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lisabon 2012 [21]
2 years ago
9

The water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day.

They would like the estimate to have a maximum error of 0.15 gallons. A previous study found that for an average family the standard deviation is 2.3 gallons and the mean is 17.6 gallons per day. If they are using a 80% level of confidence, how large of a sample is required to estimate the mean usage of water? Round your answer up to the next integer.
Mathematics
1 answer:
Alexeev081 [22]2 years ago
4 0

Answer: 387

Step-by-step explanation:

Given : Standard deviation: \sigma=2.3

Margin of error : 0.15

Critical value for 80% confidence interval = 1.2816

Required minimum sample size : n=(\dfrac{z_{\alpha/2}\cdot \sigma}{E})^2

\\\\=(\dfrac{1.2816\times2.3}{0.15})^2\\\\=386.16966144\approx387

Hence, the minimum sample size required to estimate the mean usage of water = 387

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Assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt. x2 + y2 = 25 (a) Fin
Artemon [7]

Answer:

(a) \frac{dy}{dt}=-3\frac{3}{4}

(b) \frac{dx}{dt}=3\frac{3}{4}

Step-by-step explanation:

x^{2} +y^{2}=25

Take \frac{d}{dt} of of each term.

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For Question a

2y\frac{dy}{dt}=-2x\frac{dx}{dt}\\\\\frac{dy}{dt}=\frac{-2x\frac{dx}{dt}}{2y} \\\\\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}

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2x\frac{dx}{dt}=-2y\frac{dy}{dt}\\\\\frac{dx}{dt}=\frac{-2y\frac{dy}{dt}}{2x} \\\\\frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}

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Answer:

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PS: I really do not understand the options

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