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Keith_Richards [23]
3 years ago
11

The area of a square rug is 169ft^2. How long is each side of the rug?

Mathematics
1 answer:
jek_recluse [69]3 years ago
6 0

Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{13 \: ft}}}}}

Step-by-step explanation:

\star{ \sf{ \: Area \: of \: a \: square \: rug \: ( \: A \: ) \:  =  \: 169 \:  {ft}^{2} }}

\sf{ \underline{Finding \: the \: length \: of \: a \: square \: ( \: l \: )}}

\boxed{ \sf{Area \: of \: a \: square \:  =  \:  {l}^{2} }}

\hookrightarrow{ \sf{169 =  {l}^{2} }}

\hookrightarrow{ \sf{ {l}^{2}  = 169}}

\hookrightarrow{ \sf{ \sqrt{ {(l)}^{2} }  =  \sqrt{169} }}

\hookrightarrow{ \text{length \:  =  \: 13 \: ft}}

\text{Hope \: I \: helped!}

\text{Best \: wishes !!}

~\text{TheAnimeGirl}

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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

6 0
3 years ago
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