The jonas school district gives awards to its schools based on overall student attendance. the data for attendance are shown in the table, where low represents the fewest days attended and high represents the most days attended for a single student. school low high range mean median iqr σ high school m 128 180 52 141 160 55.5 41.5 high school n 131 180 49 159 154 48.5 36.5 high school p 140 180 40 153 165 32.5 31.5 part a: if the school district wants to award the school that has the most consistent attendance among its students, which high school should it choose and why? justify your answer mathematically. (5 points) part b: if the school district wants to award the school with the highest average attendance, which school should it choose and why? justify your answer mathematically. (5 points)

The most consistent attendance is the one that has less variability (it's more regular). not necessarily the one with more students. so, the case with less variability is the one with less iq, sigma or range (all three measure the dispersion of a distribution. iq is more robust than sigma, and sigma more than the range, although in practice everyone uses sigma).so, the answer to a) is the third high school: hs pb) here one looks at the central measurement: mean, median. this example is not super easy. hs n has the highest mean value, but hs p has the highest median. the median is more robust than the mean, since it is less affected by outliers. so hs p is a good candidate.finally, looking at the low/high values, one can see that the high is the same: some day(s) when all students went and all hs have a maximum number of 180 students. however, the highest low is hs p.so, i think hs p should also be awarded for the highest rate, since its median  is the highest and the lower number of students is the highest.median means 50% of the cases have values less than the median. mean is an average.

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,                                                                    

0 = ( 2 - 4 ) ( p + q )0 = ( 2 - 4 ) ( 2 + 5 i ),   by multiplying these0 = ( 4 + 10 i - 8 - 20 i )0 = 4 - 8 + 10 i - 20 i0 = - 4 -10 i   0 = 2 ( - 2 - 5 i )0 / 2 = - 2 - 5 i- 2 - 5 i = 0- 5 i = 2i = 2 / - 5

The amount used= 13.04- 1.9 = 11.14