Mr. Pillot always rides his bicycle to work, and he begins his ride at the same time every day. If he averages 10 miles per hour, he arrives at work 2 minutes late, but, if he averages 15 miles per hour, he arrives 1 minute early. How many miles does Mr. Pillot ride to work? Express your answer as a decimal to the nearest tenth.

10 miles per hours =1/6 miles per minute

15 miles per hours=1/4 miles per minute

Now, assume he need x minute to rides  his bicycle to  work.

if he averages 10 miles per hour ,then he need (x+2) minute to arrives at work

if he averages 15 miles per hour,then he need (x-1)minute to arrives at work

distance equal velocity multiply by time. d=v*t

distance did not change,so we can set up an equation like

1/6*(x+2)=1/4*(x-1)

       x+2=1.5(x-1)

       x+2=1.5x-1.5

      0.5x=3.5

      x=7

If if he average speed is 15  miles per hour (1/4 miles per minute), he need 7-1=6 minute  to arrive at work  

6*(1/4)=3/2 miles=1.5 miles

Same,If he average speed is 10 miles per hour (1/6 miles per minute),he need 7+2=9 minute to arrive at work.

9*(1/6)=3/2 miles=1.5 miles (I just checking)

Step-by-step explanation:

step-by-step explanation:

once you graph it hit 2nd (the blue button) and graph

that takes you to a table of coordinates

on the left is x coordinates and on the right is y coordinates and an empty box

(you may have to scroll up/down in order to get ( x, 0)

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.