2 pounds depending on how much they eat.
Ex: If somebody is 500 pounds they might want to pack 10 gallons. Lol
If somebody doesn't eat alot then 1 pound wil be good.

4 0

3 years ago

4. Parking fees at IIUM are RM 5.00 for IIUM students and RM 7.50 for non-IIUM students. At the

professor190 [17]

Answer:

(a) 780 students and 960 non-students

(b) No. The maximum revenue is RM9000 from 1200 non-students.

(c). Revenue is maximum of RM9000 at 1200 non-students, decreasing by RM2.50 per student to a minimum of RM6000 at 1200 students

Step-by-step explanation:

Let x = IIUM students and

and y = non-IIUM students

You have two conditions

(a) x + y = total vehicles parked

(b) 5.00x + 7.50y = total gross receipts

(a) Wednesday

From your table,

x + y = 1740

5.00x + 7.70y = RM11 100

Solve the simultaneous equations

$\begin{array}{rrcrl}(1) & x + y & = &1740&\\(2) & 5.00x + 7.50y & = & 11 100\\(3)& 5.00x + 5.00y & = & 8700 & \text{Multiplied (1) by 5}\\&2.50 y & = &2400 &\text{Subtracted (3) from (2)}\\(4)&y& = &\mathbf{960} &\text{Divided each side by 2.50}\\& x +960& = &1740& \text{Substituted (4) into (1)}\\& x& = &\mathbf{780}& \\\end{array}\\\text{There are $\large \boxed{\textbf{780 students and 960 non-students}}$}$

(b) Can 1200 vehicles bring in RM10000?

No. Even if all the cars were from non-students, the most you could get is

1200 × 7.50 = RM9000

(c) Possible combinations for 1200 vehicles

Revenue = 5.00x + 7.50y = 5.00x + 7.50(1200 -x) = 5.00x + 9000 - 7.50x =

Revenue = 9000 - 2.50x

The maximum revenue of RM9000 occurs when there are no student cars and 1200 non-student cars.

For each student car that enters and displaces a non-student, the revenue drops by RM2.50.

Finally. when there are 1200 student cars and no non-students, the revenue has dropped to a minimum of RM6000.

3 0

3 years ago