Is x^2y^5 simplified?

Look at the following number sequence. 3, 7, 11, 15, . . . Based on the pattern, what are the next two terms?

viva [34]

So, if we look at the number pattern, we can see the numbers go up by 4.

The rule is Add 4.

So, to find the next two terms, we would add 4 to the last number shown, which is 15.

15 + 4 = 19

19 + 4 = 23

So, the next two terms are →19← and →23←.

Glad I could help, and good luck!

AnonymousGiantsFan

5 0

3 years ago

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it is known that the population proton of utha residnet that are members of the church of jesus christ 0l6 suppose a random samp

Lady_Fox [76]

Answer:

0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

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