Use the properties of exponents to enter an equivalent expression with a single exponent.

Mathematics

1 answer:

Gelneren [198K]4 years ago

5 0

$({10}^{2} ) {}^{4} = {10}^{2 \times 4} = {10}^{8}$

I hope this can help you ^_^

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The value of blank divided 40.01 is much less than one?

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Answer:400

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9x^2+6x+5=0, solve using the quadratic formula

AleksAgata [21]

Step by step explanation:

9 $x^{2}$ +6x+5=0

Uses the "a", "b", and "c" from "a $x^{2}$ + bx + c", where "a", "b", and "c" are numbers

So in this case:

a would be 9

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Then plug in from there..

Another way you can do it is using the square method.

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Public health officials claim that people living in low income neighborhoods have different Physical Activity Levels (PAL) than

Naddika [18.5K]

Answer:

The z-score for this data is Z = -0.26.

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