The cost c of opening a day school for one day is partly constant and partly varries as the number of students n it costs shs 40000 to run the school wen they are 500 students and shs 64000 when they are 900 students form am equation for de cost c and the number of student m.
treat it as if you had two ordered pairs (500,40000) and (900,64000)
slope = (64000-40000)/(900-500) = 24000/400 = 60
(it costs shs 60 per student)
cost = 60m + b , <------ relate this to y = mx + b
plug in the point (500,40000)
40000 = 60(500) + b
b = 10000
c = 60m + 10000
To form an equation for the cost c and the number of students n, we need to determine the constant and variable components.
Let's denote:
- C = constant part of the cost
- k = variable cost per student
From the given information, we have two data points:
(500, 40000) and (900, 64000)
Using the equation form y = mx + c, we can substitute the values to form two equations:
For the first data point:
40000 = C + (500 * k) ......(1)
For the second data point:
64000 = C + (900 * k) ......(2)
To solve for C and k, we can subtract equation (1) from equation (2):
(64000 - 40000) = (C + (900 * k)) - (C + (500 * k))
24000 = 900k - 500k
24000 = 400k
Now, we can solve for k:
k = 24000/400
k = 60
Substituting the value of k back into equation (1), we can solve for C:
40000 = C + (500 * 60)
40000 = C + 30000
C = 40000 - 30000
C = 10000
Therefore, the equation for the cost c and the number of students n is:
c = 10000 + (60 * n)
To form an equation for the cost (c) and the number of students (n), we need to find the relationship between the cost and the number of students.
From the given information, we know that the cost of running the school is partly constant and partly varies with the number of students. Let's consider the constant part of the cost as A and the part that varies as B.
We are given two data points:
1. When there are 500 students, the cost is shs 40,000.
So, we can form the equation: c = A + 500B
2. When there are 900 students, the cost is shs 64,000.
Therefore, we can form another equation: c = A + 900B
Now, we have a system of two equations:
1. A + 500B = 40,000
2. A + 900B = 64,000
To solve this system of equations, we can use the method of substitution or elimination.
Method 1: Substitution
1. Solve equation 1 for A: A = 40,000 - 500B
2. Substitute A in equation 2 with the value obtained from step 1: (40,000 - 500B) + 900B = 64,000
3. Simplify the equation: 40,000 + 400B = 64,000
4. Solve for B: 400B = 24,000
B = 60
Now, substitute B = 60 back into equation 1 to find A:
A + 500(60) = 40,000
A + 30,000 = 40,000
A = 10,000
So, the equation for the cost (c) and the number of students (n) is:
c = 10,000 + 500n
Therefore, the equation is c = 10,000 + 500n.