The lists below gives the miles. james drive his car in 7 days and the miles jane drove her car in 7 days. compare the means of the two data sets. show all work and explain your final answer.

james: 0, 11, 24, 69, 84, 95, 102
jane: 0, 200, 420, 760, 800, 150, 400

Step-by-step explanation:

Hi, there are actually three types of averages, the mean, the mode, and the medium for any set of numerical data.  The daily miles for both James and Jane is such a set of data  This problem seeks the mean, or the average miles driven per day by James and Jane.

So  we add up the daily miles driven by James and by Jane, separately.

In the case of James, if you do the addition, you should come up with 385 miles.  Do it and make sure we agree.

In the case of Jane, if you add the daily miles driven, you should come up with 2,730 miles. Wow that's a bunch of driving, just about the distance from California to New York if you didn't drive in circles and purposely drove the shortest route from one place to the other.

James drove 385 miles in 7 days

James drove 2,730 miles in 7 days

To get the average miles driven per day, we simply divide the total number of miles driven by the number of days of interest to us or asked of us. In this problem, we could decide to include the first day where both James and Jane could have driven some miles but didn't. This makes sense because the problem says both of them drove the car in 7 days.

So, for James we have the mean (average) = to 385/7 =  55 miles per day

And for Jane, we have the mean (average) = to  2,730/7= 390 miles per day.  

Hope this helps.

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step-by-step explanation:

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geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}