Tan socks = 4
Grey socks = 7
Total socks = 4 + 7 = 11 socks
Probability of drawing a tan sock followed by a grey sock:
P(T)*P(G) = (4/11)*(7/10) = 28/110
Probability of selecting a grey sock followed by a tan sock
P(G)*P(T) = (7/11)*(4/10) = 28/110
Probability of obtaining two differently colored socks
P(T&G)+P(G&T) = (28/110)+(28/110) = 56/110
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.
Start by letting x represent the number of Sam's pencils. Then Sari has 3x (since she has three times as many)
Together they total 28 pencils:
x + 3x = 28
4x = 28 /:4 (divide both sides by 4)
x = 7
So Sam has 7 pencils.
Sari, having three times as many, has 7 * 3 = 21.[[TAG_8]]
