Dawna walks 10 miles starting at 1: 00 p. m. every day. on average she finishes at 3: 50 p. m., with a standard deviation of 10 minutes. jeanne leaves 30 minutes later, and every day she runs 10 miles exactly twice as fast as dawna walks on that day. what is the mean time jeanne will finish and the standard deviation of the number of minutes it takes jeanne to finish? (a) mean 2: 25 p. m., standard deviation 5 minutes(b) mean 2: 25 p. m., standard deviation 20 minutes(c) mean 2: 25 p. m., standard deviation 35 minutes(d) mean 2: 55 p. m., standard deviation 5 minutes(e) mean 2: 55 p. m., standard deviation 35 minutes

(D) mean 2:55 p.m., standard deviation 5 minutes

Step-by-step explanation:

Jeanne walks twice as fast Dawna. This mean that her mean is going to be half that of Dawna, and her standard deviation is also going to be half of Dawna.

Dawna standard deviation is of 10 minutes. So Jeanne's standard deviation will be of 5 minutes.

Dawna walks 10 miles starting at 1:00 p.m. every day. On average she finishes at 3:50 p.m.

Dawna takes 2 hours and 50 minutes to walk 10 miles. That is 170 minutes.

Jeanne leaves 30 minutes later, and every day she runs 10 miles exactly twice as fast as Dawna walks on that day.

She leaves at 1:30 p.m. She takes half the time of Dawna, so 170/2 = 85 minutes. That is, 1 hour and 25 minutes. So she finishes at 2:55 p.m.

So the correct answer is:

(D) mean 2:55 p.m., standard deviation 5 minutes

d. millimeters per hour.

step-by-step explanation:

we are given that,

carlo used a rain gauge to measure the amount of water his lawn is receiving from the sprinklers.

since, according to the options, we see that,

water cannot be measured by 'miles per hour', 'meters per second' and 'grams per seconds'.

we get that, the water can be measured by 'millimeters per hour'.

hence, the appropriate rate for describing the amount of water is 'millimeters per hour'.