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olchik [2.2K]
3 years ago
9

Determine o período e o conjunto imagem, construindo o gráfico de um período completo para cada função dada.a) por f(x) = 1+ 4 s

en x.b) por f(x) = 2sen (x-3)

Mathematics
1 answer:
nataly862011 [7]3 years ago
7 0

Answer:

The given functions are

f(x)=1+4sinx

f(x)=2sin(x-3)

The sine function has a standard period of 2 \pi by definition. However, this might change if we use a factor as coefficient of the x-varible, but in this case we don't have that.

Therefore, the period of both trigonometric functions is 2 \pi.

Now, the images of each function is the y-variable set values that defines each function.

So, the function f(x)=1+4sinx has an image defined by the set [-3,5]. It's impotant to notice that the range of a standard function is [-1,1], however, in this case, the function was shifted 1 unit up and it was streched by a factor of 4, that's why the standard image changes to [-3,5].

About the second function f(x)=2sin(x-3), the image set is [-2,2], because the function was streched by a factor of 2.

Additionally, the image attached shows the graph of the given functions.

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Step-by-step explanation:

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a) 0.5.

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c) 0.8413

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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 = 6, \sigma = 0.2

(a) P(x > 6) =

This is 1 subtracted by the pvalue of Z when X = 6. So

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

Z = \frac{6-6}{0.2}

Z = 0

Z = 0 has a pvalue of 0.5.

1 - 0.5 = 0.5.

(b) P(x < 6.2)=

This is the pvalue of Z when X = 6.2. So

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

Z = \frac{6.2-6}{0.2}

Z = 1

Z = 1 has a pvalue of 0.8413

(c) P(x ≤ 6.2) =

In the normal distribution, the probability of an exact value, for example, P(X = 6.2), is always zero, which means that P(x ≤ 6.2) = P(x < 6.2) = 0.8413.

(d) P(5.8 < x < 6.2) =

This is the pvalue of Z when X = 6.2 subtracted by the pvalue of Z when X  5.8.

X = 6.2

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

Z = \frac{6.2-6}{0.2}

Z = 1

Z = 1 has a pvalue of 0.8413

X = 5.8

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

Z = \frac{5.8-6}{0.2}

Z = -1

Z = -1 has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

(e) P(x > 5.7) =

This is 1 subtracted by the pvalue of Z when X = 5.7.

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

Z = \frac{5.8-6}{0.2}

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(f) P(x > 5) =

This is 1 subtracted by the pvalue of Z when X = 5.

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

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Z = -5 has a pvalue of 0.

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A movie collector has 105 movie posters and 90 band posters if she sells 21 movie posters then how many band posters should she
kotykmax [81]

Hi there! :)

<u>Answer:</u>

In order to maintain the same ratio, she should sell 18 band posters.


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<u>Initial ratio:</u>

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72 = X ⇒ Number of band posters she should have left to maintain the same ratio.

She has 90 band posters :

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Add "X" to each side of the equation → 72 + X = 72 + X

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Subtract 72 from each side of the equation → 90 - 72 = 18

<u>18 = X</u>


There you go! I really hope this helped, if there's anything just let me know! :)

4 0
3 years ago
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