Choose the graph which represents the solution to the inequality: 1 2 x + 8 ≤ 3 A) B) C) D)

The airport security randomly selected 36 suitcases from the security line. Of these bags, they screened 8 suitcases. Based on t

castortr0y [4]

The ratio of suitcases to screened suitcases is 36:8. The second ratio is 180:?.

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180 is 5 times 36 which can be proven by dividing 180 by 36.

180 ÷ 36 = 5

Multiply 5 by 8 to find the amount of suitcases that would be screened if there were 180 total suitcases.

5 × 8 = 40

The first ratio is 36:8, and with a scale factor of 5 we get the second equivalent ratio 180:40.

<h2>Answer:</h2>

<u>Forty suitcases</u><u> will be screened.</u>

I hope this helps :)

4 0

3 years ago

Based on historical data, your manager believes that 37% of the company's orders come from first-time customers. A random sample

fomenos

Answer:

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38

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 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 $s = \sqrt{\frac{p(1-p)}{n}}$