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ExtremeBDS [4]
3 years ago
7

In automobile mileage and gasoline-consumption testing, 13 automobiles were road tested for 300 miles in both city and highway d

riving conditions. The following data were recorded for miles-per-gallon performance. City 16.2 16.7 15.9 14.4 13.2 15.3 16.8 16.0 16.1 15.3 15.2 15.3 16.2 Highway 19.1 20.3 18.0 18.3 18.9 17.1 16.9 18.3 18.7 20.8 19.1 18.2 18.4 Use the mean, median, and mode to make a statement about the difference in performance for city and highway driving. Compute the mean for the miles-per-gallon performance for city and highway driving. (Round your answers to two decimal places.) city 15.9 miles per gallon highway 15.3 miles per gallon Is the mean mileage better on the highway than in the city? Yes No Compute the median for the miles-per-gallon performance for city and highway driving. city miles per gallon highway miles per gallon Is the median mileage better on the highway than in the city? Yes No Compute the mode(s) for the miles-per-gallon performance for city and highway driving. (Enter your answers as a comma-separated list.) city miles per gallon highway miles per gallon Is the modal mileage better on the highway than in the city? Yes No Make a statement about the difference in performance for city and highway driving conditions. Miles-per-gallon performance is in highway driving condition than in city driving conditions.

Mathematics
1 answer:
Aloiza [94]3 years ago
3 0

Answer:

A. No

City driving mean requires less miles than Highway driving mean.

B. No

City driving median requires less miles than Highway driving median.

C. No

City driving mode requires less miles than Highway driving mode

D.

The driving conditions in the City is better than the driving conditions in the Highway because all measures of centers are lesser than those of Highway.

Step-by-step explanation:

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Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and t
WINSTONCH [101]

Answer:

Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and their number is increasing at the rate of 1 dP dt = rodent per month when there are P = 10 rodents.  

How long will it take for this population to grow to a hundred rodents? To a thousand rodents?

Step-by-step explanation:

Use the initial condition when dp/dt = 1, p = 10 to get k;

\frac{dp}{dt} =kp^2\\\\1=k(10)^2\\\\k=\frac{1}{100}

Seperate the differential equation and solve for the constant C.

\frac{dp}{p^2}=kdt\\\\-\frac{1}{p}=kt+C\\\\\frac{1}{p}=-kt+C\\\\p=-\frac{1}{kt+C} \\\\2=-\frac{1}{0+C}\\\\-\frac{1}{2}=C\\\\p(t)=-\frac{1}{\frac{t}{100}-\frac{1}{2}  }\\\\p(t)=-\frac{1}{\frac{2t-100}{200} }\\\\-\frac{200}{2t-100}

You have 100 rodents when:

100=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{100} \\\\2t=98\\\\t=49\ months

You have 1000 rodents when:

1000=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{1000} \\\\2t=99.8\\\\t=49.9\ months

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Step-by-step explanation: letter B is the right answer

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