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liq [111]
3 years ago
6

Which table contains data with a nonproportional relationship?

Mathematics
2 answers:
zhuklara [117]3 years ago
6 0

Answer:

Step-by-step explanation:

Natali5045456 [20]3 years ago
4 0

Answer:

The second table, counting from the top.

X      Y

2      0

4       2

6       4

Step-by-step explanation:

A proportional relationship is written as:

y = k*x

where k is called the constant of proportionality, is a real number different than zero.

Here we have different tables, if for two pairs (x, y) of the table we have the same value of k, then the relation is proportional. Also you can see that if x = 0, we have:

y = k*0 = 0

then a proportional relationship always has the point (0, 0)

A)

X       Y

1        6

2       12

3       18

Let's use the first two pairs:

(1, 6)

6 = k*1

6/1 = k = 6

(2, 12)

12 = k*2

12/2 = k = 6

in both cases we have k = 6, then this is a proportional relationship.

B)

X      Y

2      0

4       2

6       4

Here we can see the pair (2, 0)

This means that when x = 2, we have y = 0

Then this can not be a proportional relationship, because:

0 = k*2

k = 0

This is not a proportional relationship.

C)

X    Y

1     2

2    4

3    6

Let's find k for the first two points.

(1, 2)

2 = k*1

2/1 = k = 2

(2, 4)

4 = k*2

4/2 = k = 2

In both cases we have k = 2, then this is a proportional relationship.

D)

X   Y

2   -4

4   -8

6   -12

Let's find k for the first two points:

(2, -4)

-4 = k*2

-4/2 = k = -2

(4, -8)

-8 = k*4

-8/4 = k = -2

In both cases we have k = -2, then this is a proportional relationship.

 

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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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