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Yakvenalex [24]
2 years ago
12

The graph of the parent function f(x) = x3 is transformed such that g(x) = f(–2x). How does the graph of g(x) compare to the gra

ph of f(x)?
a) g(x) is stretched horizontally and reflected over the y-axis.
b) g(x) is stretched horizontally and reflected over the x-axis.
c) g(x) is compressed horizontally and reflected over the y-axis.
d) g(x) is compressed horizontally and reflected over the x-axis.
Mathematics
2 answers:
Anni [7]2 years ago
8 0

Answer:

b) g(x) is stretched horizontally and reflected over the x-axis.

Step-by-step explanation:

The given parent function is

f(x) =  {x}^{3}

The transformed function is

g(x) = f (- 2x)

We want to see how the transformed graph compares with the parent graph.

The negation inside means all x-coordinates we're negated.

This means, there is a reflection in the y-axis.

The factor of 2 within the function means a horizontal stretch by a factor of 1/2.

The correct answer is B.

Sophie [7]2 years ago
5 0

Answer:

It's C. g(x) is compressed horizontally and reflected over the y-axis.

Step-by-step explanation:

I just got it correct on e2020

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A brewery produces cans of beer that are supposed to contain exactly 12 ounces. But owing to the inevitable variation in the fil
MArishka [77]

Answer:

T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)

P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solutio to the problem

Let X the random variable that represent the amount of beer in each can of a population, and for this case we know the distribution for X is given by:

X \sim N(12,0.3)  

Where \mu=12 and \sigma=0.3

For this case we select 6 cans and we are interested in the probability that the total would be less or equal than 72 ounces. So we need to find a distribution for the total.

The definition of sample mean is given by:

\bar X = \frac{\sum_{i=1}^n X_i}{n} = \frac{T}{n}

If we solve for the total T we got:

T= n \bar X

For this case then the expected value and variance are given by:

E(T) = n E(\bar X) =n \mu

Var(T) = n^2 Var(\bar X)= n^2 \frac{\sigma^2}{n}= n \sigma^2

And the deviation is just:

Sd(T) = \sqrt{n} \sigma

So then the distribution for the total would be also normal and given by:

T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)

And we want this probability:

P(T\leq 72)

And we can use the z score formula given by:

z = \frac{x-\mu}{\sigma}

P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z

6 0
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First, list all the given information:
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The easiest approach to use here is the dimensional analysis. Cancel out like units if they appear both in the numerator and denominator side. Solve first the original cost. The solution is as follows:

100 miles/week * 1 gal/25 miles * $4/gal = $16/week

The reduced cost would be:
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New average miles/week = 68.75 miles/week
8 0
2 years ago
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