(a) 1 gallon = 8.35 pounds
So, 5 gallons = 5 x 8.35 = 41.75 pounds
So, 41.75 pounds of honey will be required to fill 5 gallon jug.
(b) Shear stress = 1.50 N/m^2 = 1.50 Pa
As we now, for water, dynamic viscosity = 10^-3 Pa*s at 20 deg C.
Therefore, the velocity gradient required = Shear stress / Dynamic viscosity
= (1.50 Pa) / (10^-3 Pa*s) = 1.50 x 10^3 /s
2. Estimate the number of pounds of honey it would take to fill a 5 gallon...
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Bob Nale is the owner of Nale’s Quick Fill. Bob would like to estimate the mean number of gallons of gasoline sold to his customers. Assume the number of gallons sold follows the normal distribution with a population standard deviation of 2.50 gallons. From his records, he selects a random sample of 70 sales and finds the mean number of gallons sold is 7.80. What is the point estimate of the population mean? (Round your answer to 2 decimal places.)...
Bob Nale is the owner of Nale’s Quick Fill. Bob would like to estimate the mean number of gallons of gasoline sold to his customers. Assume the number of gallons sold follows the normal distribution with a population standard deviation of 2.30 gallons. From his records, he selects a random sample of 70 sales and finds the mean number of gallons sold is 8.40. What is the point estimate of the population mean? (Round your answer to 2 decimal places.)...
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Bob Nale is the owner of Nale’s Quick Fill. Bob would like to estimate the mean number of gallons of gasoline sold to his customers. Assume the number of gallons sold follows the normal distribution with a population standard deviation of 1.80 gallons. From his records, he selects a random sample of 80 sales and finds the mean number of gallons sold is 6.70. A)What is the point estimate of the population mean? (Round your answer to 2 decimal places.)...