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2 years ago

Using the relation in Part B, what is the value of x if f(x) = 99?

Mathematics

1 answer:

Sergeu [11.5K]2 years ago

4 0

I hope this answer helps

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A certain large manufacturing facility produces 20,000 parts each week. The manager of the facility estimates that about 1% of t

Slav-nsk [51]

Answer:

198 is the variance for the number of defective parts made each week.

Step-by-step explanation:

We are given the following in the question:

Number of parts produced each week = 20,000

Percentage of defective parts = 1%

We have to calculate the variance for the number of defective parts made each week.

We treat defective part as a success.

P(Defective part) = 1% = 0.01

$p = 0.01$

Then the number defective parts follows a binomial distribution .

Formula for variance =

$\sigma^2 = np(1-p) = 20000(0.01)(1-0.01) = 198$

Thus, 198 is the variance for the number of defective parts made each week.

8 0

3 years ago

Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

5 0

2 years ago

What are the zeros of the function below? Check all that apply.

nordsb [41]

F, A.

-3, 1

use a graphing calc

or do

x+3 = 0 or x-1 = 0 to get x = -3 or x = 1

8 0

2 years ago

Please help me with this!

GREYUIT [131]

Answer:

p=-7

Step-by-step explanation:

-11+10(p+10)=4-5(2p+11)

Firstly, expand the bracket

-11+10p+100=4-10p-55

Next, you collect like terms

10p+10p=4-55-100+11

20p=-140

Divide through by the coefficient of p

20p/20=-140/20

p=-7

5 0

1 year ago

Plssss help me my test is due is 30 minutes.

Anon25 [30]

Answer:i think its 1.75

Step-by-step explanation

6 0

2 years ago

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