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PIT_PIT [208]
3 years ago
11

How do you find the parameterization of a helix?

Mathematics
1 answer:
Marina CMI [18]3 years ago
7 0

Answer:

\left\{\begin{array}{l}x=r\cos t\\ \\ y=r\sin t\\ \\z=at\end{array}\right.

Step-by-step explanation:

A <u>helix</u> is a curve for which the tangent makes a constant angle with a fixed line. The shortest path between two points on a cylinder (one not directly above the other) is a fractional turn of a helix, as can be seen by cutting the cylinder along one of its sides, flattening it out, and noting that a straight line connecting the points becomes helical upon re-wrapping (see attached diagram for helix).

The parametrization for the helix is

\left\{\begin{array}{l}x=r\cos t\\ \\ y=r\sin t\\ \\z=at\end{array}\right.,

where r is the radius of the helix, a is the constant, and t\in [0,2\pi ) is the parameter

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Find the perimeter of angle ABC with verticals A(1,-6), B(1,-3) and C(-4,-6)
disa [49]

Answer:

its 13.83

Step-by-step explanation:

Use the distance formula and plug them in

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3 years ago
I need the answer fast please​
Rainbow [258]
Substitute each value of x in y=2x-3

so,

first box (-2)
y= 2(-2)-3
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y= -7

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y= 2(0)-3
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6 0
2 years ago
Read 2 more answers
The lengths of pregnancies are normally distributed with a mean of days and a standard deviation of days. a. Find the probabilit
Alik [6]

Answer:

a) The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of Z = \frac{X - \mu}{\sigma}, in which \mu is the mean and \sigma is the standard deviation.

b) We have to find X when Z has a p-value of \frac{a}{100}, and X is given by: X = \mu - Z\sigma, in which \mu is the mean and \sigma is the standard deviation.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

In this question:

Mean \mu, standard deviation \sigma

a. Find the probability of a pregnancy lasting X days or longer.

The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of Z = \frac{X - \mu}{\sigma}, in which \mu is the mean and \sigma is the standard deviation.

b. If the length of pregnancy is in the lowest a​%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

We have to find X when Z has a p-value of \frac{a}{100}, and X is given by: X = \mu - Z\sigma, in which \mu is the mean and \sigma is the standard deviation.

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3 years ago
Volumes of Pyramids and Cones: Please help!
Elanso [62]

Answer:

Step-by-step explanation:256

8 0
3 years ago
The number of tickets purchased by an individual for Beckham College's holiday music festival is a uniformly distributed random
Zanzabum

1) Mean: 6.5

2) Standard deviation: 3.2

Step-by-step explanation:

1)

A uniform random distribution is the distribution of a variable that can assume n different values, and each of these values is equally likely to occur.

This means that the probability of occurring of each of these values is the same:

p(x_1)=p(x_2)=...=p(x_n)

For a uniform random distribution, the mean of the distribution is simply equal to the median value of the distribution, which is exactly at the center of the distribution itself. It can be calculated as

\bar x = \frac{1}{2}(a+b)

where

a is the lowest value of the distribution

b is the highest value of the distribution

Here the distribution consists of all values ranging from 1 to 12, so:

a = 1

b = 12

Therefore the mean value is

\bar x=\frac{1}{2}(1+12)=6.5

2)

For a uniform random distribution, it can be dimostrated that the variance of the distribution is

\sigma^2 = \frac{1}{12}(b-a)^2

For the distribution in this problem, we have:

a = 1

b = 12

Therefore the variance is

\sigma^2 = \frac{1}{12}(12-1)^2=10.08

The standard deviation is the square root of the variance, therefore for this distribution it is:

\sqrt{\sigma^2}=\sqrt{10.08}=3.2

Learn more about mean and standard deviation:

brainly.com/question/6073431

brainly.com/question/8799684

brainly.com/question/4625002

#LearnwithBrainly

3 0
2 years ago
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