Change c of a telephone company is partly constant and partly varies as the number of units call y the cost of 50 unti is 2500 and the cost of120 units 3000find the formular of connecting c and y when =4000
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From what I can tell, the cost for 50 units is 2500, and cost of 120 units is 3000.
let c = my + b
for (50,2500), 2500 = 50m + b
for(120,3000), 3000 = 120m + b
subtract these .....
70m = 500
m = 50/7
and in 50m + b = 2500
50(50/7) + b = 2500
b = 15000/7
c = (50/7)y + 15000/7
once you have decided what "when =4000" means, plug into the appropriate variable, (my guess: you meant ... when c = 4000)
To find the formula connecting c and y, we need to determine the constant part and the variable part of c.
Let's denote the constant part as A and the variable part as By.
Given that the cost of 50 units is 2500, we can write it as:
A + 50B = 2500 --------(1)
Similarly, the cost of 120 units is 3000, so we have:
A + 120B = 3000 --------(2)
To find A and B, we can solve these equations simultaneously. Here's how:
1. Multiply equation (1) by 12 and equation (2) by 5 to eliminate A:
12A + 600B = 30000 --------(3)
5A + 600B = 15000 --------(4)
2. Subtract equation (4) from equation (3) to cancel out B:
(12A - 5A) + (600B - 600B) = 30000 - 15000
7A = 15000
3. Solve for A:
A = 15000 / 7
A ≈ 2142.86
4. Plug the value of A back into equation (1) to solve for B:
2142.86 + 50B = 2500
Subtract 2142.86 from both sides:
50B = 2500 - 2142.86
50B = 357.14
Divide both sides by 50:
B = 357.14 / 50
B ≈ 7.14
Now we have found the constant part (A ≈ 2142.86) and the variable part (B ≈ 7.14) of the cost.
The formula connecting c and y, where c is the cost and y is the number of units called, is:
c = A + By
Substituting the values we found:
c ≈ 2142.86 + 7.14y
So, when y is 4000, the cost c would be approximately:
c ≈ 2142.86 + 7.14(4000)
c ≈ 2142.86 + 28560
c ≈ 30702.86