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Effectus [21]
3 years ago
9

Plz help/ Trigonometric models and sine functions: plz show how you got and the GRAPH! Thx to anyone that can help

Mathematics
2 answers:
Shtirlitz [24]3 years ago
7 0

Answer:

y=-2\sin(2x)-2 - The graph is attached below.

Step-by-step explanation:

Given : A sine function has the following key features:

Period = π , Amplitude = 2,  Midline: y = -2 , y-intercept: (0,-2)

The function is a reflection of its parent function over the x-axis.

To plot the graph :

Solution :

General form of sin function is y=A sin(Bx)+C

Where A is the amplitude

B=\frac{2\pi}{\text{Period}}

C is the midline    

Substitute the given key features  

A=2 , C=-2  

B=\frac{2\pi}{\pi}=2

The sin function is  y=2\sin(2x)-2

The function is a reflection of its parent function over the x-axis.

i.e, The coordinate became (x,y)→(x,-y)

The sin function is y=-2\sin(2x)-2

Now, We plot the graph of the sin function.

The graph is attached below.  

First point is midline i.e, (0,-2)

Minimum point is (\frac{\pi}{4},-4)

Vaselesa [24]3 years ago
5 0

Answer:

Step-by-step explanation:

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egoroff_w [7]

Supposing that the stock increases in 37 days, the 95% confidence interval for the proportion of days JMJ stock increases is: (0.484, 0.7292)

  • The lower bound is of 0.484.
  • The upper bound is of 0.7292.
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In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

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The lower limit of this interval is:

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The upper limit of this interval is:

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The interpretation is that we are <u>95% sure that the true proportion</u> of all days in which the JMJ stock increases <u>is between 0.484 and 0.7292.</u>

A similar problem is given at brainly.com/question/16807970

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Hi

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3 years ago
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Answer:

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