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oksano4ka [1.4K]
3 years ago
9

If G is the midpoint of FH find FGfg=11x-7 gh= 3x+9

Mathematics
1 answer:
gavmur [86]3 years ago
3 0

Solution:

we are given that

If G is the midpoint of FH,  it mean that

FG=GH

we are also given that

FG=11x-7 ,GH= 3x+9

So we can write

11x-7=3x+9\\
\\
11x-3x=7+9\\
\\
8x=16\\
\\
x=2

So FG=11x-7=11*\frac{1}{2}-7=11*2-7=15


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A courier service company wishes to estimate the proportion of people in various states that will use its services. Suppose the
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Answer:

95.44% probability that the sample proportion will differ from the population proportion by less than 0.03.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this question:

p = 0.05, n = 212, \mu = 0.05, s = \sqrt{\frac{0.05*0.95}{212}} = 0.015

What is the probability that the sample proportion will differ from the population proportion by less than 0.03?

This is the pvalue of Z when X = 0.03 + 0.05 = 0.08 subtracted by the pvalue of Z when X = 0.05 - 0.03 = 0.02. So

X = 0.08

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

By the Central Limit Theorem

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

Z = \frac{0.08 - 0.05}{0.015}

Z = 2

Z = 2 has a pvalue of 0.9772

X = 0.02

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

Z = \frac{0.02 - 0.05}{0.015}

Z = -2

Z = -2 has a pvalue of 0.0228

0.9772 - 0.0228 = 0.9544

95.44% probability that the sample proportion will differ from the population proportion by less than 0.03.

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