The following data for a random sample of banks in two cities represent the ATM fees for using another bank's ATM. Compute the range and sample standard deviation for ATM fees for each city. Which city has the most dispersion based on range? Which city has more dispersion based on the standard deviation? City A 1.001.00 1.001.00 1.501.50 1.251.25 1.501.50 City B 2.502.50 1.001.00 1.501.50 0.000.00 2.002.00 The range for city A is $0.50.5. The range for city B is $2.52.5. The standard deviation for city A is $0.250.25. (Round to the nearest cent as needed.) The standard deviation for city B is $0.960.96. (Round to the nearest cent as needed.) Which city has the most dispersion based on range? A. City Upper BB, because it has a higher range. B. City Upper AA, because it has a higher range. C. City A, because it has a lower range. D. City B, because it has a lower range. Which city has the most dispersion based on standard deviation?

Option A, City B has more dispersion than City A based on range

City B has more dispersion than City A based on standard deviation

Step-by-step explanation:

City A $1.00     $1.50     $1.25     $1.50     $1.25

Range = highest - lowest = $1.50 - $1.00

The range for city A is $0.50

The standard deviation for city A is $0.25

City B $2.00     $1.50     $1.50     $0.00     $2.00

The range for city B is $2.5

The standard deviation for city B is $0.96

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step-by-step explanation:

i took the test: )

x + y = 90 5x – 10y = 90 90 – 10y = 5x 90 + 10y = 5x

step-by-step explanation:

last month maria hiked the 5-mile mountain trail a number of times and she hiked the 10-mile canal trail several times. let x represent the number of times she hiked the 5-mile trail, and let y represent the number of times she hiked the 10-mile trail. if she hiked a total of 90 miles, which equation can be used to find the number of times maria hiked each trail? x + y = 90 5x – 10y = 90 90 – 10y = 5x 90 + 10y = 5x