Question

The distance to the first delivery is 12 1/3miles. The distance between the next three delivery is is 8 3/4 miles, 17 2/8 miles, and 23 2/3 miles respectiviely. The distance from the final delivery to the shop is 10 5/10 miles. What is average distance for all segments of this trip?

Answer 1

the average value of something, is the (sum of the data elements) / (how many elements).

for example, in this case we have 5 elements, 5 delivery distances, so the average distance for them will be, their sum divided by 5, so let's do so them, firstly changing the mixed fractions to "improper" fractions,


`\bf \stackrel{mixed}{12(1)/(3)}\implies \cfrac{12\cdot 3+1}{3}\implies \stackrel{improper}{\cfrac{37}{3}} \\\\\\ \stackrel{mixed}{8(3)/(4)}\implies \cfrac{8\cdot 4+3}{4}\implies \stackrel{improper}{\cfrac{35}{4}} \\\\\\ \stackrel{mixed}{17(2)/(8)}\implies \cfrac{17\cdot 8+2}{8}\implies \stackrel{improper}{\cfrac{138}{8}} \\\\\\ \stackrel{mixed}{23(2)/(3)}\implies \cfrac{23\cdot 3+2}{3}\implies \stackrel{improper}{\cfrac{71}{3}}`


`\bf \stackrel{mixed}{10}(5)/(10)\implies \cfrac{10\cdot 10+5}{10}\implies \stackrel{improper}{\cfrac{105}{10}}\\\\ -------------------------------\\\\\\`

now, let's add them up, and divide by 5,


`\bf \cfrac{(37)/(3)+(35)/(4)+(138)/(8)+(71)/(3)+(105)/(10)}{5}\impliedby \textit{so our LCD is 120 above} \\\\\\ \cfrac{(1480+1050+2070+2840+1260)/(120)}{5}\implies \cfrac{(8700)/(120)}{5}\implies \cfrac{(145)/(2)}{5}\implies \cfrac{(145)/(2)}{(5)/(1)} \\\\\\ \cfrac{145}{2}\cdot \cfrac{1}{5}\implies \cfrac{145}{10}\implies \cfrac{29}{2}\implies 14(1)/(2)`