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

add all the numbers and check the ones place.

the answer is 486. as the ones place is more than 4, the estimation will be the next 10 ie, 490

this is not the answer this is just a question. is this middle school math?

step-by-step explanation: