Reber and kotovsky (1997), in a study of problem solving, described one of their results comparing a specific group of participants within their overall control condition as follows: "this group took an average of 179 moves to solve the puzzle, whereas the rest of the control participants took an average of 74 moves, t1192 = 3.31, p 6 .01" (p. 183). explain this result to a person who has never had a course in statistics. (focus on the meaning of this result in terms of the general logic of hypothesis testing and statistical significance.)

question

what is the value of the third quartile of the data set represented by this box plot? a. 19 b. 21 c. 26 d. 29

answer

c

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!

answer: the proof is mentioned below.

step-by-step explanation:

since, given:   lines bc = ef

prove:   lines ac > ef

          statements                                           reason

1. bc=ef                                           1. given

2. ac= ab+bc                                 2. because b is the point between the  

                                                            line segment ac

3.   ac > bc                                       3. because if a line segment is equal to

                                                          the sum of two line segments then it is

                                                          always greater then each of these

                                                          individual line segments

4. ac > ef                                         4. by equality property.