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Alexandra [31]
3 years ago
10

HELP PLS 100PTS!!! Evaluate s(-19) s(n)=7-4n

Mathematics
2 answers:
Len [333]3 years ago
5 0

Answer: Therefore, the recursive formular is A(n) = A(n - 1) + 9

Step-by-step explanation: (n) = 2 + 9(n - 1) = 2 + 9n - 9  

(n + 1) = 2 + 9(n + 1 - 1) = 2 + 9n = (2 + 9n - 9) + 9 = A(n) + 9

Alika [10]3 years ago
4 0
S(-19)=7-4(-19)
In this step, I inserted the value into the equation.
Next, evaluate.
s(-19)=83
(Multiply the -4 times -19, which gives 76, then add the 7, which gives 83)
Hope this helps!!
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Answer:

A sample of 499 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

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In this question, we have that:

\pi = 0.21

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So \alpha = 0.1, z is the value of Z that has a pvalue of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 90% confidence level with an error of at most 0.03

We need a sample of n, which is found when M = 0.03. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.645\sqrt{\frac{0.21*0.79}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.21*0.79}

\sqrt{n} = \frac{1.645\sqrt{0.21*0.79}}{0.03}

(\sqrt{n})^2 = (\frac{1.645\sqrt{0.21*0.79}}{0.03})^2

n = 498.81

Rounding up

A sample of 499 is needed.

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