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dedylja [7]
4 years ago
12

Please tell me what x is 3(x+6)=48

Mathematics
1 answer:
Alekssandra [29.7K]4 years ago
7 0

Answer

X=10

Step by step explanation

48÷3 = 16

16-6 = 10

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An unbiased estimator is one for which: Group of answer choices its value always equals the value of the parameter it estimates.
m_a_m_a [10]

Answer:

its sampling distribution is centered exactly at the parameter it estimates.

Step-by-step explanation:

An unbiased estimator is an estimator that shows the correct statistic applied for the parameter of the population

Here correct represents that it should not be overestimated or underestimated. In the case when there is an underestimate or overestimate this means that there is a bias

So as per the given options the above option is correct

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3 years ago
Find the roots of the equation f(x) = x3 - 0.2589x2 + 0.02262x -0.001122 = 0
devlian [24]

Answer:

The root of the equation x^3-0.2589x^{2}+0.02262x-0.001122=0 is x ≈ 0.162035

Step-by-step explanation:

To find the roots of the equation x^3-0.2589x^{2}+0.02262x-0.001122=0 you can use the Newton-Raphson method.

It is a way to find a good approximation for the root of a real-valued function f(x) = 0. The method starts with a function f(x) defined over the real numbers, the function derivative f', and an initial guess x_{0} for a root of the function. It uses the idea that a continuous and differentiable function can be approximated by a straight line tangent to it.

This is the expression that we need to use

x_{n+1}=x_{n} -\frac{f(x_{n})}{f(x_{n})'}

For the information given:

f(x) = x^3-0.2589x^{2}+0.02262x-0.001122=0\\f(x)'=3x^2-0.5178x+0.02262

For the initial value x_{0} you can choose x_{0}=0 although you can choose any value that you want.

So for approximation x_{1}

x_{1}=x_{0}-\frac{f(x_{0})}{f(x_{0})'} \\x_{1}=0-\frac{0^3-0.2589\cdot0^2+0.02262\cdot 0-0.001122}{3\cdot 0^2-0.5178\cdot 0+0.02262} \\x_{1}=0.0496021

Next, with x_{1}=0.0496021 you put it into the equation

f(0.0496021)=(0.0496021)^3-0.2589\cdot (0.0496021)^2+0.02262\cdot 0.0496021-0.001122 = -0.0005150, you can see that this value is close to 0 but we need to refine our solution.

For approximation x_{2}

x_{2}=x_{1}-\frac{f(x_{1})}{f(x_{1})'} \\x_{1}=0-\frac{0.0496021^3-0.2589\cdot 0.0496021^2+0.02262\cdot 0.0496021-0.001122}{3\cdot 0.0496021^2-0.5178\cdot 0.0496021+0.02262} \\x_{1}=0.168883

Again we put x_{2}=0.168883 into the equation

f(0.168883)=(0.168883)^3-0.2589\cdot (0.168883)^2+0.02262\cdot 0.168883-0.001122=0.0001307 this value is close to 0 but again we need to refine our solution.

We can summarize this process in the following table.

The approximation x_{5} gives you the root of the equation.

When you plot the equation you find that only have one real root and is approximate to the value found.

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