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Norma-Jean [14]
3 years ago
7

Ed & Heath won the grand

Mathematics
1 answer:
mestny [16]3 years ago
6 0

Answer:

224

Step-by-step explanation:

(100+12)*2

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Step-by-step explanation:

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40.51x+12.45y=666.64
irga5000 [103]

Answer:

The graph of the equation 40.51x+12.45y=666.64 is attached with the answer where the horizontal axis represents the X axis and the vertical axis represents Y axis.

To plot the graph physically just find two points lying on the line. Mark the points on the graph sheet and then join them. This will give you the line represented by the equation.

To find points on the line assume the value of any one variable, substitute it in the equation, then solve the equation to find the value of other variable. For example : assume y = 1; substitute the value of y in the equation;

⇒ 40.51x + 12.45×1 = 666.64

⇒ 40.51x = 666.64 - 12.45

⇒ 40.51x = 654.19

⇒ x = \frac{654.19}{40.51}

⇒ x ≈ 16.149

Therefore point ( 16.149 , 1 ) lie on the graph of the equation.

***Only two points are required to plot this graph just because it represents a straight line, that we can conclude just by observing the equation. If in an equation the power of x is 1 or 0 and power of y is 1 or 0 then only it will represent a straight line in 2-D plane.***

5 0
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

6 0
3 years ago
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