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3 years ago

Geometry math question please help

Mathematics

1 answer:

Triss [41]3 years ago

6 0

A square is a figure with four equal sides and four right angles.In a square the diagonals bisect at right angles .

In the above given options option A is the right answer.

If the diagonals of any parallelogram are perpendicular then the figure is not necessarily a square.

Diagonals are perpendicular in a rhombus too.

A square and Rhombus both have diagonals that are perpendicular.

So it is not a necessary condition for a parallelogram to be a square .

The other options are right .

So option B is the right option that if diagonals are perpendicular is not sufficient to prove the figure to be a square.

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What are the coordinates of point B on AC such that the rario of AB to BC is 2:3

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Answer:

Step-by-step explanation:

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3 years ago

Why can't the product of two digit numbers ever be two digits

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4 years ago

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3 years ago

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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3 years ago

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Zarrin [17]

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3 years ago

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