A linear equation representing the relationship between two entities can be expressed as:
y = ax + b
Here, x denotes the number of passengers while y represents the weight (in tons). Based on the initial two scenarios,
10 = 60(a) + b
13 = 84(a) + b
The calculated values for a and b are 0.125 and 2.5 respectively.
For a bus designed for 50 passengers,
y = (50)(0.125) + 2.5
This results in y being equal to 8.75.
It amounts to 96. I hope this was helpful.
-1, 1, 5
-1, -1, -5
1, 1, -5
We will create the equations for this scenario:
(1) 1100*y + z = 113
(2) 1500*y + z = 153
Find z: Monthly administration fee is represented by z, which is the question of this problem.
The amounts of kilowatt hours consumed are 1100 and 1500 respectively.
The cost for each kilowatt hour is denoted by y, although its value is not required for this math problem, we can compute it regardless.
This results in a system of two equations with two unknowns, which can be solved using the substitution method:
(1) 1100*y + z = 113
(2) 1500*y + z = 153
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(1) z = 113 - 1100*y [substituting z (right side) into equation (2) instead of z]:
(2) 1500*y + (113 - 1100*y) = 153
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(1) z = 113 - 1100*y
(2) 1500*y + 113 - 1100*y = 153
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(1) z = 113 - 1100*y
(2) 400*y + 113 = 153
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(1) z = 113 - 1100*y
(2) 400*y = 153 - 113
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(1) z = 113 - 1100*y
(2) 400*y = 40
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(1) z = 113 - 1100*y
(2) y = 40/400
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(1) z = 113 - 1100*y
(2) y = 1/10
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by placing the calculated value of y back into equation (1), we can find z:
(1) z = 113 - 1100*(1/10)
(1) z = 113 - 110
(1) z = 3 dollars serves as the monthly fee.
