25 months and n = 1, on substituting this parameters into the formula, we will have;

25 = S/√1

25 = S/1

cross multiply

S = 25*1

S = 25 months

<em>Hence the standard deviation based on the sample is 25 months</em>

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2.9 mi

Step-by-step explanation:

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The plane x + y + z = 12 intersects paraboloid z = x^2 + y^2 in an ellipse.(a) Find the highest and the lowest points on the ell
emmasim [6.3K]

Answer:

a)

Highest (-3,-3)

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b)

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Step-by-step explanation:

To solve this problem we will be using Lagrange multipliers.

a)

Let us find out first the restriction, which is the projection of the intersection on the XY-plane.

From x+y+z=12 we get z=12-x-y and replace this in the equation of the paraboloid:

\bf 12-x-y=x^2+y^2\Rightarrow x^2+y^2+x+y=12

completing the squares:

\bf x^2+y^2+x+y=12\Rightarrow (x+1/2)^2-1/4+(y+1/2)^2-1/4=12\Rightarrow\\\\\Rightarrow (x+1/2)^2+(y+1/2)^2=12+1/2\Rightarrow (x+1/2)^2+(y+1/2)^2=25/2

and we want the maximum and minimum of the paraboloid when (x,y) varies on the circumference we just found. That is, we want the maximum and minimum of  

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

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\bf g(x,y)=(x+1/2)^2+(y+1/2)^2-25/2=0

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\bf \nabla f=(\displaystyle\frac{\partial f}{\partial x},\displaystyle\frac{\partial f}{\partial y})=(2x,2y)\\\\\nabla g=(\displaystyle\frac{\partial g}{\partial x},\displaystyle\frac{\partial g}{\partial y})=(2x+1,2y+1)

Let \bf \lambda be the Lagrange multiplier.

The maximum and minimum must occur at points where

\bf \nabla f=\lambda\nabla g

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\bf (2x,2y)=\lambda(2x+1,2y+1)\Rightarrow 2x=\lambda (2x+1)\;,2y=\lambda (2y+1)

we can assume (x,y)≠ (-1/2, -1/2) since that point is not in the restriction, so

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<em> </em>

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(-3,-3) is the farthest from the origin

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