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mash [69]
3 years ago
5

A restaurant has one type of lemonade that is 12% sugar and another that is 7% sugar. How many gallons of each type does that re

staurant need to make 20 gallons of a lemonade mixture that is 10% sugar? A.12 gallons of the 12% lemonade and 8 gallons of the 7% lemonade. B. 10 gallons of the 12% lemonade and 10 gallons of the 7% lemonade. C. 8 gallons of the 12% lemonade and 12 gallons of the 7% lemonade. D. 2 gallons of the 12% lemonade and 18 gallons of the 7% lemonade.
Mathematics
1 answer:
alexira [117]3 years ago
7 0
For the restaurant to make 20 gallons of a lemonade mixture that is 10% sugar, they need 12 gallons of the 12% lemonade and 8 gallons of the 7% lemonade. The correct answer is A. 
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Answer:

Therefore the  95% confidence interval is (25,707.480 < E < 26,744.920)

Step-by-step explanation:

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let E = true mean

let A = test statistic

Find 95% Confidence Interval

so

let  A =  (u - E) *  (\sqrt{n}  / s)   be the test statistic

we want      P( average_l <  A  < average_u )  = 95%

look for  lower 2.5%  and the upper 97.5%  Because I think this is a 2-tail test

average_l =  -1.96  which corresponds to the 2.5%

average_u = 1.96

P(  -1.96  <  A  <  1.96)  =  95%

P(  -1.96  <  (u - E) *  (\sqrt{n}  / s)  <  1.96)  =  95%

Solve for the true mean E ok

E   <   u + 1.96* (s  / \sqrt{n})

from  -1.96  <  (u - E) *  (\sqrt{n}  / s)

E < 26,226.2 +  1.96*( 2,322.32 / \sqrt{77} )

E < 26,226.2 +  1.96*( 2,322.32 / \sqrt{77} )

E < 26,226.2 +  518.7197348105429466

upper bound is 26,744.9197

or

u - 1.96* (s  / \sqrt{n})  < E

26,226.2 -  518.7197348105429466  < E

25,707.48026519  < E

lower bound is 25,707.48026519

Therefore the  95% confidence interval is (25,707.480 < E < 26,744.920)

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