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sergejj [24]

3 years ago

class="latex-formula"> ill give brainliest help quickkkkkkkkkkkkkkkk

Mathematics

1 answer:

lawyer [7]3 years ago

4 0

Answer:B

Step-by-step explanation:

You might be interested in

Part A: Find a rational number that is between 5.2 and 5.5. Explain why it is rational. (2 points) Part B: Find an irrational nu

max2010maxim [7]

A rational number can always be represented as a fraction.
5.2 = 52/10
5.5 = 55/10

A number between them would be 54/10 since 54 is between 52 and 55.

An irrational number cannot be represented as a fraction. A good example of an irrational number is the square root of a number.
sqrt(5.2^2) = 5.2 ----> 5.2^2 = 27.04
sqrt(5.5^2) = 5.5 ----> 5.5^2 = 30.25

The square root of a number between 27.04 and 30.25 will be an irrational number between 5.2 and 5.5
sqrt(28) = 5.2915...

3 0

3 years ago

A ski lift is designed with a total load limit of 20,000 pounds. It claims a capacity of 100 persons. An expert in ski lifts thi

Yanka [14]

Answer:

0.5 = 50% probability that a random sample of 100 independent persons will cause an overload

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 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 the sum of n values of a distribution, the mean is $\mu \times n$ and the standard deviation is $\sigma\sqrt{n}$

An expert in ski lifts thinks that the weights of individuals using the lift have expected weight of 200 pounds and standard deviation of 30 pounds. 100 individuals.

This means that $\mu = 200*100 = 20000, \sigma = 30\sqrt{100} = 300$