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amid [387]
3 years ago
14

I would appreciate the help!

Mathematics
2 answers:
pochemuha3 years ago
8 0

Answer:

its b I'm pretty sure. it makes the most sense i

djverab [1.8K]3 years ago
3 0

Answer:

A

Step-by-step explanation:

Multiply 15 by 8 to get 120. 120 + 15 × 11 = 285. 285 + 15 × 22 = 615. 615 + 20 × 7 = 755. 755 + 20 × 6 = 875. 875 + 20 × 14 = 1,155. 1,155 - 292.58 = 862.42. 862.42 - 292.58 = 569.84. Remember PEMDAS.

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Can you help me simplify the expression 1.452/0.6
Anastasy [175]

Answer:

1.452/0.6 simplified is 1.000/1.0

8 0
3 years ago
I need help plz help
ivanzaharov [21]

Answer:

B. 70%

Step-by-step explanation:

7/10 of the numbers are more than 5

7/10= 70%

5 0
2 years ago
Read 2 more answers
Express 120 as a product of its prime factors.
raketka [301]
By the Fundamental Theorem of Arithmetic, all number can be expressed as a product of prime numbers.

So naturally, lets divide 120 by an easy prime number.
We know that 120 is even, so lets try 2 120/2 = 60 lets keep dividing it by two until it becomes odd or prime 60/2 = 30 30/2 = 15 now lets see, what are some factors of 15?

Well the obvious ones are 3 and 5, both of which are prime. So now we can just count up how many times we divided it by 2
 
120/2 = 60 60/2 = 30 30/2 = 15 and 15 is just 3 x 5, so: <span>
120=(<span>23</span>)×(3)×(5)</span>
or <span><span>
120 = 2 × 2 × 2 × 3 × 5</span></span>
7 0
3 years ago
Read 2 more answers
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
Number 14 please and thank you
Fed [463]
You can do a notebook a door , a frame or a mirror, mostly anything that has angles
4 0
3 years ago
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