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Zielflug [23.3K]
3 years ago
5

Parametric Equations? How do you do them? I don't even know how to graph them and its so confusing because all the equations I p

ut in say they're not written correctly???
Mathematics
1 answer:
mote1985 [20]3 years ago
4 0

Answer:

A rectangular equation, or an equation in rectangular form is an equation composed of variables like x and y which can be graphed on a regular Cartesian plane.

Parametric equations are a set of equations that express a set of quantities as explicit functions of a number of independent variables, known as "parameters." For example, while the equation of a circle in Cartesian coordinates can be given by r^2=x^2+y^2, one set of parametric equations for the circle are given by

x=r cost

y=r sin

Note that parametric representations are generally nonunique, so the same quantities may be expressed by a number of different parameterizations. A single parameter is usually represented with the parameter t, while the symbols  u and v are commonly used for parametric equations in two parameters.Parametric equations provide a convenient way to represent curves and surfaces, as implemented, for example, in the Wolfram Language commands ParametricPlot[{x, y}, {t, t1, t2}] and ParametricPlot3D[{x, y, z}, {u, u1, u2}, {v, v1, v2}]. Unsurprisingly, curves and surfaces obtained by way of parametric equation representations are known as parametric curves and parametric surfaces, respectively.

Also if you dont know how to graph them you can Use DESMOS graphing calculator

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Construct a 95% confidence interval for the population mean, µ. Assume the population has a normal distribution. A sample of 25
Alex787 [66]

Answer:

Step-by-step explanation:

We want to determine a 95% confidence interval for the mean mean test score of students.

Number of sample, n = 25

Mean, u = 81.5

Standard deviation, s = 10.2

For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z score ×standard deviation/√n

It becomes

81.5 ± 1.96 × 10.2/√25

= 81.5 ± 1.96/× 2.04

= 81.5 ± 3.9984

The lower end of the confidence interval is 81.5 - 3.9984 =77.5

The upper end of the confidence interval is 81.5 + 3.9984 = 85.5

Therefore, with 95% confidence interval, the mean test score of students is between 77.5 and 85.5

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Pls help me, this is past due and every day it’s pasted due I get points deducted:( (I have an F in math)
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Answer:

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Step-by-step explanation:

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