The complete question is
"The cafeteria staff made sandwiches. Each sandwich had either rye or white bread, either ham or turkey, and either cheese or no cheese. The staff made an equal number of each type of sandwich.
The sandwiches were placed on a tray. Without looking, Mary will choose a sandwich.
What are the chances that Mary will get a sandwich with cheese?"
The chances of Mary getting a sandwich with cheese is 1/2.
<h3>How to calculate the number of different ways?</h3>
The number of different sandwiches is calculated by multiplying all of the possibilities for each material being used.
rye or white bread: 2 options
ham or turkey: 2 options
cheese or no cheese: 2 options
So the number of different sandwiches = 2*2*2 = 8.
4 have cheese and 4 have no cheese, as the staff made an equal number of each type of sandwich.
So, The chances of Mary getting a sandwich with cheese is 4/8 = 1/2.
Learn more about possibilities ;
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Answer:
what do you mean bro ?
Step-by-step explanation:
Im pretty sure that it would be 600%. If it was 600% that would be 600 over 100 which would be 6
Answer:
a) The interval for those who want to go out earlier is between 43.008 and 46.592
b) The interval for those who want to go out later is between 47.9232 and 51.9168
Step-by-step explanation:
Given that:
Sample size (n) =128,
Margin of error (e) = ±4% =
a) The probability of those who wanted to get out earlier (p) = 35% = 0.35
The mean of the distribution (μ) = np = 128 * 0.35 = 44.8
The margin of error = ± 4% of 448 = 0.04 × 44.8 = ± 1.792
The interval = μ ± e = 44.8 ± 1.792 = (43.008, 46.592)
b) The probability of those who wanted to start school get out later (p) = 39% = 0.39
The mean of the distribution (μ) = np = 128 * 0.39 = 49.92
The margin of error = ± 4% of 448 = 0.04 × 49.92 = ± 1.9968
The interval = μ ± e = 44.8 ± 1.792 = (47.9232, 51.9168)
The way for those who want to go out earlier to win if the vote is counted is if those who do not have any opinion vote that they want to go earlier
Answer:
C and D
Step-by-step explanation:
Equating the line A and the parabola, we get
-3x + 2 = x² - 3x + 4
0 = x² - 3x + 4 +3x - 2
0 = x² + 2
-2 = x²
which has no real solutions. Then, the line A and the parabola don't intersect each other.
Equating the line B and the parabola, we get
-3x + 3 = x² - 3x + 4
0 = x² - 3x + 4 + 3x - 3
0 = x² + 1
-1 = x²
which has no real solutions. Then, the line B and the parabola don't intersect each other.
Equating the line C and the parabola, we get
-3x + 5 = x² - 3x + 4
0 = x² - 3x + 4 + 3x - 5
0 = x² - 1
1 = x²
√1 = x
which has 2 solutions, x = 1 and x = -1. Then, the line C and the parabola intersect each other.
Equating the line D and the parabola, we get
-3x + 6 = x² - 3x + 4
0 = x² - 3x + 4 + 3x - 6
0 = x² - 2
2 = x²
√2 = x
which has 2 solutions, x ≈ 1.41 and x ≈ -1.41. Then, the line D and the parabola intersect each other.