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3 years ago

Find the domain of f(x)= 10x+1 / 8x + 3 and put it in interval notation?

Mathematics

2 answers:

vesna_86 [32]3 years ago

4 0

$f(x) = \frac{10x + 1}{8x + 3}$

8x + 3 ≠ 0

8x ≠ -3

$\frac{8x}{8} \neq \frac{-3}{8}$

x ≠ $\frac{-3}{8}$

Interval Notation: (∞, $\frac{-3}{8}$ ) U ( $\frac{-3}{8}$ , ∞)

Alja [10]3 years ago

3 0

(- ∞, - $\frac{3}{8}$ ) ∪ (- $\frac{3}{8}$ , ∞ )

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be.

solve : 8x + 3 = 0 ⇒ x = - $\frac{3}{8}$

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Read 2 more answers

Let X1 and X2 be independent random variables with mean μand variance σ².

My name is Ann [436]

Answer:

a) $E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu$

So then we conclude that $\hat \theta_1$ is an unbiased estimator of $\mu$

$E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu$

So then we conclude that $\hat \theta_2$ is an unbiased estimator of $\mu$

b) $Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}$

$Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}$

Step-by-step explanation:

For this case we know that we have two random variables:

$X_1 , X_2$ both with mean $\mu = \mu$ and variance $\sigma^2$

And we define the following estimators:

$\hat \theta_1 = \frac{X_1 + X_2}{2}$

$\hat \theta_2 = \frac{X_1 + 3X_2}{4}$

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

$E(\hat \theta_i) = \mu , i = 1,2$