Please need help I suck at math ‼️‼️

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

Help will name Brainliest

11111nata11111 [884]

Answer:

x = $\frac{\sqrt{22} }{2}$

Step-by-step explanation:

Using the Pythagorean Theorem we can solve this problem.

The legs of the triangle will be the same (the dashes in the middle of the line indicates they are the same).

The Pythagorean Theorem will look like this:

$a^{2}+a^{2} = (\sqrt {11})^{2}$

Now, let's find a:

$2a^{2} = 11$ The square of a radical will be whatever is under the arch

$a^{2}=\frac{11}{2}$ Divide 11 by 2

$\sqrt{a^{2} } = \sqrt{\frac{11}{2} }$ Square root both sides

a = $\frac{\sqrt{22} }{2}$

Therefore, the length of side x is equal to $\frac{\sqrt{22} }{2}$ .

Hope this helps!!

8 0

2 years ago

in the last election candidate b received twice as many votes as candidate candidate c received 15,000 fewer votes than candidat

Alenkinab [10]

This is the best i can do

5 0

4 years ago

Convert to length 2.1 cm to mm

Anastasy [175]

Here we need to convert 2. 1 cm to mm .

And for that we will use the following conversion .

1cm = 10 mm

So for 2.1 cm, we will multiply 2.1 and 10 to convert it to mm. Therefore 2.1 cm = 2.1* 10mm = 21mm

And that's the required answer .

7 0

4 years ago

Matt and Ming are selling fruit for a school fundraiser. Customers can buy boxes of oranges and boxes of apples. Matt sold 3 box

ANEK [815]

Answer:

The Answer is that one box of Oranges costs $7.13 and one box of Apples costs $12.99.

Step-by-step explanation:

Let o = oranges and a = apples

Equation 1:

3o + 14a = 203

Solve for o:

3o = 203 - 14a

o = 203/3 - 14/3a

o = 67.66 - 4.66a

Equation 2:

11o + 11a = 220

Substitute:

11(67.66 - 4.66a) + 11a = 220

744.26 - 51.26a + 11a = 220

-40.26a = 220 - 744.26

-40.36a = -524.26

a = -524.26/-40.36 = $12.99 for a box of Apples

Solve for oranges, o:

o = 67.66 - 4.66a

o = 67.66 - 4.66(12.99)

o = 67.66 - 60.53 = $7.13 for a box of Oranges

Hope this helps!! Have an Awesome day!!! :-)

7 0

4 years ago