Plz help plz I am already failing I dont want to fail more

Find the percent increase. Round to the nearest percent. From 85 books to 143 books The percent increase is

klasskru [66]

We are given :

Initial = 86
Final = 143

Let's find the percent increase with the help of formula:

$\boxed{ \sf \: formula : 100 \times \frac{(final - initial)}{initial} }$

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$\tt \% \: increase = 100 \times \frac{(143 - 85)}{85}$

$\tt \% \: increase = 100 \times \frac{58}{85}$

$\tt\% \: increase = 20 \times \frac{58}{17}$

$\tt\% \: increase = \frac{1160}{17}$

$\tt\% \:increase = 6823.53$

3 0

1 year ago

Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distribut

defon

Answer:

0.8665 = 86.65% probability that the sample mean would be at least $39000

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 $s = \frac{\sigma}{\sqrt{n}}$ .