A four-lane freeway (two lanes in each direction) is located on rolling terrain and has 12-ft lanes, no lateral obstructions wit
hin 6 ft of the pavement edges, and there are 2 ramps within 3 miles upstream of the segment midpoint and 3 ramps within 3 miles downstream of the segment midpoint. A weekday directional peak-hour volume of 1800 vehicles (familiar users) is observed, with 700 arriving in the most congested 15-min period. If a level of service no worse than C is desired, determine the maximum number of heavy vehicles that can be present in the peak-hour traffic stream.
Consider the freeway and traffic conditions described in Problem 6.3. If 180 of the 1800 vehicles observed in the peak hour were heavy vehicles (assume a 70%/30% SUT/TT split), what would the level of service of this freeway be on a 5-mi, 6% downgrade?
Consider the freeway and traffic conditions described in Problem 6.3. If 180 of the 1800 vehicles observed in the peak hour were heavy vehicles (assume a 70%/30% SUT/TT split), what would the level of service of this freeway be on a 5-mi, 6% downgrade?
1 answer:
Answer:
Check the explanation
Explanation:
FFS = Free-Flow Speed (km/h) BFFS = the Base Free-Flow Speed (km/h) fLS = the Adjustment for lane and shoulder widths less than 3.65 m and 1.80 m, respectively. fAPD = Adjustment for access points density (km/h) fm = Adjustment for proportion (km/h) 3.0 Results
Kindly check the below image for the step by step explanation to the above question.
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Answer:
a) ≈ 30 mins
b) 8 vpm
Explanation:
<u>a) Determine how long after the first vehicle arrival will the queue dissipate</u>
The time after the arrival of the first vehicle for the queue to dissipate
= 29.9 mins ≈ 30 mins
<u>b) Determine the average service rate at the parking lot gate </u>
U = A / t
where : A = 240 vehicles , t = 30
U = 240 / 30 = 8 Vpm
attached below is a detailed solution of the given problems above
Answer:
a. 0.11
b. 110 students
c. 50 students
d. 0.46
e. 460 students
f. 540 students
g. 0.96
Explanation:
(See attachment below)
a. Probability that a student got an A
To get an A, the student needs to get 9 or 10 questions right.
That means we want P(X≥9);
P(X>9) = P(9)+P(10)
= 0.06+0.05=0.11
b. How many students got an A on the quiz
Total students = 1000
Probability of getting A = 0.11 ---- Calculated from (a)
Number of students = 0.11 * 1000
Number of students = 110 students
So,the number of students that got A is 110
c. How many students did not miss a single question
For a student not to miss a single question, then that student scores a total of 10 out of possible 10
P(10) = 0.05
Total Students = 1000
Number of Students = 0.05 * 1000
Number of Students = 50 students
We see that 5
d. Probability that a student pass the quiz
To pass, a student needed to get at least 6 questions right.
So we want P(X>=6);
P(X>=) =P(6)+P(7)+P(8)+P(9)+P(10)
=0.08+0.12+0.15+0.06+0.05=0.46
So, the probability of a student passing the quiz is 0.46
e. Number of students that pass the quiz
Total students = 1000
Probability of passing the quiz = 0.46 ----- Calculated from (d)
Number of students = 0.46 * 1000
Number of students = 460 students
So,the number of students that passed the test is 460
f. Number of students that failed the quiz
Total students = 1000
Total students that passed = 460 ----- Calculated from (e)
Number of students that failed = 1000 - 460
Number of students that failed = 540
So,the number of students that failed is 540
g. Probability that a student got at least one question right
This means that we want to solve for P(X>=1)
Using the complement rule,
P(X>=1) = 1 - P(X<1)
P(X>=1) = 1 - P(X=0)
P(X>=1) = 1 - 0.04
P(X>=1) = 0.96
