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murzikaleks [220]
2 years ago
5

The tides around Cherokee Bay range between a low of one foot to a high of five feet. The tide is at its lowest point when time,

t, is 0 and completes a full cycle over a 24 hour period. What is the amplitude, period, and midline of a function that would model this periodic phenomenon?
amplitude = 4 feet; period = 24 hours; midline: y = 3

amplitude = 2 feet; period = 12 hours; midline: y = 2

amplitude = 2 feet; period = 24 hours; midline: y = 3

amplitude = 4 feet; period = 24 hours; midline: y = 2
Mathematics
1 answer:
olga_2 [115]2 years ago
3 0
The answer is <span>amplitude = 2 feet; period = 24 hours; midline: y = 3 

x1 - the lowest point
x2 - the highest point
x1 = 1 ft
x2 = 5 ft
t1 = 0
t2 = 24

The amplitude is: (x2 - x1)/2 = (5 - 1)/2 = 4/2 = 2 ft
The period is: t2 - t1 = 24 - 0 = 24 h
The midline is: (x1 + x2)/2 = (5 + 1)/2 = 6/2 = 3 ft</span>
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2400 lbs

Step-by-step explanation:

To find the best estimate, you have to round the numbers to the nearest ten.

39 -> 40

58 -> 60

40 x 60 = 2400

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The points plotted below are on the graph of a polynomial. In what range of x-values must the polynomial have a root?
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A local factory uses a manufacturing process in which 70% of the final products meet quality standards and 30% are found to be d
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8 0
2 years ago
Suppose babies born in a large hospital have a mean weight of 4095 grams, and a standard deviation of 569 grams. If 130 babies a
Butoxors [25]

Answer:

P =0.3998

Step-by-step explanation:

Let {\displaystyle {\overline{x}}} be the average of the sample, and the population mean will be \mu

We know that:

\mu = 4095 gr

Let \sigma be the standard deviation and n the sample size, then we know that the standard error of the sample is:

E=\frac{\sigma}{\sqrt{n}}

Where

\sigma=569

n=130

In this case we are looking for:

P(|{\displaystyle{\overline{x}}}- \mu|>42)

This is:

{\displaystyle{\overline{x}}}- \mu>42 or {\displaystyle{\overline{x}}}- \mu

P=P({\displaystyle{\overline{x}}}- \mu>42)+ P({\displaystyle{\overline{x}}}- \mu

Now we get the z score

Z=\frac{{\displaystyle{\overline{x}}}-\mu}{\frac{\sigma}{\sqrt{n}}}

P=P(z>\frac{42}{\frac{569}{\sqrt{130}}}) + P(z

P=P(z>0.8416) + P(z

Looking at the tables for the standard nominal distribution we get

P =0.1999+0.1999

P =0.3998

6 0
1 year ago
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