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Ivenika [448]
3 years ago
14

You are making a scale model of the solar system. The radius of Earth is 6378 kilometers. The radius of the Sun is 695,500 kilom

eters. Is it reasonable to choose a baseball as a model of Earth? Explain your reasoning.
It_____ reasonable to use a baseball as a model for Earth because the model for the Sun would be about the size of ______.
Mathematics
1 answer:
jeka943 years ago
5 0

Express the radius of the Sun as a proportion to radius of Earth:

695500/6378 = 109.04...

Now if a baseball would represent Earth, the model for Sun would have to be about 110 times larger than a baseball. A baseball is about 3" in diameter, so the model for Sun would have to be about 330" (or 27 feet) in diameter.

The answer should be no it is not reasonable to use baseball as a model for the Earth because the Sun would need to be about the size of a hot air balloon!

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Of the total population of the United States, 20% live in the northeast. If 200 residents of the United States are selected at r
Snezhnost [94]

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0.0465 = 4.65% probability that at least 50 live in the northeast.

Step-by-step explanation:

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

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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:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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 = 200, p = 0.2

So

\mu = E(X) = np = 200*0.2 = 40

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.2*0.8} = 5.65685

Approximate the probability that at least 50 live in the northeast.

Using continuity correction, this is P(X \geq 50 - 0.5) = P(X \geq 49.5, which is 1 subtracted by the pvalue of Z when X = 49.5. So

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

Z = \frac{49.5 - 40}{5.65685}

Z = 1.68

Z = 1.68 has a pvalue of 0.9535

1 - 0.9535 = 0.0465

0.0465 = 4.65% probability that at least 50 live in the northeast.

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3 years ago
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