Regardless of the value of M, it will correspond to 18m squared.
Answer:
(C) 10% to 70%(
Step-by-step explanation:
Given that at least 40% of the students are learning German, the upper limit of those who might be enrolled in English but not in German is 60%. However, since a minimum of 70% study English, it leads to the conclusion that at least 10% of students must be taking both German and English.
If we consider that at least 30% of students are learning Italian, and assuming that no student is studying all three languages simultaneously, then there is a maximum of 70% of students who could potentially be registered in both English and German.
This means the possible percentage for students enrolled in both English and German ranges from 10% to 70%
A. Mean and standard deviation.
The sampling distribution’s mean closely matches the population mean. Since the population mean is 174.5, the sampling distribution’s mean equals this value.
The standard deviation of the sampling distribution is:
σₓ̄ = σ / √n
Plugging in values:
σₓ̄ = 6.9 / √25 = 1.38
b. Calculate z-scores for both values:
z = (value - mean) / standard deviation
For 172.5:
z = (172.5 - 174.5) / 1.38 = -1.49
Corresponding probability ≈ 0.068
For 175.8:
z = (175.8 - 174.5) / 1.38 = 0.94
Corresponding probability ≈ 0.83
The difference between these probabilities is 0.762.
Approximately 0.762 × 200 = 152 sample means lie between 172.5 and 175.8.
c. Z-score for 172 cm:
z = (172 - 174.5) / 1.38 = -1.81
Probability ≈ 0.03
Hence, samples with means below 172 cm equal 0.03 × 200 = 6.
Answer:
It could either be 455 or 680, based on assumptions.
Step-by-step explanation:
Assuming the three choices are distinct, we can calculate...
15C3 = 15·14·13/(3·2·1) = 35·13 = 455
ways to create the pizza.
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In the case where two or more of the toppings may be identical, this would lead to...
2(15C2) + 15C1 = 2·105 + 15 = 225
additional combinations, resulting in a grand total of...
455 + 225 = 680
unique pizza varieties.
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There is a multiplication factor of 2 for the two-topping selections, since it allows for variations like double anchovies and tomatoes or double tomatoes and anchovies when the topping choices are anchovies and tomatoes.
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nCk = n!/(k!(n-k)!)
