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Viefleur [7K]
3 years ago
13

Julius needs to determine how much information needs to be given about an isosceles triangle to be able to draw an isosceles tri

angle. He is looking at six different cases and is trying to determine if that amount of detail will allow him to construct just one isosceles triangle or if it allows for more than one isosceles triangle.
Mathematics
1 answer:
Brrunno [24]3 years ago
3 0

Answer:

Six cases are enough to construct more than one isosceles triangle

Step-by-step explanation:

An Isosceles triangle is a triangle in which at least two of it's sides are congruent.

If two of the three sides of a triangle are congruent, but the third side is of a different length, the triangle is isosceles.

If all three sides of the triangle are congruent, even though this is an equilateral triangle, it satisfies the condition of an isosceles, and it is ok to still it isosceles.

Julius might decide to split information for a better explanation, it doesn't matter the number of cases he has, but six cases should be enough to draw an isosceles triangle.

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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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I'll go ahead and assume you meant to say that <em>S</em> is the surface given by

S = \left\{(x,y,z) \mid z = \cos(y)\text{ with } 0\le y\le \pi\text{ and }0\le x\le4\right\}

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\vec r(x, y) = \left\langle x, y, \cos(y)\right \rangle

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\vec n = \dfrac{\partial\vec r}{\partial x} \times \dfrac{\partial\vec r}{\partial y} = \left\langle0,\sin(y),1\right\rangle

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