Station A and Station B are 120 miles apart. A train traveling from station A to station B travels exactly 1/3 of the distance when it is held up and stops for 16 minutes. When the train continues its journey, the engineer increases its avarage speed by 15 mph so that the train will arive at station B on schedule. Find the avarage speed of the train before it stopped for minutes.

88.64mph

Step-by-step explanation:

Distance, d from A to B = 120

Initial speed up to 40miles distance = v

1/3 of 120 = 40 miles

To arrive on schedule :

New speed = v + 15

Distance traveled = (120 - 40) = 80 miles

Distance train should have covered during the stop time :

Initial speed * stop time = v * (16/60) = 0.2666666v

Speed increase to compensate for the delay = 15 mph over (120 - 40) = 80

v + 15 mph over 80 miles

v + 15 = 80 + 0.2666666v

v - 0.2666666v = 80 - 15

0.7333333v = 65

v = 88.636363

b. 6

step-by-step explanation:

we are given the table representing an exponential function.

we can see from the table that, as x is increasing, the value of y is increasing in the power of 6.

i.e. the following relation is obtained,

x = 1, y = 6

x = 2, y = 36 =

x = 3, y = 216 =

x = 4, y = 1296 =

x = 5, y = 7776 =

so, the geometric series obtained is .

thus, the common ratio is given by = = 6.

hence, the common ratio is 6.