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daser333 [38]
3 years ago
12

A recent drug survey showed an increase in the use of drugs and alcohol among local high school seniors as compared to the natio

nal percent. Suppose that a survey of 100 local seniors and 100 national seniors is conducted to see if the proportion of drug and alcohol use is higher locally than nationally. Locally, 69 seniors reported using drugs or alcohol within the past month, while 60 national seniors reported using them. Conduct a hypothesis test at the 5% level of significance.
Mathematics
1 answer:
murzikaleks [220]3 years ago
5 0

Answer:

There is evidence to claim that the proportion of drug and alcohol use is higher locally than nationally.

Step-by-step explanation:

Given that a recent drug survey showed an increase in the use of drugs and alcohol among local high school seniors as compared to the national percent.

Let p1 be the proportion of local seniors and p2 national seniors.

H_0: p_1=p_2\\H_a: p_1 >p_2

(right tailed test at 5% significance level)

Group I II  

Success 69 60 129

Total 100 100 200

p 0.69 0.6 0.645

q 0.31 0.4 0.355

se 0.046249324 0.048989795 0.033836002

p diff 0.09  

Std error 0.03384  

Z 2.659574468 0  

p 0.00391  

we find that p value is 0.00391 <0.05

Hence we reject null hypothesis

There is evidence to claim that the proportion of drug and alcohol use is higher locally than nationally.

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Answer:

  37,528 containers of cookies

Step-by-step explanation:

a) <u>Cost Function</u>

For the fixed annual demand of 500,000 containers, the manufacturing cost of $0.53 each will total ...

  manufacturing cost = 500,000×$0.53 = $265,000 annually.

For a production batch size of x containers, those x containers will be put into storage, and withdrawn at the uniform rate of 500,000 containers per year. On average, (1/2)x containers will be in storage, so storage costs will be ...

  storage cost = (1/2)($0.36x) = $0.18x . . . . annually

For annual demand of 500,000 containers, and a production batch size of x, there will be 500,000/x production batches each year. The setup cost is $507 for each of those, so the annual setup cost is ...

  setup cost = (500,000/x)($507) = $253,500,000/x . . . . annually

The total cost of producing 500,000 containers annually will be ...

  C = setup cost + storage cost + manufacturing cost

  C(x) = 253,500,000/x + 0.18x + 265,000

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b) <u>Minimal-Cost Batch Size</u>

Cost will be minimized when its derivative with respect to x is zero.

  dC/dx = -253,500,000/x^2 +0.18 = 0

  x^2 = 253,500,000/0.18 . . . . . . . solve for x^2

  x ≈ 37,527.8 ≈ 37,528 . . . . . . . . . take the square root

The size of the production run that will minimize production cost is 37,528 boxes.

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<em>Comment on the solution</em>

You may notice that the equation we finally solve for batch size is equivalent to one that sets setup cost equal to storage cost:

  253,500,000/x = 0.18x

This relation (setup cost = storage cost) is the general solution for this sort of problem regarding batch size.

8 0
3 years ago
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IrinaVladis [17]

Answer:

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Step-by-step explanation:

Given

As the expression is given by

3(5+4y)+8-2y

To determine

Simplify the expression

Let us simplify the expression

3(5+4y)+8-2y

Expand 3(5 + 4y) = 15 + 12y

3\left(5+4y\right)+8-2y=15+12y+8-2y

grouping like terms

                              =12y-2y+15+8

                              =10y+23

Therefore,

3\left(5+4y\right)+8-2y=10y+23

4 0
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