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Fiesta28 [93]
3 years ago
10

The equation for the cost in dollars of producing automobile tires is c = 0.000015x2 - 0.03x + 35, where x is the number of tire

s produced. find the number of tires that minimizes this cost. what is the cost for that number of tires?
Mathematics
1 answer:
statuscvo [17]3 years ago
3 0
The cost function is
c = 0.000015x² - 0.03x + 35
where x = number of tires.

To find the value of x that minimizes cost, the derivative of c with respect to x should be zero. Therefore
0.000015*2x - 0.03 = 0
0.00003x = 0.03
x = 1000

Note:
The second derivative of c with respect to x is positive (= 0.00003), so the value for x will yield the minimum value.

The minimum cost is
Cmin = 0.000015*1000² - 0.03*1000 + 35
          = 20

Answer:
Number of tires = 1000
Minimum cost = 20
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3 years ago
(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each
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Following are the solution parts for the given question:

For question A:

\to (n) = 16

\to (\bar{X}) = 410

\to (\sigma) = 40

In the given question, we calculate 90\% of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}

\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\

Using the t table we calculate t_{ \frac{\alpha}{2}} = 1.753  When 90\% of the confidence interval:

\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53

So 90\% confidence interval for the mean weight of shipped homemade candies is between 392.47\ \ and\ \ 427.53.

For question B:

\to (n) = 500

\to (X) = 155

\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31

Here we need to calculate 90\% confidence interval for the true proportion of all college students who own a car which can be calculated as

\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}

\to C.I= 0.90

\to (\alpha) = 0.10

\to \frac{\alpha}{2} = 0.05

Using the Z-table we found Z_{\frac{\alpha}{2}} = 1.645

therefore 90\% the confidence interval for the genuine proportion of college students who possess a car is

\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344

So 90\% the confidence interval for the genuine proportion of college students who possess a car is between 0.28 \ and\ 0.34.

For question C:

  • In question A, We are  90\% certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
  • In question B, We are  90\% positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:  

brainly.in/question/16329412

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3 years ago
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Peter has built a gazebo, whose shape is a regular heptagon, with a side length of $1$ unit. He has also built a pathway around
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Answer:

Area of pathway = 10.375 unit²

Step-by-step explanation:

Given - Peter has built a gazebo, whose shape is a regular heptagon, with

             a side length of 1 unit. He has also built a pathway around the

             gazebo, of constant width 1 unit, as shown below.

To find -  Find the area of the pathway.

Proof -

Central angle = \frac{360}{7} = 51.43°

Now,

Central angle / 2 = \frac{360}{14} = 25.71°

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height = \frac{1}{2}.\frac{1}{tan25.41} = 1.038

Now,

Inner area of gazebo = 7·\frac{1}{2}·1·(1.038) = 3.634 unit²

Outer area of gazebo = (\frac{2.038}{1.038})²·(3.364) = 14.009 unit²

∴ we get

Area of pathway = Outer area - Inner area

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⇒Area of pathway = 10.375 unit²

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