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olya-2409 [2.1K]

3 years ago

10

If we fold a rectangular piece of paper in half multiple times and count the number of rectangles created, what type of sequence

are we creating? Write an explicit formula for this situation.

2 answers:

Luda [366]3 years ago

8 0

Answer:

this and that, I ain't even gonna cap I don't know the answer

Step-by-step explanation:

Serggg [28]3 years ago

3 0

Answer:

<h2>We are creating a geometric sequence because each time we fold, we double the number of rectangles.✍️✍️✍️</h2>

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iren [92.7K]

The Correct answer is A. i hope this helps :)

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3 years ago

Please answer the attachment for me. The answer that is circled is wrong.

Inessa05 [86]

Answer:

The answer is 46 units

Step-by-step explanation:

You were correct in multiplying 12 by 4. However, you forgot to notice that the corners of every square were only shaded halfway. All you have to do then is subtract 48 by 2.

7 0

4 years ago

A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement

dsp73

Answer:

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

$P(Same) = \frac{67}{210}$

Step-by-step explanation:

Given (Omitted from the question)

$Red = 7$

$White = 9$

$Blue = 5$

Solving (a): $P(First\ White\ and\ Second\ Blue)$

This is calculated using:

$P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)$

$P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}$

<em>We used Total - 1 because it is a probability without replacement</em>

So, we have:

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}$

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}$

$P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}$

$P(First\ White\ and\ Second\ Blue) = \frac{45}{420}$

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

Solving (b) $P(Same)$

This is calculated as:

$P(Same) = P(First\ Blue\ and Second\ Blue)\$ $or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)$

$P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})$

$P(Same) = (\frac{5}{21} * \frac{4}{20})+(\frac{7}{21} * \frac{6}{20})+(\frac{9}{21} * \frac{8}{20})$

$P(Same) = \frac{20}{420}+\frac{42}{420} +\frac{72}{420}$

$P(Same) = \frac{20+42+72}{420}$

$P(Same) = \frac{134}{420}$

$P(Same) = \frac{67}{210}$

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3 years ago

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neonofarm [45]

Answer:

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3 years ago

Read 2 more answers