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Deffense [45]
2 years ago
10

If you vertically compress the absolute value patent function, f(x)=|x|, by a factor of 5, what is the equation of the new funct

ion?

Mathematics
1 answer:
Hunter-Best [27]2 years ago
5 0

A vertical compression implies a reduction in the y-values of the absolute value function. Therefore, the term that we use to accomplish the reduction cannot be attached to the x in the equation. If so, we would end up with a horizontal stretch, not a vertical compression.

Compression implies a reduction, or dividing, so our term will be 1/5. This should be placed outside of the absolute value bars so that the compression affects only the y-values.

Correct answer: B. g(x) = 1/5|x|

Hope this helps! :)

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a=-30

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3 years ago
Let U1, ..., Un be i.i.d. Unif(0, 1), and X = max(U1, ..., Un). What is the PDF of X? What is EX? Hint: Find the CDF of X first,
Kryger [21]

Answer:

E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}

Step-by-step explanation:

A uniform distribution, "sometimes also known as a rectangular distribution, is a distribution that has constant probability".

We need to take in count that our random variable just take values between 0 and 1 since is uniform distribution (0,1). The maximum of the finite set of elements in (0,1) needs to be present in (0,1).

If we select a value x \in (0,1) we want this:

max(U_1, ....,U_n) \leq x

And we can express this like that:

u_i \leq x for each possible i

We assume that the random variable u_i are independent and P)U_i \leq x) =x from the definition of an uniform random variable between 0 and 1. So we can find the cumulative distribution like this:

P(X \leq x) = P(U_1 \leq 1, ...., U_n \leq x) \prod P(U_i \leq x) =\prod x = x^n

And then cumulative distribution would be expressed like this:

0, x \leq 0

x^n, x \in (0,1)

1, x \geq 1

For each value x\in (0,1) we can find the dendity function like this:

f_X (x) = \frac{d}{dx} F_X (x) = nx^{n-1}

So then we have the pdf defined, and given by:

f_X (x) = n x^{n-1} , x \in (0,1)  and 0 for other case

And now we can find the expected value for the random variable X like this:

E(X) =\int_{0}^1 s f_X (x) dx = \int_{0}^1 x n x^{n-1}

E(X)= n \int_{0}^1 x^n dx = n [\frac{1}{n+1}- \frac{0}{n+1}]=\frac{n}{n+1}

6 0
3 years ago
The thumb length of fully grown females of a certain type of frog is normally distributed with a mean of 8.59 mm and a standard
PilotLPTM [1.2K]

Answer:

21.77% probability that a randomly selected frog of this type has thumb length longer than 9.08 mm.

Step-by-step explanation:

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.

In this problem, we have that:

\mu = 8.59, \sigma = 0.63

Calculate the probability that a randomly selected frog of this type has thumb length longer than 9.08 mm.

This is 1 subtracted by the pvalue of Z when X = 9.08. So

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

Z = \frac{9.08 - 8.59}{0.63}

Z = 0.78

Z = 0.78 has a pvalue of 0.7823

1 - 0.7823 = 0.2177

21.77% probability that a randomly selected frog of this type has thumb length longer than 9.08 mm.

8 0
2 years ago
A) Compare the y-intercepts of f(x) and g(x). Use complete sentences.
Lady bird [3.3K]

Answer:


Step-by-step explanation:

a) y-intercept is when x=0


f(0)


=1/(0-3)


=-1/3


g(0)=0 from the graph


b) vertical asymptotes are vertical lines that f(x) and g(x) get close but do not exist there


8 0
3 years ago
Read 2 more answers
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