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3 years ago

Simplify the following using PEMDAS: [3−(3⋅3)+33÷3] × 4⋅5/10−2⋅3

Mathematics

1 answer:

kompoz [17]3 years ago

8 0

Answer: 34

Step-by-step explanation:

Given expression

[3 - (3 · 3) + 33 ÷ 3] × 4 · 5 / 10 - 2 · 3

Simplify by parentheses in the bracket

=[3 - 9 + 33 ÷ 3] × 4 · 5 / 10 - 2 · 3

Simplify by multiplication

=[3 - 9 + 33 ÷ 3] × 20 / 10 - 6

Simplify by division

=[3 - 9 + 11] × 10 - 6

Simplify by bracket (addition/subtraction)

=[-6 + 11] × 10 - 6

=[4] × 10 - 6

Simplify by multiplication

=40 - 6

Simplify by subtraction

= $\boxed{34}$

Hope this helps!! :)

Please let me know if you have any questions

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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.

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

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.

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

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Total load of more than 20,000 pounds, which is 1 subtracted by the pvalue of Z when X = 20000. So

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

$Z = \frac{20000 - 20000}{300}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

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3 years ago

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