On a highway a wreck occurred and cause a 10 mile traffic jam. The average car is 13 feet in length, the average truck is 20 feet in length and the average 18 wheeler

is 75 feet in length. If ($2% of the traffic jam consists of cars, 29% of trucks and 9%

of 18 wheelers, how many vehicles are stuck in the traffic jam if the average

distance between vehicles is 3 feet?

81 cars

756 trucks

63 wheelers

Step-by-step explanation:

1 mile = 5280 feet

So the jam is 10 miles = 52800 feet

Cars' length = 2 % of the 52800 = 1056 feet

Trucks' length =  29%  of the 52800= 15312 feet

Wheelers' length = 9%of the 52800= 4752 feet

Average car is 13 feet in length

The number of cars = 1056 feet/13= 81.23 almost 81 cars

The average truck is 20 feet in length

Number of trucks in 15312 feet=15312 feet/20= 765.6= 756 trucks

The average 18 wheeler is 75 feet in length.

Number of wheelers in 4752 feet= 4752 feet/75= 63.36= 63 wheelers

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92