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Whitepunk [10]
3 years ago
14

A university council wants to determine the need for a new staff room. The council conducted a survey of the staff by randomly s

peaking to 860 staff members out of 2,844 staff members. The council found that 70% of those surveyed were in favor of building a new staff room. Assuming a 90% confidence level, which statement holds true?
As the sample size is appropriately large, the margin of error is 0.036.


As the sample size is appropriately large, the margin of error is 0.026.


As the sample size is too small, the margin of error is 0.026.


As the sample size is too small, the margin of error cannot be trusted.

To find a margin of error for a sample proportion, these assumptions must be met, where n is the sample size and is the sample proportion.




First rewrite the proportion of positive responses as a decimal number.




Now substitute the sample proportion and the sample size, 860, into the formulas for the assumptions to determine whether they are met.




Both values are greater than 10, so the assumptions are met, and the margin of error can be calculated.


Find the margin of error using this formula, where z is the critical value.




For a 90% confidence level, . Substitute the values of z, n, and into the formula for margin of error and evaluate.
Mathematics
1 answer:
iogann1982 [59]3 years ago
7 0

Answer:

As the sample size is appropriately large, the margin of error is 0.026

Step-by-step explanation:

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3 years ago
Factor 125x^3-1000. Show your work.
Salsk061 [2.6K]

The left term is (5x)³.

The right term is 10³.

So, you can use the formula for the factorization of the difference of cubes.

... a³ - b³ = (a-b)(a² +ab +b²)

Here, you have a=5x, b=10, so the factorization is

... 125x³ -1000 = (5x-10)(25x² +50x +100)

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The decimal 0.3 represents 1/3. What type of number best describes 0.9, which is 3 times 0.3? Explain.
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8 0
3 years ago
Read 2 more answers
What is the sum of the first 51 consecutive odd positive integers?
Angelina_Jolie [31]
We call:

a_{n} as the set of <span>the first 51 consecutive odd positive integers, so:

</span>a_{n} = \{1, 3, 5, 7, 9...\}

Where:
a_{1} = 1
a_{2} = 3
a_{3} = 5
a_{4} = 7
a_{5} = 9
<span>and so on.

In mathematics, a sequence of numbers, such that the difference between two consecutive terms is constant, is called Arithmetic Progression, so:

3-1 = 2
5-3 = 2
7-5 = 2
9-7 = 2 and so on.

Then, the common difference is 2, thus:

</span>a_{n} = \{ a_{1} , a_{1} + d, a_{1} + d + d,..., a_{1} + (n-2)d+d\}
<span>
Then:

</span>a_{n} = a_{1} + (n-1)d
<span>
So, we need to find the sum of the members of the finite series, which is called arithmetic series:

There is a formula for arithmetic series, namely:

</span>S_{k} = ( \frac{a_{1} +  a_{k}}{2}  ).k
<span>
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Given that a_{1} = 1, then:

a_{n} = a_{1} + (n-1)d = 1 + (n-1)(2) = 2n-1

Thus:
a_{k} = a_{51} = 2(51)-1 = 101

Lastly:

S_{51} = ( \frac{1 + 101}{2} ).51 = 2601 

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Step-by-step explanation:

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