The cost of a carton of milk is a) $0.35.
To find the cost of a carton of milk, we can set up a system of equations based on the given information.
Let's assume the cost of a hamburger is "h" and the cost of a carton of milk is "m".
From the information given, we can create the following equations:
Equation 1: 2h + 1m = 2.85 (from Monday's purchase)
Equation 2: 3h + 2m = 4.45 (from Tuesday's purchase)
We can solve this system of equations to find the value of "m", the cost of a carton of milk.
Multiplying Equation 1 by 2 and Equation 2 by 1, we can eliminate "h" and solve for "m":
4h + 2m = 5.70
3h + 2m = 4.45
Subtracting Equation 2 from Equation 1, we get:
(4h + 2m) - (3h + 2m) = 5.70 - 4.45
h = 1.25
Now, we can substitute the value of "h" back into Equation 1 or Equation 2 to find the value of "m":
2(1.25) + 1m = 2.85
2.50 + m = 2.85
m = 2.85 - 2.50
m = 0.35
Therefore, the cost of a carton of milk is $0.35.
The correct answer is option a) $0.35.
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A marker is randomly selected from a drawer that contains 20 green, 44 orange, and 30 blue markers. Which statement is true? P(blue)≈0. 41 P(green)≈0. 21 P(orange)≈0. 53.
none of the provided approximations for the probabilities are accurate.To determine which statement is true, we need to calculate the probabilities of selecting each color marker.
Total number of markers = 20 green + 44 orange + 30 blue = 94 markers.
P(blue) = Number of blue markers / Total number of markers = 30 / 94 ≈ 0.319.
P(green) = Number of green markers / Total number of markers = 20 / 94 ≈ 0.213.
P(orange) = Number of orange markers / Total number of markers = 44 / 94 ≈ 0.468.
Based on the calculations, none of the given statements are true. The actual probabilities are approximately:
P(blue) ≈ 0.319,
P(green) ≈ 0.213,
P(orange) ≈ 0.468.
Therefore, none of the provided approximations for the probabilities are accurate.
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