Mathematics, 03.01.2020 00:31 BraedynM2089
What price do farmers get for their watermelon crops? in the third week of july, a random sample of 37 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. assume that ? is known to be $1.94 per 100 pounds.
(a) find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop. what is the margin of error? (round your answers to two decimal places.) lower limit $ upper limit $ margin of error $
(b) find the sample size necessary for a 90% confidence level with maximal error of estimate e = 0.25 for the mean price per 100 pounds of watermelon. (round up to the nearest whole number.) farming regions
(c) a farm brings 15 tons of watermelon to market. find a 90% confidence interval for the population mean cash value of this crop. what is the margin of error?
hint: 1 ton is 2000 pounds. (round your answers to two decimal places.) lower limit $ upper limit $ margin of error $
Answers: 2
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Refer to the table below if needed. second quadrant third quadrant fourth quadrant sin(1800- - cos(180° -) tan(180°-e) =- tane cot(1800-0) 10 it to solo 888 sin(180° +c) = - sine cos(180° +) =- cose tan(180° +c) = tane cot(180° +o) = cote sec(180° + c) = - seco csc(180° +2) = - csce sin(360° -) =- sine cos(360° -) = cose tan(360° - e) =- tane cot(360° -) = -cote sec(360° -) = seco csc(360° -) = csco sec(180° -) = csc(180° -) = csca 1991 given that sine = 3/5 and lies in quadrant ii, find the following value. tane
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Can anyone explain to me how to do this? anything will .
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You use technology ans find a sum of 6.712e-8. write this sum in standard form
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What price do farmers get for their watermelon crops? in the third week of july, a random sample of...
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