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3 years ago

Is 15 over 50 equivalent to 0.3

Mathematics

2 answers:

alexdok [17]3 years ago

6 0

$Yes,\ because\\\\\frac{15}{50}=\frac{15:5}{50:5}=\frac{3}{10}=\boxed{0.3}$

dsp733 years ago

3 0

$\dfrac{15}{50}=\dfrac{3}{10}=0.3$

YES

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Write these as normal numbers.

netineya [11]

Answer:

Step-by-step explanation:

5.1 × 108 = 550.8

9.6 × 10^0

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3 years ago

Suppose 44% of the children in a school are girls. If a sample of 727 children is selected, what is the probability that the sam

Harman [31]

Answer:

0.9484 = 94.84% probability that the sample proportion of girls will be greater than 41%

Step-by-step explanation:

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

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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 $s = \sqrt{\frac{p(1-p)}{n}}$

Suppose 44% of the children in a school are girls.

This means that $p = 0.44$

Sample of 727 children

This means that $n = 727$

Mean and standard deviation:

$\mu = p = 0.44$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.44*0.56}{727}} = 0.0184$

What is the probability that the sample proportion of girls will be greater than 41%?

This is 1 subtracted by the p-value of Z when X = 0.41. So

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

By the Central Limit Theorem

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

$Z = \frac{0.41 - 0.44}{0.0184}$

$Z = -1.63$

$Z = -1.63$ has a p-value of 0.0516

1 - 0.0516 = 0.9884

0.9484 = 94.84% probability that the sample proportion of girls will be greater than 41%

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3 years ago

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irina1246 [14]

Answer:

8.40

Step-by-step explanation:

Add them all up, which equals to 41.98 you then divide by 5 since you want 20%. you would get 8.396 which rounded to the next penny is 8.40

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