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-BARSIC- [3]
3 years ago
7

Sylvie folds a large piece of paper in half. The fold divides the paper into two equal parts. She folds it in half again. When s

he unfolds it, the folds divide the paper into four equal parts. She contines to fold and unfold the paper until the folds divide the paper into 64 equal parts. How many times altogether has Sylvie folded the paper?
A. 5 times
B. 6 times
C. 7 times
D. 8 times
Mathematics
1 answer:
Nikolay [14]3 years ago
6 0
Each time you fold a paper it will double the amount of parts. When you think about it, it's just multiplying by 2 a bunch of times.

One fold is 2 times 1, which is 2.
The second fold is 2 times 2, which is 4.
The third fold is 4 times 2, which is 8.

Using this process, we can simplify that into exponents. If the amount of times you fold is x and the parts double for each fold, then the amount of parts can be represented by:

2^{x}

So if the amount of parts is given and we need the amount of folds, just keep doubling until you get to 64. The amount of times you doubled is the number of times Sylvie folded.

Later you will learn that the opposite of an exponent is a logarithm, which would look like this:

log_{2}(64) = ?

But don't worry about that yet.
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Answer:

see explanation

Step-by-step explanation:

(1)

(a)

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(b)

cos(a) = \frac{adjacent}{hypotenuse} = \frac{8}{17}

(c)

tan(a) = \frac{opposite}{adjacent} = \frac{15}{8}

----------------------------------------------------

(2)

(a)

sin(b) = \frac{opposite}{hypotenuse} = \frac{8}{17}

(b)

cos(b) = \frac{adjacent}{hypotenuse} = \frac{15}{17}

(c)

tan(b) = \frac{opposite}{adjacent} = \frac{8}{15}

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Answer:

1. The required information are

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The average annual bonuses, \bar {x}_2 received by employees from company B

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The standard deviation, σ₂, of the average annual bonuses for employees from company A

The number of employees in company A, n₁

The number of employees in company B, n₂

2. The null hypothesis is H₀: \bar {x}_1 - \bar {x}_2 ≤ 100

The alternative hypothesis is Hₐ: \bar {x}_1 - \bar {x}_2 > 100

Step-by-step explanation:

1. The required information are

The average annual bonuses, \bar {x}_1 received by employees from company A

The average annual bonuses, \bar {x}_2 received by employees from company B

The standard deviation, σ₁, of the average annual bonuses for employees from company A

The standard deviation, σ₂, of the average annual bonuses for employees from company A

The number of employees in company A, n₁

The number of employees in company B, n₂

2. The null hypothesis is H₀: \bar {x}_1 - \bar {x}_2 ≤ 100

The alternative hypothesis is Hₐ: \bar {x}_1 - \bar {x}_2 > 100

The z value for the hypothesis testing of the difference between two means is given as follows;

z=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\frac{\sigma_{1}^{2} }{n_{1}}-\frac{\sigma _{2}^{2}}{n_{2}}}}

At 0.5 level of significance, the critical z_\alpha = ± 0

The rejection region is z > z_\alpha and z < -z_\alpha

Therefore, the value of z obtained from the relation above more than or less than 0, we reject the null hypothesis, and we fail to reject the alternative hypothesis.

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