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3 years ago

(33x3+6x-9)+(-33x3+9x2-10)

Mathematics

2 answers:

34kurt3 years ago

5 0

(33x3+6x-9)+(-33x3+9x2-10) = -46

JulijaS [17]3 years ago

4 0

Answer:

9x² + 6x -19

Step-by-step explanation:

= 33x³ + 6x - 9 - 33x³ + 9x² -10

= 33x³ - 33x³ + 9x² + 6x -10 -9

= 9x² + 6x -19

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A pair of fair dice is cast. Let Edenote the event that the number landing uppermost on the first die is a 1 and let Fdenote the

iris [78.8K]

Answer:

They are not independent

Step-by-step explanation:

Given

E = Occurrence of 1 on first die

F = Sum of the uppermost occurrence in both die is 5

Required

Are E and F independent

First, we need to list the sample space of a roll of a die

$Event\ 1 = \{1,2,3,4,5,6\}$

Next, we list out the sample space of F

$Event\ 2 = \{2,3,4,5,6,7,3,4,5,6,7,8,4,5,\$

$6,7,8,9,5,6,7,8,9,10,6,7,8,9,10,11,7,8,9,10,11,12\}$

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$E = \{1\}$

So:

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$P(E) = \frac{1}{6}$

In (2): the sample space of F is:

$F = \{5,5,5,5\}$

So:

$P(F) = \frac{n(F)}{n(Event\ 2)}$

$P(F) =\frac{4}{36}$

$P(F) =\frac{1}{9}$

For E and F to be independent:

$P(E\ and\ F) = P(E) * P(F)$

Substitute values for P(E) and P(F)

This gives:

$P(E\ and\ F) = \frac{1}{6} * \frac{1}{9}$

$P(E\ and\ F) = \frac{1}{54}$

However, the actual value of P(E and F) is 0.

This is so because $E = \{1\}$ and $F = \{5,5,5,5\}$ have 0 common elements:

So:

$P(E\ and\ F) = 0$

Compare $P(E\ and\ F) = \frac{1}{54}$ and $P(E\ and\ F) = 0$ .

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Hence: the two events are not independent

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Answer:

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In this case, 3/12 which is 1/4

Step 2: Plug in known variables into equation

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Step 3: Solve/Plug in calc

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