6.4 Consider the freeway and traffic conditions

described in Problem 6.3. If 180 of the 1800 vehicles

observed in the peak hour were large trucks and

buses, what would the level of service of this freeway

be on a 5-mi, 6% downgrade?

(a). the peak hour factor, PHF= 1800/700 × 4 = 0.6429.

(b). level of service of this freeway = LOS D => 2110 pc/h/ln.

Explanation:

Step one: Calculate the proportion of the buses and the trucks. Which is given as; h= 180/1800 = 0.1

Step two: determine the heavy vehicle adjustment factor. Thus can be calculated by using the formula below;

The heavy vehicle adjustment factor,fh = 1/ (1 +h)(b -1) + c (a - 1). = 1/ (1+ 0.1) (4 - 1) + 0 = 0.769.

Step three : Calculate the peak hour factor.

Thus, the peak hour factor, PHF= 1800/700 × 4 = 0.6429.

Step four: Calculate the 15mins passenger car equivalent flow rate by using the formula below;

V/PHF × N × fh × fp.= 1800/ 0.6429 × 2 × 0.769 × 1 = 1820 pc/h/in.

Step 5 : determine the estimated free flow speed by using the formula below;

75.4 - Flw - Flc - 3.22TRD^0.84 = 75.4 - 0 - 0 - 3.22(5/6)^0.84.

= 70.64 mi/h.

(b). To determine the LOS type one need to check the LOS table. The level of service of freeway = LOS D since our values lies between LOS C and LOS D, we will take LOS D which is 2110 pc/h/ln.

answer:

i am not 100% sure but i think it is a.)

dont take my word for it

explanation:

explanation:

which ordered pair is not in the solution

y > -1/2x + 5

y ≤ 3x - 2

a) (5,3) b) (4,3)

c) (3,4) d) (4,4)