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iren2701 [21]
3 years ago
10

A scout troop is collecting aluminum cans to raise money. They have collected 150 lbs. of aluminum for which they can get 80 cen

ts per pound. If they continue to collect, they estimate they can get an additional 5 pounds of aluminum per day. The price of aluminum is decreasing 2 cents per day. If they can only make 1 trip to the recycling center, when should they make it? How much can they expect to collect for their cans? ( 5 days , $122.50 )
Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
4 0
Since the price of aluminum is dropping at some point the extra 5 pounds you can collect per day does not give you more money and you should stop collecting. Let us find out how much money we will get for each extra day of collecting. Remember each day will give us 5 pounds more but the amount paid is 2 cents less per pound.

NOW: (150)(.80) = 120
1 day more of collecting: (155)(78)=120.90
2 days more (160)(76)=121.60
3 days more (165)(.74)=122.10
4 days more (170)(.72)=122.40
5 days more (175)(.70)=122.50
6 days more (180)(.68)=122.40

At 6 days more you see that the extra aluminum is not enough to make more money than the prior day because the price has dropped too much. This will continue. So, the best choice is to go 5 days more and make $122.50

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A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex
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Answer:

a

 The  90% confidence interval that  estimate the true proportion of students who receive financial aid is

     0.533  <  p <  0.64

b

   n = 1789

Step-by-step explanation:

Considering question a

From the question we are told that

      The sample size is  n = 200

      The number of student that receives financial aid is k = 118

Generally the sample proportion is  

      \^ p = \frac{k}{n}

=>   \^ p = \frac{118}{200}

=>   \^ p = 0.59

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

Generally from the normal distribution table the critical value  of \frac{\alpha }{2}  is  

   Z_{\frac{\alpha }{2} } =  1.645

Generally the margin of error is mathematically represented as  

     E =  Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }

 =>E =  1.645 * \sqrt{\frac{0.59 (1- 0.59)}{200} }

=>  E = 0.057

Generally 90% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

  =>  0.533  <  p <  0.64  

Considering question b

From the question we are told that

    The margin of error  is  E = 0.03

From the question we are told the confidence level is  99% , hence the level of significance is    

      \alpha = (100 - 99 ) \%

=>   \alpha = 0.01

Generally from the normal distribution table the critical value  of   is  

   Z_{\frac{\alpha }{2} } = 2.58

Generally the sample size is mathematically represented as      

        [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )

=>      n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )

=>      n = 1789

8 0
3 years ago
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