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salantis [7]
3 years ago
10

Normally I don’t ask for help but I’m very lost on question 4. I know I need to first find the derivative of f(x) but I’m not su

re how to go about it. Any help would be greatly appreciated

Mathematics
1 answer:
Daniel [21]3 years ago
8 0

Answer:

\displaystyle f'(\frac{\pi}{4})=2

Step-by-step explanation:

<u>Trigonometric Derivatives</u>

We have the function

\displaystyle f(x)=\sin x \csc x-\frac{1}{\tan x}

We must recall that the sine and the cosecant are reciprocal functions, i.e.:

\displaystyle \sin x =\frac{1}{\csc x}

Also, the reciprocal of the tangent is the cotangent:

\displaystyle \cot x =\frac{1}{\tan x}

Thus:

\displaystyle f(x)=1-\cot x

Now it's easier to take the derivative

\displaystyle f'(x)=0+\csc^2 x= \csc^2 x

Evaluating for x=π/4:

\displaystyle f'(\frac{\pi}{4})= \csc^2 \frac{\pi}{4}

Since csc π/4=\sqrt{2}:

\displaystyle f'(\frac{\pi}{4})= \sqrt{2}^2

\displaystyle f'(\frac{\pi}{4})=2

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The process standard deviation is 0.27, and the process control is set at plus or minus one standard deviation. Units with weigh
mr_godi [17]

Answer:

a) P(X

And for the other case:

tex] P(X>10.15)[/tex]

P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z

So then the probability of being defective P(D) is given by:

P(D) = 0.159+0.159 = 0.318

And the expected number of defective in a sample of 1000 units are:

X= 0.318*1000= 318

b) P(X

And for the other case:

tex] P(X>10.15)[/tex]

P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z

So then the probability of being defective P(D) is given by:

P(D) = 0.00135+0.00135 = 0.0027

And the expected number of defective in a sample of 1000 units are:

X= 0.0027*1000= 2.7

c) For this case the advantage is that we have less items that will be classified as defective

Step-by-step explanation:

Assuming this complete question: "Motorola used the normal distribution to determine the probability of defects and the number  of defects expected in a production process. Assume a production process produces  items with a mean weight of 10 ounces. Calculate the probability of a defect and the expected  number of defects for a 1000-unit production run in the following situation.

Part a

The process standard deviation is .15, and the process control is set at plus or minus  one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces  will be classified as defects."

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(10,0.15)  

Where \mu=10 and \sigma=0.15

We can calculate the probability of being defective like this:

P(X

And we can use the z score formula given by:

z=\frac{x-\mu}{\sigma}

And if we replace we got:

P(X

And for the other case:

tex] P(X>10.15)[/tex]

P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z

So then the probability of being defective P(D) is given by:

P(D) = 0.159+0.159 = 0.318

And the expected number of defective in a sample of 1000 units are:

X= 0.318*1000= 318

Part b

Through process design improvements, the process standard deviation can be reduced to .05. Assume the process control remains the same, with weights less than 9.85 or  greater than 10.15 ounces being classified as defects.

P(X

And for the other case:

tex] P(X>10.15)[/tex]

P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z

So then the probability of being defective P(D) is given by:

P(D) = 0.00135+0.00135 = 0.0027

And the expected number of defective in a sample of 1000 units are:

X= 0.0027*1000= 2.7

Part c What is the advantage of reducing process variation, thereby causing process control  limits to be at a greater number of standard deviations from the mean?

For this case the advantage is that we have less items that will be classified as defective

5 0
2 years ago
A new amusement park presold discounted tickets for the opening day as well as upon arrival at the park the opening day ticket s
julia-pushkina [17]

Answer:

-64 tickets per hour

Step-by-step explanation:

4 0
3 years ago
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AleksAgata [21]
Maybe after the dog walked the street
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What is the product (9t-4)(-9-4)
Scrat [10]

Answer:

-81t^2 +16

Step-by-step explanation:

(9t-4)(-9t-4)

FOIL

first: -9t*9t = -81t^2

outer : -4*9t = -36t

inner : -4 *-9t = 46t

last = -4*-4 = 16

Add them together

-81t^2 -36t+36t+16

-81t^2 +16

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2 years ago
Please help quick pleaseeeeeeeeeeee
aniked [119]

Answer:

think this one

Step-by-step explanation:

the answers there

i think

8 0
2 years ago
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