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My name is Ann [436]
2 years ago
7

Martin has a bag of marbles. He removed one marble at random, recorded the color and then placed it back in the bag. He repeated

this process several times and recorded his results in the table. Find the experimental probability of drawing each color.
Red: 12
Blue: 10
Green: 15
Yellow: 13

What are two different ways you could find the experimental probability of the event that Martin does not draw a red marble?
Mathematics
1 answer:
dybincka [34]2 years ago
6 0

Answer:

38/50 ; or 19/25

Step-by-step explanation:

One way could be that you add all the marbles together to get your total marbles; 12 + 10 + 15 + 13 = 50.

Then you add all the colours expect red; 10 + 15 + 13 = 38.

Which would give me 38/50; which simplifed would be 19/25

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The sun of three consecutive integers is 60 find the integers
prohojiy [21]

Answer:

<em>let </em><em>the </em><em>three</em><em> </em><em>consecutive</em><em> Integers</em><em> </em><em>be</em>

<em>x </em><em>,</em><em> </em><em>x+</em><em>1</em><em> </em><em>,</em><em> </em><em>x+</em><em>2</em>

<em>their </em><em>sum </em><em>is </em><em>6</em><em>0</em><em>,</em>

<em>so,</em>

<em>x </em><em>+</em><em> </em><em>x </em><em>+</em><em> </em><em>1</em><em> </em><em>+</em><em> </em><em>x </em><em>+</em><em> </em><em>2</em><em> </em><em>=</em><em> </em><em>6</em><em>0</em>

<em>→</em><em> </em><em>3</em><em>x</em><em> </em><em>+</em><em> </em><em>3</em><em> </em><em>=</em><em> </em><em>6</em><em>0</em>

<em>→</em><em> </em><em>3</em><em>x</em><em> </em><em>=</em><em> </em><em>6</em><em>0</em><em> </em><em>-</em><em> </em><em>3</em><em> </em><em>=</em><em> </em><em>5</em><em>7</em>

<em>→</em><em> </em><em>x </em><em>=</em><em> </em><em>5</em><em>7</em><em>/</em><em>3</em><em> </em><em>=</em><em> </em><em>1</em><em>9</em>

<em>therefore</em><em>,</em><em> </em><em>1</em><em>s</em><em>t</em><em> </em><em>integer </em><em>=</em><em> </em><em>1</em><em>9</em>

<em>2</em><em>n</em><em>d</em><em> </em><em>integer </em><em>=</em><em> </em><em>x </em><em>+</em><em> </em><em>1</em><em> </em><em>=</em><em> </em><em>1</em><em>9</em><em> </em><em>+</em><em> </em><em>1</em><em> </em><em>=</em><em> </em><em>2</em><em>0</em>

<em>3</em><em>r</em><em>d</em><em> </em><em>integer </em><em>=</em><em> </em><em>x </em><em>+</em><em> </em><em>2</em><em> </em><em>=</em><em> </em><em>1</em><em>9</em><em> </em><em>+</em><em> </em><em>2</em><em> </em><em>=</em><em> </em><em>2</em><em>1</em>

<em>so,</em>

<em>the </em><em>three </em><em>consecutive</em><em> Integers</em><em> </em><em>are </em><em>1</em><em>9</em><em> </em><em>,</em><em> </em><em>2</em><em>0</em><em> </em><em>and </em><em>2</em><em>1</em><em>.</em>

<em><u>hope</u></em><em><u> this</u></em><em><u> answer</u></em><em><u> helps</u></em><em><u> you</u></em><em><u> dear</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u> take</u></em><em><u> care</u></em><em><u> and</u></em><em><u> may</u></em><em><u> u</u></em><em><u> have</u></em><em><u> a</u></em><em><u> great</u></em><em><u> </u></em><em><u>day </u></em><em><u>ahead</u></em><em><u>!</u></em>

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Step-by-step explanation:

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