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Temka [501]
3 years ago
5

Stacy rolls a pair of six-sided fair dice.

Mathematics
1 answer:
Nutka1998 [239]3 years ago
5 0

Answer:

The probability that the sum of the numbers rolled is either a multiple of 3 or an even number is 24/36 = 2/3

The two events are mutually exclusive

Step-by-step explanation:

we first need to create a table of the sample space from the experiment, the sum of the numbers rolled. Find the attached for a depiction of the possible outcomes.

The probability that the sum of the numbers rolled is either a multiple of 3 or an even number will be given by counting the numbers that are either even or multiples of 3 and then dividing by the number of possible outcomes, 36.

In our case this will be;

24/36 = 2/3

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What does x equal in this equation -5x + 8 = -2
AnnyKZ [126]

Answer:

x = 2

Step-by-step explanation:

-5x + 8 = -2

-5x + 8 - 8 = -2 -8

-5x = -10

-5x/5 = -10/5

-x = -2

x=2

8 0
2 years ago
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Step-by-step explanation:

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3 years ago
In what order would we solve for n?<br> 3n - 4 = 14
Blizzard [7]
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation;

3*n-4-(14)=0

Pull out like factors :

3n - 18 = 3 • (n - 6)

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This equation has no solution.
A a non-zero constant never equals zero.

Solve : n-6 = 0

Add 6 to both sides of the equation :
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3 0
3 years ago
Use trig identities to transform the left side of the equation to the right side.
Taya2010 [7]

<u>Answer and explanation</u>

(1+sinθ)(1-sinθ)=cos²θ

We are to prove that the left hand side is equal to the right hand side.

(1+sinθ)(1-sinθ) = 1(1-sinθ) + sinθ(1-sinθ)

                         = 1 - sinθ + sinθ - sin²θ

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From the trigonometric identity sin²θ + cos²θ = 1,

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7 0
3 years ago
Find the simplified product b-5/2b x b^2+3b/b-5
White raven [17]

Answer:

The product \frac{b-5}{2b}\times\frac{b^2+3b}{b-5}=\frac{b+3}{2}

Step-by-step explanation:

Given expression \frac{b-5}{2b} and \frac{b^2+3b}{b-5}

We have to find the product of  \frac{b-5}{2b}\times\frac{b^2+3b}{b-5}

   

Consider the given expression  \frac{b-5}{2b}\times\frac{b^2+3b}{b-5}

Multiply fractions, we have,

\frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d}

=\frac{\left(b-5\right)\left(b^2+3b\right)}{2b\left(b-5\right)}

Cancel common factor ( b - 5 )

we have, =\frac{b^2+3b}{2b}

Apply exponent rule,

\:a^{b+c}=a^ba^c

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=\frac{b\left(b+3\right)}{2b}

Cancel common factor b , we have,

=\frac{b+3}{2}

Thus, the product  \frac{b-5}{2b}\times\frac{b^2+3b}{b-5}=\frac{b+3}{2}

8 0
3 years ago
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