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Luden [163]
4 years ago
5

H(x)=3x-1; Find h(-5)

Mathematics
1 answer:
Taya2010 [7]4 years ago
5 0

Answer:

I'm not sure but I got -16

Step-by-step explanation:

I just plugged in the -5 in place for the x

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4 years ago
A state insurance commission estimates that 13% of all motorists in its state areuninsured. Suppose this proportion is valid. Fi
umka21 [38]

Answer:

79.95% probability that in a randomsample of 50 motorists, at least 5 will be uninsured.

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

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 = 50, p = 0.13

So

\mu = E(X) = np = 50*0.13 = 6.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{50*0.13*0.87} = 2.38

Find the probability that in a randomsample of 50 motorists, at least 5 will be uninsured.

This is 1 subtracted by the pvalue of Z when X = 4. So

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

Z = \frac{4 - 6.5}{2.38}

Z = -0.84

Z = -0.84 has a pvalue of 0.2005

1 - 0.2005 = 0.7995

79.95% probability that in a randomsample of 50 motorists, at least 5 will be uninsured.

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4 years ago
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Answer:

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Step-by-step explanation:

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