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adoni [48]
3 years ago
7

Describe this scatter plot

Mathematics
2 answers:
saul85 [17]3 years ago
7 0

Answer:

It is a negative correlation there is one outlier, it is going from the top left to the bottom right. a line could be drawn through it. it could be linear but i would  say non linear.

Step-by-step explanation:

timofeeve [1]3 years ago
3 0

Answer:

Negative correlation

Step-by-step explanation:

The trend shown by the data points appear to be downward sloping, indicating negative correlation between the variables

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10v2 + 50
Svetach [21]

Answer:

what is the question? probably describe the polynomial. okay. the polynomial 10v^2 + 50 is a quadratic (it is to the 2nd power) and a binomial (has 2 terms).

7 0
3 years ago
1/3into a percent and a decimal
jeka57 [31]

Answer:

33% or .33

Step-by-step explanation:

Divide 1/3 = .3333 (repeating). I would round to the nearest hundreths place:

1/3 = .33

For the precentage, multiple 0.33 by 100:

100 * 0.33 = 33%

3 0
3 years ago
I need some help with geometry?​
kramer
39=9x+7-3x+20

Subtract 27 from each side

12=6x
2=x

Now plug in 2 for x to find AB and BC

9x+7= AB
9(2)+7=AB
18+7= AB
25= AB

-3x+20= BC
-3(2)+20= BC
-6+20= BC
14= BC
3 0
3 years ago
1/3(3x - 12) = x - 4
PIT_PIT [208]

Step-by-step explanation:

the formula is too long so the link is below:

https://www.geteasysoloution.com/1/3x+12=x-4

3 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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