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horsena [70]
3 years ago
9

Using the graph of f(x)=x^2 as a guide, describe the transformations, and then graph each function.

Mathematics
2 answers:
leonid [27]3 years ago
8 0
Multiplying the whole function by -1 reflects the function across the x axis
so f(x) to -f(x) would be a reflection across the x axis


multiplying the whole function by a fraction is a vertical transformation, if you multiply by a value x, such that 0<x<1, then it is a vertical shrink, if x>1, then it is a vertical stretch
so like f(x) to 2f(x) is a vertical stretch by a factor of 2

adding a value to the whole function moves it up by that number



ok

so f(x)=(-1)(\frac{1}{2})(x^2)+2
we have multiplied the whole thing by -1 and 1/2 and then added 2 to the whole function

that is a reflection across the x axis, a vertical shrink by a factor of 1/2, and translated up by 2 units in that order

nordsb [41]3 years ago
3 0

Answer:

hi

Step-by-step explanation:

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Yes, it would be unusual.

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 normally distributed samples are solved using 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.

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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}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 22, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1

Would it be unusual for this sample mean to be less than 19 days?

We have to find Z when X = 19. So

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

By the Central Limit Theorem

Z = \frac{19 - 22}{1}

Z = -3

Z = -3 \leq -2, so yes, the sample mean being less than 19 days would be considered an unusual outcome.

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