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tankabanditka [31]
3 years ago
6

Need help!

Mathematics
1 answer:
chubhunter [2.5K]3 years ago
8 0

The correct equation is P(t) = Po*e^(kt). Please enclose "kt" inside parentheses and use the " ^ " symbol to indicate exponentiation.


"k" is the growth constant, which here is 0.017. "t" is the number of yeasrs. P(7) is the population after 7 years. Po is the initial population, which in this case is 91 million.


The initial value, when t = 0, is 91 million.


After 7 years, the population, P(7), was P(7) = (91 million)*e^(0.017*7). This evaluates to P(7) = (91 million)*(1.070) = 96 million

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Answer:

99.74% probability that the sample proportion will be less than 0.1

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

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Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

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The standard deviation of the binomial distribution is:

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Normal probability distribution

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Z = \frac{X - \mu}{\sigma}

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When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 276, p = 0.06

So

\mu = E(X) = np = 276*0.06 = 16.56

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{276*0.06*0.94} = 3.9454

What is the probability that the sample proportion will be less than 0.1

This is the pvalue of Z when X = 0.1*276 = 27.6. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{27.6 - 16.56}{3.9454}

Z = 2.8

Z = 2.8 has a pvalue of 0.9974

99.74% probability that the sample proportion will be less than 0.1

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Answer:

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