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krok68 [10]
3 years ago
13

Let the random variable X be the outcome of rolling a 6-sided die. Assuming the die is fair, the probability distribution is as

follows: X 1 2 3 4 5 6 Probability 1/6 1/6 1/6 1/6 1/6 1/6 Using the above table to answer the question: What is the mean of X?
Mathematics
2 answers:
vitfil [10]3 years ago
8 0

Answer: The mean is 3.5

Step-by-step explanation: For a discrete random variable probability, the mean is given by the formula: μ = ∑xiPi, where:

xi is the outcome and Pi is its probability.

In this case,

μ = 1.\frac{1}{6} + 2·\frac{1}{6} + 3·\frac{1}{6} + 4.\frac{1}{6} + 5.\frac{1}{6} + 6.\frac{1}{6}

μ = \frac{1+2+3+4+5+6}{6}

μ =\frac{21}{6}

μ = 3.5

Thus, the mean of X is 3.5

gayaneshka [121]3 years ago
6 0
The answer would be 1.  

1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 6.  
Divide by 6 and you have 1 for the answer.  
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Given the position of the particle, what the position(s) of the particle when it’s at rest
choli [55]

The position function of a particle is given by:

X\mleft(t\mright)=\frac{2}{3}t^3-\frac{9}{2}t^2-18t

The velocity function is the derivative of the position:

\begin{gathered} V(t)=\frac{2}{3}(3t^2)-\frac{9}{2}(2t)-18 \\ \text{Simplifying:} \\ V(t)=2t^2-9t-18 \end{gathered}

The particle will be at rest when the velocity is 0, thus we solve the equation:

2t^2-9t-18=0

The coefficients of this equation are: a = 2, b = -9, c = -18

Solve by using the formula:

t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}

Substituting:

\begin{gathered} t=\frac{9\pm\sqrt[]{81-4(2)(-18)}}{2(2)} \\ t=\frac{9\pm\sqrt[]{81+144}}{4} \\ t=\frac{9\pm\sqrt[]{225}}{4} \\ t=\frac{9\pm15}{4} \end{gathered}

We have two possible answers:

\begin{gathered} t=\frac{9+15}{4}=6 \\ t=\frac{9-15}{4}=-\frac{3}{2} \end{gathered}

We only accept the positive answer because the time cannot be negative.

Now calculate the position for t = 6:

undefined

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1 year ago
If the level of significance of a hypothesis test is raised from 0.005 to 0.2, the probability of a type ii error will:_________
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The probability of type II error will decrease if the level of significance of a hypothesis test is raised from 0.005 to 0.2.

<h3 /><h3>What is a type II error?</h3>

A type II error occurs when a false null hypothesis is not rejected or a true alternative hypothesis is mistakenly rejected.

It is denoted by 'β'. The power of the hypothesis is given by '1 - β'.

<h3>How the type II error is related to the significance level?</h3>

The relation between type II error and the significance level(α):

  • The higher values of significance level make it easier to reject the null hypothesis. So, the probability of type II error decreases.
  • The lower values of significance level make it fail to reject a false null hypothesis. So, the probability of type II error increases.
  • Thus, if the significance level increases, the type II error decreases and vice-versa.

From this, it is known that when the significance level of the given hypothesis test is raised from 0.005 to 0.2, the probability of type II error will decrease.

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7 0
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Consider the system of equations:
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Answer:

(x, y) = (5, 1)

Step-by-step explanation:

To <em>eliminate</em> x, you can double the second equation and subtract the first.

... 2(x +4y) -(2x -3y) = 2(9) -(7)

...11y = 11 . . . . . simplify

... y = 1 . . . . . . divide by 11

Using the second equation to find x, we have ...

... x + 4·1 = 9

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<u>Check</u>

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(Since we used the second equation to find x, we know it will check.)

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