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MArishka [77]
3 years ago
13

Two types of barrel units were in use in the 1920s in the United States. The apple barrel had a legally set volume of 7056 cubic

inches; the cranberry barrel, 5826 cubic inches. If a merchant sells 33 cranberry barrels of goods to a customer who thinks he is receiving apple barrels, what is the discrepancy in the shipment volume in liters (L)? Give your answer as a positive number.
Mathematics
1 answer:
babymother [125]3 years ago
7 0

Answer:

Discrepancy = 665,15 L

Step-by-step explanation:

Data: 1 Apple Barrel: 7056 cubic inches

         1 Cranberry Barrel: 5826 cubic inches

The merchant sells 33 cranberry barrels, so we need to find out the total volume of that. So we multiply our cranberry volume (data) 33 times:

33 x 5826 cubic inches = 192258 cubic inches

Now, the customer thinks that he is receiveing apple barrels. To calculate this volume, we need to multiply the apple volume (data) 33 times:

33 x 7056 cubic inches = 232848 cubic inches

(You can see that 33 apple barrels have more volume that 33 cranberry barrels, so the customer will receive less volume than he is expecting)

The problem is asking the discrepancy in the shipment in liters (L). First we calculate the discrepancy (difference) in cubic inches.

Discrepancy (cubic inches) = 232848 cubic inches - 192258 cubic inches = 40590 cubic inches

Finally we need to transform the units. As a general rule, we know that:

1 litre (L) = 61,0237 cubic inches. Using a simple rule of three we can solve it:

Discrepancy (L) = \frac{40590 cubic inches}{61,0237 cubic inches\\} x 1 L

Discrepancy (L) = 665,15 L

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From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean leng
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Complete Question

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean length of bass caught in a small lake. A normal distribution was assumed. Using the 90% confidence interval obtain:

a. A point estimate of \mu and its 90% margin of error.

b. A 95% confidence interval for \mu.

Answer:

a

\= x  = 13.4   .   E = 2.3

b

10.7 <  \mu < 16.1

Step-by-step explanation:

From the question we are told that

  The sample size is  n = 18

  The 90% confidence interval is  (11.1, 15.7)

Generally the point estimate of  \mu is mathematically  evaluated  as

       \= x  = \frac{11.1 + 15.7 }{2}

=>    \= x  = 13.4

Generally the margin of error is mathematically evaluated  as

     E = \frac{15.7 - 11.1}{2 }

=> E = 2.3

  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 equation for the lower limit of the confidence interval is  

      \= x  -  Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{18} } = 11.1

=> 13.4   -  0.3877 s  = 11.1

=>  s = 5.932

  From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

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

   Z_{\frac{\alpha }{2} } =  1.96

Generally the margin of error is mathematically represented as  

      E = Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }

=>    E =  1.96 *  \frac{5.932}{\sqrt{18} }

=>    E =  2.7      

Generally 95% confidence interval is mathematically represented as  

      \= x -E <  \mu <  \=x  +E

=>    13.4  -  2.7  <  \mu < 13.4  +   2.7

=>    10.7 <  \mu < 16.1

3 0
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