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BigorU [14]
3 years ago
8

An experiment consists of randomly drawing a cube from a bag containing three red and two blue cubes What is the sample space of

this experiment
Mathematics
1 answer:
lawyer [7]3 years ago
4 0

Answer:

{Red,Blue}

Step-by-step explanation:

Since the experiment consists of randomly drawing just a cube from a bag containing three red and two blue cubes, the outcome could be either a red or blue cube. Therefore the sample space is:

{Red,Blue}

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Yuliya22 [10]
The range for this set of numbers is 38
8 0
3 years ago
Read 2 more answers
How do you solve this
NNADVOKAT [17]
By using numbers and solving it how your teacher teached you lol XD
3 0
3 years ago
1. Type an equation in the equation editor that uses 2 fractions with parentheses around one of them. Example: <img src="https:/
avanturin [10]

Answer:

i) \frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}    \Rightarrow \frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}

ii)4^{3} + 8^{2} + \sqrt{9}   \Rightarrow  4^{3} + 8^{2} + \sqrt{9}

iii) (\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}} \Rightarrow \hspace{0.2cm}    (\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}}

Step-by-step explanation:

i) \frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}    \Rightarrow \frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}

ii)4^{3} + 8^{2} + \sqrt{9}   \Rightarrow  4^{3} + 8^{2} + \sqrt{9}

iii) (\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}} \Rightarrow \hspace{0.2cm}    (\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}}

7 0
4 years ago
How to solve this question<br><br><br>b/5-1=10
Marizza181 [45]
Our goal is to isolate the variable, which is b. What you do to one side must also be done to the other side since we need to maintain equality.

Add 1 to both sides
b/5=11

Then multiply both sides by 5
b=55

Final answer: b=55
3 0
3 years ago
Pentagon ABCDE and pentagon A″B″C″D″E″ are shown on the coordinate plane below:
erastova [34]

The transformations that are applied to pentagon ABCDE to create A"B"C"D"E" are:

1) Translation (x, y) → (x + 8, y + 2)

2) Reflection across the x-axis (x, y) → (x, -y)

So, the overall transformation given in the graph is (x, y) → {(x + 8), -(y + 2)}.

<h3>What are the transformation rules?</h3>

The transformation rules are:

  • Reflection across x-axis: (x, y) → (x, -y)
  • Reflection across y-axis: (x, y) → (-x, y)
  • Translation: (x, y) → (x + a, y + b)
  • Dilation: (x, y) → (kx, ky)

<h3>Calculation:</h3>

The pentagons in the graph have vertices as

For the pentagon ABCDE: A(-4, 5), B(-6, 4), C(-5, 1), D(-2, 2), and (-2, 4)

For the pentagon A"B"C"D"E": A"(4, -7), B"(2, -6), C"(3, -3), D"(6, -4), and E"(6, -6)

Consider the vertices A(-4, 5) from the pentagon ABCDE and A"(4, -7) from the pentagon A"B"C"D"E".

Applying the Translation rule for the pentagon ABCDE:

The rule is (x, y) → (x + a, y + b)

So, the variation is

-4 + a = 4

⇒ a = 4 + 4 = 8

5 + b = 7

⇒ b = 7 - 5 = 2

So, the pentagon ABCDE is translated by (x + 8, y + 2).

Applying the Reflection rule for the translated pentagon:

The translated pentagon has vertices (x + 8, y + 2).

When applying the reflection across the x-axis,

(x + 8, y + 2) → {(x + 8), -(y + 2)}

Therefore, the complete transformation of the pentagon ABCDE to the pentagon A"B"C"D"E" is (x, y) → {(x + 8), -(y + 2)}

Verification:

A(-4, 5) → ((-4 + 8), -(5 + 2)) = (4, -7)A"

B(-6, 4) → ((-6 + 8), -(4 + 2)) = (2, -6)B"

C(-5, 1) → ((-5 + 8), -(1 + 2)) = (3, -3)C"

D(-2, 2) → ((-2 + 8), -(2 + 2)) = (6, -4)D"

E(-2, 4) → ((-2 + 8), -(4 + 2)) = (6, -6)E"

Learn more about transformation rules here:

brainly.com/question/4289712

#SPJ1

5 0
2 years ago
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