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anzhelika [568]
2 years ago
13

In a game the player wins if he rolls a 6 on a number cube. If the number cube is rolled 18 times, then what is a reasonable pre

diction for the number of unsuccessful rolls? Show work
Mathematics
1 answer:
SVETLANKA909090 [29]2 years ago
8 0

If the number cube is rolled 18 times, then the number of unsuccessful rolls = 90 unsuccessful rolls.

<u>Step-by-step explanation</u>:

<u>step 1</u>: A cube has six faces and each face is numbered from 1 to 6.

<u>step 2</u>: The win in the game requires the number 6, when a cube is rolled.

<u>step 3</u>: On rolling a cube, the outcomes are (1,2,3,4,5, and 6).

<u>step 4</u>: Out of these outcomes, 5 are unsuccessful rolls (1,2,3,4,5) and 1 is a successful roll (6).

<u>step 5</u>: If the cube is rolled 18 times,

then the number of unsuccessful rolls = 18 * 5 unsuccessful rolls

                                                                  = 90 unsuccessful rolls.

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Answer:

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Step-by-step explanation:

Area = \pi r^{2}

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3 years ago
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What are the types of roots of the equation below?<br> - 81=0
Tju [1.3M]

Option B, that is Two Complex and Two Real which are x + 3, x - 3, x + 3i and x - 3i, are the types of roots of the equation x⁴ - 81 = 0. This can be obtained by finding root of the equation using algebraic identity.    

<h3>What are the types of roots of the equation below?</h3>

Here in the question it is given that,

  • the equation x⁴ - 81 = 0

By using algebraic identity, (a + b)(a - b) = a² - b², we get,  

⇒ x⁴ - 81 = 0                      

⇒ (x² +  9)(x² - 9) = 0

⇒ (x² + 9)(x² - 9) = 0

  1. (x² -  9) = (x² - 3²) = (x - 3)(x + 3) [using algebraic identity, (a + b)(a - b) = a² - b²]
  2. x² + 9 = 0 ⇒ x² = -9 ⇒ x = √-9 ⇒ x= √-1√9 ⇒x = ± 3i

⇒ (x² + 9) = (x - 3i)(x + 3i)

Now the equation becomes,

[(x - 3)(x + 3)][(x - 3i)(x + 3i)] = 0

Therefore x + 3, x - 3, x + 3i and x - 3i are the roots of the equation

To check whether the roots are correct multiply the roots with each other,

⇒ [(x - 3)(x + 3)][(x - 3i)(x + 3i)] = 0

⇒ [x² - 3x + 3x - 9][x² - 3xi + 3xi - 9i²] = 0

⇒ (x² +0x - 9)(x² +0xi - 9(- 1)) = 0

⇒ (x² - 9)(x² + 9) = 0

⇒ x⁴ - 9x² + 9x² - 81 = 0

⇒ x⁴ - 81 = 0

Hence Option B, that is Two Complex and Two Real which are x + 3, x - 3, x + 3i and x - 3i, are the types of roots of the equation x⁴ - 81 = 0.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: What are the types of roots of the equation below?

x⁴ - 81 = 0

A) Four Complex

B) Two Complex and Two Real

C) Four Real

Learn more about roots of equation here:

brainly.com/question/26926523

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1 year ago
10.<br> JKLM is a rhombus KLN is a triangle. Find MKN.
rjkz [21]

Answer:

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Step-by-step explanation:

JM = JK (all sides of a rhombus are equal)

Angle JKM = 25° (isosceles triangle)

Angle JKL = 50° (consecutive angles of rhombus)

Angle MKL = 25° (angle subtraction)

Angle MLK = 130° (opposite angles of a rhombus)

Angle KLN = 50° (angles on a straight line)

Angle LKN = 40° (angle sum of triangle)

Angle MKN = 65° (angle addition)

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andre [41]

Answer:

For x^2 + 5x - 14 = 0

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For x^2 + 5x + 6 = 0

x1 = -3, x2 = -2

Step-by-step explanation:

Pls see attached....

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Norma-Jean [14]
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