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mylen [45]
2 years ago
6

Jonathon has a bag that contains exactly one red marble (r), one yellow marble (y), and one green marble (g). He chooses a marbl

e from the bag without looking. Without replacing that marble, he chooses a second marble from the bag without looking. Which outcomes would be included in the sample space for Jonathan’s experiment? Select three options.
Mathematics
1 answer:
wariber [46]2 years ago
7 0

Answer:

          [ry, yr, gy, yg, rg, gr]-Any three picks from this sample space.

Step-by-step explanation:

-The bag contains exactly 1r, \ 1y, \ 1g \ marbles.

-He picks one ball without looking, and chooses a second ball without replacing the first:

  • If the first is red, the the second is either yellow or green,
  • If the first is a yellow, then the second can be red or green,
  • If the first is green, then the second can be a red or yellow.

The sample space of the second pick is therefore:

                               [ry, yr, gy, yg, rg, gr]

Hence, any three picks from the sample space are correct possible outcomes.

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G = x-c/x solve for x
Elan Coil [88]

Answer:

x=\frac{c}{1-g}

Step-by-step explanation:

Step 1: Multiply both sides by x,

 gx=−c+x

Step 2: Add -x to both sides.

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gx−x=−c

Step 3: Factor out variable x.

x(g−1)=−c

Step 4: Divide both sides by g-1.

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Hope this helps!

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A group of eight grade 11 and five grade 12 students wish to be on the senior prom committee. The committee will consist of thre
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2 years ago
Coach De Leon purchases sports equipment. Basketballs cost $20.00 each, and soccer balls cost $18.00 each. He had a budget of $1
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Answer:

18s+20b<_150

Step-by-step explanation:

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8 0
1 year ago
Show one way to count from $82 to $512
Sedaia [141]
<span><u><em>First way:</em></u>
The easiest and simplest way is to <u>count by 1</u> starting from 82 till you reach 512.
<u>This will go as follows:</u>
82, 83, 84, 85, ........... , 510, 511, 512

<u><em>Second way:</em></u>
We can note that the two given numbers are even numbers. This means that the two numbers are divisible by 2.
Therefore, we can <u>count by 2</u> starting from 82 till we reach 512.
<u>This will go as follows:</u>
82, 84, 86, 88, ................... , 508, 510, 512

<u><em>Third way:</em></u>
We can note that the units digit in both numbers is the same (the digit is 2). This means that we can count from 82 till 512 by <u>adding 10 each time</u>.
<u>This will go as follows:</u>
82, 92, 102, 112, ......................, 492, 502, 512

Hope this helps :)</span>
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