Can someone help me with this? pleasealguien me puede ayudar con esto? porfavorr

Mathematics

1 answer:

soldi70 [24.7K]2 years ago

8 0

See attachment for the number line that compares the numbers 5.3, 5 1/5 and 5

<h3>How to plot the numbers on the number line?</h3>

The numbers are given as:

5.3, 5 1/5 and 5

We start by representing the numbers as decimals (i.e. real numbers).

So, we have:

5.3, 5.2 and 5

Reorder the numbers in ascending number (i.e. from least to greatest)

So, we have

5, 5.2 and 5.3

Next, we plot the numbers on a number line using a scale of 0.1

This can be visualized as

5 5.2 5.3

See attachment for the number line that compares the numbers 5.3, 5 1/5 and 5

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A door delivery florist wishes to estimate the proportion of people in his city that will purchase his flowers. Suppose the true

kvv77 [185]

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

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .