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blsea [12.9K]
4 years ago
9

Help please, someone.

Mathematics
1 answer:
Lapatulllka [165]4 years ago
8 0

Answer:

117,649a^8c^2

Step-by-step explanation:

(7^3 a^4 c) ^2

= 7^(3*2) a^(4*2) c^2

= 7^6a^8c^2

= 117,649a^8c^2

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If six times a certain number is added to eight, the result is 32.which of the following equations could be used to solve the pr
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3 years ago
Tammy completed the following steps to simplify the expression
Brrunno [24]

Answer:

the third term wasn't distributed properly the -2 has to multiply by +2 = -4

Step-by-step explanation:

-5x^2y - 3xy^2 - 2(4y+2) - 2.5x^2y

-5x^2y - 3xy^2 - 8y - 4 - 2.5x^2y

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3 0
4 years ago
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}}

This is based on knowledge that in the U.S., the mean PAL is 1.65 and the standard deviation is 0.55.

This means that \mu = 1.65, \sigma = 0.55

A study took a random sample of 51 people who lived in low income neighborhoods and found their mean PAL to be 1.63.

This means that n = 51, X = 1.63

Using a one-sample z test, what is the z-score for this data

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{1.63 - 1.65}{\frac{0.55}{\sqrt{51}}}

Z = -0.26

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

4 0
3 years ago
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