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Mathematics

1 answer:

algol [13]3 years ago

8 0

Answer:

3k^2 -k+5

reason:

when adding two equations together, you combine like terms.
First, add -2 to 7 to get 5
Then, add k, or 1k, to -2k, which gives you -1k, or just -k
3k^2 stays the same because there are no like terms in the other equation to combine it with

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In a survey of 1,300 women who gave birth in Maine in 2003, a total of 588 reported taking a multivitamin at least 4 times a wee

vesna_86 [32]

Using it's concept, it is found that the prevalence of frequent multivitamin use is of 0.4523 = 45.23%.

<h3>What is the prevalence of a sample?</h3>

The prevalence of a sample is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

In this problem, 588 out of 1300 women reported taking a multivitamin at least 4 times a week during the month before pregnant, hence the prevalence is given by:

p = 588/1300 = 0.4523 = 45.23%.

More can be learned about prevalence at brainly.com/question/14398287

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5 0

3 years ago

A researcher reports survey results by stating that the standard error of the mean is 25 the population standard deviation is 40

bezimeni [28]

Answer:

a) A sample of 256 was used in this survey.

b) 45.14% probability that the point estimate was within ±15 of the population mean

Step-by-step explanation:

This question is solved using the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 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 $s = \frac{\sigma}{\sqrt{n}}$ .