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Arlecino [84]
2 years ago
8

Can anyone help me please

Mathematics
1 answer:
Lapatulllka [165]2 years ago
5 0

Answer:

x= 3 and x= -3

Step-by-step explanation:

x^2 -9 = 0

Factor x^2-9

(x+3)(x-3)

If x+3=0, then x= -3

If x-3=0, then x= 3

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0.2 · 100% = 20%
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Laura's school is holding an outdoor activities celebration event for its 20th anniversary. Laura decides to participate in the
Pachacha [2.7K]

Probabilities are used to determine the chances of Laura winning an activity

  • The value of p is 1/12
  • The expected score of the game is -1/6
  • The probability that she has a total score of 5 after two rounds is 0

<h3>How to calculate the value of p</h3>

To calculate the value of p, we make use of the following probability formula

\sum P(x) = 1

So, we have:

\frac 3{12} + \frac 1{12} + \frac 4{12} + \frac 2{12} + p + \frac 1{12} = 1

Collect like terms

\frac 3{12} + \frac 1{12} + \frac 4{12} + \frac 2{12} + \frac 1{12}  + p= 1

Evaluate the sums, on the left-hand side

\frac{11}{12} + p = 1

Subtract 11/12 from both sides

p = \frac 1{12}

Hence, the value of p is 1/12

<h3>(b) How to calculate the expected score</h3>

This is calculated using the following expected value formula

E(x) = \sum x * P(x)

So, we have:

E(x) = -3 * \frac{3}{12} -1 * \frac 1{12} + 0 * \frac 4{12} + 1 * \frac 2{12} + 2 * \frac 1{12} + 4 * \frac 1{12}

Evaluate the products

E(x) =  -\frac{9}{12} - \frac 1{12}  + \frac 2{12} + \frac 2{12} + \frac 4{12}

E(x) =  -\frac 2{12}

Simplify

E(x) =  -\frac 1{6}

Hence, the expected score of the game is -1/6

<h3>(c) The probability that she has a total score of 5 after two rounds</h3>

From the table, we have:

P(5) = 0

For after two rounds, we make use of the following equation

P(5) * P(5) = 0 * 0

P(5) * P(5) = 0

Hence, the probability that she has a total score of 5 after two rounds is 0

Read more about probabilities at:

brainly.com/question/25870256

4 0
2 years ago
Simplify 6 - 4[x + 1 - (y + 3)] + 5y.
o-na [289]
6-4[x-1-(y+3)]+5y
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6-4+12-4x+4y+5y
14-4x+9y


6 0
3 years ago
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