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SSSSS [86.1K]
3 years ago
11

The latest poll of 400 randomly selected residents showed that 288 of them are hoping that the old town cinema stays open on Mai

n Street.
With a desired confidence interval of 95%, which has a z*-score of 1.96, what is the approximate margin of error for this polling question?

A) 4%
B) 5%
C) 7%
D) 8%

Mathematics
2 answers:
Varvara68 [4.7K]3 years ago
8 0
The approximate margin of error is 4%.

Using the formula given and our information, we have:

E=1.96\times \sqrt{\frac{\frac{288}{400}(1-\frac{288}{400})}{400}}
\\
\\=1.96\times \sqrt{\frac{0.72(1-0.72)}{400}}=1.96\times \sqrt{\frac{0.72(0.28)}{400}}
\\
\\=0.044\approx 0.04=4%
Sauron [17]3 years ago
8 0

Answer:

A

Step-by-step explanation:

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Which pair of numbers below have 4 and 6 as common factors?
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explanations
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7 0
3 years ago
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Step-by-step explanation:

4 0
3 years ago
Given the declaration, where is the value 97 stored in the numbers array? int numbers[]={83, 62, 77, 97, 88};
Zepler [3.9K]
Let me help you!

Ok, so before I give you the answer, which is very simple, I will first show you how to answer questions like this in the future.

In the variable declaration, the array "numbers[]" has been declared as an integer. The values of the array "numbers[]" are: <span>83, 62, 77, 97, 88; respectively, the values are equivalent to: 0, 1, 2, 3, 4 which makes the range of array "numbers[]": 5.

To make it clearer:
</span>numbers[0] = 83
numbers[1] = 62
numbers[2] = 77
numbers[3] = 97 <---- This is what we are looking for!
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*** By the way, when counting the range, we don't start at 1, we start at 0. That's why even if the range is 5, the last value is 4.

Therefore, the answer is: 3 or numbers[3].
3 0
3 years ago
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