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Katarina [22]
3 years ago
10

Over a 24-hour period, the tide in a harbor can be modeled by one period of a sinusoidal function. The tide measures 4.35 ft at

midnight, rises to a high of 8.3 ft, falls to a low of 0.4 ft, and then rises to 4.35 ft by the next midnight.
What is the equation for the sine function f(x), where x represents time in hours since the beginning of the 24-hour period, that models the situation?

f(x)=
Mathematics
1 answer:
netineya [11]3 years ago
8 0

Answer:

  f(x) = 4.35 +3.95·sin(πx/12)

Step-by-step explanation:

For problems of this sort, a sine function is used that is of the form ...

  f(x) = A + Bsin(2πx/P)

where A is the average or middle value of the oscillation, B is the one-sided amplitude, P is the period in the same units as x.

It is rare that a tide function has a period (P) of 24 hours, but we'll use that value since the problem statement requires it. The value of A is the middle value of the oscillation, 4.35 ft in this problem. The value of B is the amplitude, given as 8.3 ft -4.35 ft = 3.95 ft. Putting these values into the form gives ...

  f(x) = 4.35 + 3.95·sin(2πx/24)

The argument of the sine function can be simplified to πx/12, as in the Answer, above.

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Scorpion4ik [409]

Answer:

A

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8 0
2 years ago
Compare the perimeters of the two regular figures. Which statement is true?
adell [148]

Answer:

(A). The perimeter of the octagon is greater than that of the hexagon.

Step-by-step explanation:

Since, hexagon consists of 6 sides and 6 angles, thus the measure of one angle of the hexagon will be=\frac{(n-2){\timeS}180^{\circ}}{6}

=\frac{(6-2){\timeS}180^{\circ}}{6}

=\frac{(4){\timeS}180^{\circ}}{6}

=120^{\circ}

Now, since MQ is the angle bisector of the one of the angle of the hexagon, therefore ∠QMP=60°.

Now, from ΔQMP. we have

\frac{MP}{MQ}=cos60^{\circ}

MP=\frac{1}{2}

Thus, the perimeter of the hexagon is:

P=12{\times}MP

P=12{\times}\frac{1}{2}

P=6 units

Thus, the perimeter of hexagon is 6 units.

Also, Since, octagon consists of 8 sides and 8 angles, thus the measure of one angle of the octagon will be=\frac{(n-2){\timeS}180^{\circ}}{8}

=\frac{(8-2){\timeS}180^{\circ}}{8}

=\frac{(6){\timeS}180^{\circ}}{8}

=135^{\circ}

Now, since AP is the angle bisector of the one of the angle of the octagon, therefore {\angle}PAC=cos\frac{135}{2}.

From ΔAPC, we have

AC=cos\frac{135}{2}

Now, Perimeter of octagon is:

P=16{\times}cos\frac{135}{2}

P=16{\times}0.382

P=6.122 units

Thus, the perimeter of octagon is 6.122 units.

Now, the perimeter of octagon is greater than perimeter of the hexagon, thus option A is correct that is The perimeter of the octagon is greater than that of the hexagon.

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3 years ago
What is the surface area of a right circular cylinder with a diameter of 14 meters and a height of 9 meters
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Answer:

This is your answer ☺️☺️☺️

4 0
3 years ago
Suppose a geyser has a mean time between eruptions of 72 minutes. Let the interval of time between the eruptions be normally dis
nikitadnepr [17]

Answer:

(a) The probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is 0.3336.

(b) The probability that a random sample of 13-time intervals between eruptions has a mean longer than 82 ​minutes is 0.0582.

(c) The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is 0.0055.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) The population mean must be more than 72​, since the probability is so low.

Step-by-step explanation:

We are given that a geyser has a mean time between eruptions of 72 minutes.

Also, the interval of time between the eruptions be normally distributed with a standard deviation of 23 minutes.

(a) Let X = <u><em>the interval of time between the eruptions</em></u>

So, X ~ N(\mu=72, \sigma^{2} =23^{2})

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

Now, the probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is given by = P(X > 82 min)

       P(X > 82 min) = P( \frac{X-\mu}{\sigma} > \frac{82-72}{23} ) = P(Z > 0.43) = 1 - P(Z \leq 0.43)

                                                           = 1 - 0.6664 = <u>0.3336</u>

The above probability is calculated by looking at the value of x = 0.43 in the z table which has an area of 0.6664.

(b) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{13} } } ) = P(Z > 1.57) = 1 - P(Z \leq 1.57)

                                                           = 1 - 0.9418 = <u>0.0582</u>

The above probability is calculated by looking at the value of x = 1.57 in the z table which has an area of 0.9418.

(c) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 34

Now, the probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{34} } } ) = P(Z > 2.54) = 1 - P(Z \leq 2.54)

                                                           = 1 - 0.9945 = <u>0.0055</u>

The above probability is calculated by looking at the value of x = 2.54 in the z table which has an area of 0.9945.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) If a random sample of 34-time intervals between eruptions has a mean longer than 82 ​minutes, then we conclude that the population mean must be more than 72​, since the probability is so low.

6 0
3 years ago
What type of triangle has exactly two acute angles?
Rufina [12.5K]
The correct answer is right.

An acute one would have all three acute, while an obtuse one would have at least one obtuse angle.
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3 years ago
Read 2 more answers
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