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QveST [7]
3 years ago
14

Find the point P on the line y=2x that is closest to the point (20,0) .what is the least distance between pans (20,0)?

Mathematics
1 answer:
Vsevolod [243]3 years ago
6 0
The closest point on a line, to another point, will be a point that's on a normal of that line, or a line that is perpendicular to it, notice picture below

so, y = 2x, has a slope of 2, a perpendicular line to it, will have a slope of negative reciprocal that, or \bf 2\qquad negative\implies -2\qquad reciprocal\implies \cfrac{1}{-2}\implies -\cfrac{1}{2}

so, we know that line passes through the point 20,0, and has a slope of -1/2

if we plug that in the point-slope form, we get \bf y-0=-\cfrac{1}{2}(x-20)\implies y=-\cfrac{1}{2}x+10

now, the point that's on 2x and is also on that perpendicular line, is the closest to 20,0 from 2x, thus, is where both graphs intersect, as you can see in the graph

thus  \bf 2x=-\cfrac{1}{2}x+10  solve for "x'

------------------------------------------------------------

not sure on the 2nd part, but sounds like, what's the distance from that point to 20,0, well, if that's the case, just use the distance equation

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ \square }}\quad ,&{{ \square }})\quad 
%  (c,d)
&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}


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A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and
mr Goodwill [35]

Answer:

Bias for the estimator = -0.56

Mean Square Error for the estimator = 6.6311

Step-by-step explanation:

Given - A normally distributed random variable with mean 4.5 and standard deviation 7.6 is sampled to get two independent values, X1 and X2. The mean is estimated using the formula (3X1 + 4X2)/8.

To find - Determine the bias and the mean squared error for this estimator of the mean.

Proof -

Let us denote

X be a random variable such that X ~ N(mean = 4.5, SD = 7.6)

Now,

An estimate of mean, μ is suggested as

\mu = \frac{3X_{1} + 4X_{2}  }{8}

Now

Bias for the estimator = E(μ bar) - μ

                                    = E( \frac{3X_{1} + 4X_{2}  }{8}) - 4.5

                                    = \frac{3E(X_{1}) + 4E(X_{2})}{8} - 4.5

                                    = \frac{3(4.5) + 4(4.5)}{8} - 4.5

                                    = \frac{13.5 + 18}{8} - 4.5

                                    = \frac{31.5}{8} - 4.5

                                    = 3.9375 - 4.5

                                    = - 0.5625 ≈ -0.56

∴ we get

Bias for the estimator = -0.56

Now,

Mean Square Error for the estimator = E[(μ bar - μ)²]

                                                             = Var(μ bar) + [Bias(μ bar, μ)]²

                                                             = Var( \frac{3X_{1} + 4X_{2}  }{8}) + 0.3136

                                                             = \frac{1}{64} Var( {3X_{1} + 4X_{2}  }) + 0.3136

                                                             = \frac{1}{64} ( [{3Var(X_{1}) + 4Var(X_{2})]  }) + 0.3136

                                                             = \frac{1}{64} [{3(57.76) + 4(57.76)}]  } + 0.3136

                                                             = \frac{1}{64} [7(57.76)}]  } + 0.3136

                                                             = \frac{1}{64} [404.32]  } + 0.3136

                                                             = 6.3175 + 0.3136

                                                              = 6.6311

∴ we get

Mean Square Error for the estimator = 6.6311

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f(x)=5x^2+9x-2

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f(x)=5x^2+9x-2

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