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Mariulka [41]
3 years ago
7

If there is 36 teachers how much students will there be

Mathematics
2 answers:
Anastasy [175]3 years ago
7 0

Answer:

About 720 students

Step-by-step explanation:

There will always be about 20 students in a class room so what I did was a multiplied 20 by 36 and got the answer

dsp733 years ago
6 0

Answer: uh  1152?

Step-by-step explanation: 32 people in each classroom and multiply by 36?

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(10 pts) A four-lane freeway (two lanes in each direction) is located on rolling terrain and has 12-ft lanes, no lateral obstruc
dimulka [17.4K]

Answer:

309

Step-by-step explanation:

To determine the estimated free-flow speed

FFS = 75.4 -f_{LW}-f_{LC} - 3.22 TRD^{0.84}

From the table "Adjustment for lane width," which corresponds to a lane width of 12 feet. As a result, f_{LW} equals 0 mi/h.

Take the meaning from the table "Adjustment for right-shoulder lateral clearance," which corresponds to 6 feet of right shoulder lateral clearance and two lanes in one direction.

The f_{LC} = 0 mi/h

FFS = 75.4 -0-0-3.22 ( \dfrac{5}{6})^{0.84}

= 72.64 \ mi/h

\simeq  73 \ mi/h

Determine the peak-hour factor

PHF = \dfrac{V}{V_{15}\times 4} \\ \\ =\dfrac{1800}{700\times 4}\\ \\ = 0.6429

Now, Find the heavy-vehicle adjustment factor.

v_P = \dfrac{V}{PHF\times N \times f_{HV}\times f_P} --- (1)

Take the value for the 15-minute passenger car equivalent flow rate from the table "LOS requirements for freeway segments" for the FFS and LOS C conditions. The free-flow speed is estimated to be 73 miles per hour

v_p = 1735+ \dfrac{73-70}{75-70}(1775-1735)

v_p = 1759 \ pc/h/In

Think about for familiar users the value of f_p = 1.00

Replace all of the values obtained in (1)

v_p = \dfrac{V}{PHF \times N\times f_{HV}\times f_p}

1759 = \dfrac{1800}{0.6429\times 2 \times f_{HV}\times 1}

f_{HV} = \dfrac{1800}{0.6429\times 2 }\times \dfrac{1}{1759}

= 0.795

Calculate the percentage of trucks and buses in the flow of traffic stream.

f_{HV} = \dfrac{1}{1+P_{\tau}(E_{\tau}-1) + P_R(E_R-1)}

Take the values from the table "Passenger car equivalent for extended highway segments" referring to trucks and buses and rolling terrains for passenger car equivalent for trucks and buses and recreational vehicles. As a result, E_T has a value of 2.5 and E_R has a value of 2.0.

Since P_R is zero and there is no recreational vehicle.

Then;

0.759 = \dfrac{1}{1+ P_T(2.5 -1) +0}

P_T = 0.1719

Finally, to estimate the maximum number of large trucks and buses; we have:

= V\times P_T = 1800 \times 0.1719

Maximum number of large trucks and buses = 309

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