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1 year ago

6

find the area under the standard normal distribution curve to the right of . use a ti-83 plus/ti-84 plus calculator and round th

e answer to at least four decimal places.

1 answer:

xxTIMURxx [149]1 year ago

4 0

using calculator

Area right to the right of the 2=2.12

P(Z > 2.12)

• P(Z < -2.12)

= 0.0170

Learn more about Standard normal distribution curve:

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Consider the surface defined by the equation x3 + y3 + z3 = 3xyz. Use implicit differ- ∂z entiation to find ∂x. (Note z is not a

Aloiza [94]

Answer:

$\frac{∂z}{∂x}=\frac{(3yz-3x^{2})}{3z^{2} -3xy}$

Step-by-step explanation:

Given equation is $x^{3}+y^{3}+z^{3}=3xyz$ ………………(1)

we use derivative formula $\frac{d}{dx}(x^{n}) = n x^{n-1}$

$\frac{d}{dx}(x^{3)} = 3 x^{2}$

$\frac{d}{dx}(z^{3)} = 3 z^{2}$

And also apply 'u v' formula

$\frac{d}{dx}(uv}) = u\frac{d}{dx}(v)+v\frac{d}{dx}(u)$

Differentiating equation (1) partially with respective to 'x' , we treated 'y' as constant.

$(3x^{2} +0+3z^{2}\frac{∂z}{∂x}) =3y(x\frac{∂z}{∂x}+z(1))$ ( here y treated as constant so the derivative of constant function is zero in addition but in multiplication the constant is keep as like 'y').

on simplification , we get

$(3x^{2} +0+3z^{2}\frac{∂z}{∂x}) =(3yx\frac{∂z}{∂x}+3yz)$

again simplification, we get

$3z^{2}\frac{∂z}{∂x}- 3yx\frac{∂z}{∂x}=3yz-3x^{2}$

taking common ' $\frac{∂z}{∂x}$ on left on side , we get

$(3z^{2}- 3yx)\frac{∂z}{∂x}=3yz-3x^{2}$

dividing ' $(3z^{2}- 3yx)$ on both sides, we get

$\frac{∂z}{∂x}=\frac{(3yz-3x^{2})}{3z^{2} -3xy}$

<u>Final answer</u>:-

$\frac{∂z}{∂x}=\frac{(3yz-3x^{2})}{3z^{2} -3xy}$

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Let U= U= Universal set ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9}={0, 1, 2, 3, 4, 5, 6, 7, 8, 9} , A={1, 3, 4, 5, 7, 9} A={1, 3, 4, 5, 7,

blagie [28]

Answer:

a) $A' = [0,2,6,8]$

b) $(AUB)' = [0,2,6]$

c) $(AUB')' = [9]$

d) A∩B′ $= [3,5,9]$

Step-by-step explanation:

Assuming this problem: "Let U= U= Universal set ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9} , A={1, 3, 4, 5, 7, 9} , and B={1, 4, 7, 8} . List the elemetns of the following sets in the increasing order: a) A′= b) (A∪B)′={ , , }} c) (A∪B′)′={ }} d) A∩B′={ , , }}"

Part a

For this case we just need to find the elements in the universal set that are not in A. And we see that:

$A' = [0,2,6,8]$

And that represent the complement for A

Part b

For this case we need to find first the Union AUB who are the elements on A or B without repetition and we got:

$AUB = [1,3,4,5,7,8,9]$

And now the complement for (AUB)' are the elements that are not in AUB but are on the universal set and we got:

$(AUB)' = [0,2,6]$

Part c

For this case we need to find B' who are the elements on the universal set that are not in B

$B' = [0,2,3,5,6,9]$

Then we can find the union between AUB' and we got:

$AUB' = [0,1,2,3,4,5,6,7,9]$

And then the complment is just:

$(AUB')' = [9]$

Part d

For this case we just need to see the elements in common between A and B' and we got:

A∩B′ $= [3,5,9]$

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