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OleMash [197]
3 years ago
5

Determine whether the distribution is a probability distribution. x 0 1 2 3 4 5 ​P(x) StartFraction 1 Over 25 EndFraction one fi

fth one half three fourths StartFraction 1 Over 50 EndFraction StartFraction 1 Over 100 EndFraction Is the probability distribution a discrete​ distribution? Why? Choose the correct answer below. A. No comma because some of the probabilities have values greater than 1 or less than 0. B. Yes comma because the distribution is symmetric. C. Yes comma because the probabilities sum to 1 and are all between 0 and 1 comma inclusive. D. No comma because the total probability is not equal to 1. Click to select your answer and then click Check Answer.
Mathematics
1 answer:
Citrus2011 [14]3 years ago
6 0

Answer:

D. No comma because the total probability is not equal to 1.

Step-by-step explanation:

Given the distribution:

\left|\begin{array}{c|cccccc}x&0&1&2&3&4&5\\ P(x)&1/25&1/5&1/2&3/4& 1/50&1/100\end{array}\right|

The sum of the probabilities

\sum P(x)=1/25+1/5+1/2+3/4+ 1/50+1/100\\\sum P(x) =1.52 \neq 1

Therefore, the distribution is not a probability distribution because the total probability is not equal to 1.

The correct option is D.

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Easy math, but not sure what I did wrong.
Ierofanga [76]

Answer:

8

Step-by-step explanation:

In his 5th year, he took 3 times as many exams as the first year.  So the number of exams taken in the 5th year must be a multiple of 3.

If a₁ = 1, then a₅ = 3.  However, this isn't possible because we need 4 integers between them, and a sum of 31.

If a₁ = 2, then a₅ = 6.  Same problem as before.

If a₁ = 3, then a₅ = 9.  This is a possible solution.

If a₁ = 4, then a₅ = 12.  If we assume a₂ = 5, a₃ = 6, and a₄ = 7, then the sum is 34, so this is not a possible solution.

Therefore, Alex took 3 exams in his first year and 9 exams in his fifth year.  So he took 19 exams total in his second, third, and fourth years.

3 < a₂ < a₃ < a₄ < 9

If a₂ = 4, then a₃ = 7 and a₄ = 8.

If a₂ = 5, then a₃ = 6 and a₄ = 8.

If a₂ = 6, then there's no solution.

So Alex must have taken 8 exams in his fourth year.

4 0
3 years ago
Suppose that the polynomial function A(x) = - 0.0082x2 + 0.06x approximates a person's blood-alcohol level x hours after drinkin
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Answer:

<h2>the answer will be B</h2>

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Algebraic expressions are expressions that use variables and combination of terms.

<h3>How to determine the equivalent expressions</h3>

<u>1. Addition</u>

The addition expression is:

5x^2 - \frac 13x + \frac 52 - \frac 12x^2 + \frac 12x - \frac 13 -2x^2 + \frac 15x - \frac 16

Collect like terms

5x^2 - \frac 12x^2 -2x^2 - \frac 13x + \frac 12x + \frac 15x + \frac 52 - \frac 13  - \frac 16

Evaluate the like terms

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<u>2. Subtraction</u>

The subtraction expression is:

7x^2 -2x + 10 - (-2x^2 + \frac 12x - 3)

Open the bracket

7x^2 -2x + 10 + 2x^2 - \frac 12x +3

Collect like terms

7x^2 + 2x^2 -2x - \frac 12x+ 10   +3

Evaluate the like terms

9x^2- \frac 52x+ 13

<u>3. Simplify</u>

The expression is given as:

(\frac 13x^2 - \frac 47x + 11) - (\frac 17x - 3 -2x^2) - (\frac 27x - \frac 23x^2 + 2)

Rewrite as:

(\frac 13x^2 - \frac 47x + 11) - ( -2x^2 + \frac 17x - 3) - (- \frac 23x^2 + \frac 27x + 2)

Open the brackets

\frac 13x^2 - \frac 47x + 11 + 2x^2 - \frac 17x + 3+ \frac 23x^2 - \frac 27x - 2

Collect like terms

\frac 13x^2 + 2x^2 + \frac 23x^2 - \frac 47x - \frac 17x - \frac 27x + 11   + 3 - 2

Evaluate the like terms

3x^2 - x + 12

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The expression is given as:

(x-\frac 1x)(x + \frac 1x)(x^2 + \frac 1{x^2})(x^4 + \frac 1{x^4})

Apply the difference of two squares

(x^2-\frac 1{x^2})(x^2 + \frac 1{x^2})(x^4 + \frac 1{x^4})

Apply the difference of two squares

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Apply the difference of two squares

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Apply the difference of two squares

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<u>7. Identity</u>

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Rewrite as:

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<u></u>

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The expression is given as:

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Apply the difference of two squares

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5^2 - 2 * 6

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<u>10. Coefficient</u>

The expression is given as:

\frac x2 +\frac y2 - xy

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Read more about algebraic expressions at:

brainly.com/question/4344214

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Step-by-step explanation:

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