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Kruka [31]
3 years ago
14

Here's an experiment you can try. Start by folding an ordinary sheet of typing paper exactly in half. Then fold it in half over

and over again. You will find that, try as you might, you can't fold it more than 6 times. If a single sheet of paper is .01 cm thick, how thick is the last, 6-fold wad of paper?
a. 0.64 cm
b. 1.28 cm
c. 0.6 cm
d. 0.064 cm
e. 0.32 cm
Mathematics
2 answers:
Katyanochek1 [597]3 years ago
7 0
First fold = .02
2nd fold =   .04
3rd fold =   .08
4th fold =    .16
5th fold =     .32
6th fold = .64 cm
goldfiish [28.3K]3 years ago
3 0

Answer:

a. 0.64 cm

Step-by-step explanation:

To understand what happened here, the best thing to do is doing it so you can count the layers.

First, you have a one layer paper 0.1cm thick.

With the first fold, you end with a two-layer paper or 0.2cm thick.

Now, with the second you are folding two-layers over two-layers, this made a four-layer paper so, you have a 2x2 paper thick or 0.4cm thick.

When you fold the third time, the same happened.

For layer paper over for layer paper so you have 2x(4x0.1) paper layers or 0.08cm thick.

Continuing whit this, at the sith fold, you'll have

Fourth time: 2x(8x0.1) layers or 0.16cm thick.

Fifth time: 2x(16x0.1) layers or 0.32cm thick.

Sixth time: 2x(32x0.1) layers or 0.64cm thick.

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8 0
3 years ago
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The zookeeper needed a new uniform for work. He bought his hat for $26.92, his shirt for $54.73, and his pants for $33.28. How m
Mumz [18]
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Identify the equation in slope-intercept form for the line containing the point (-4,1) and (2,3)
oksano4ka [1.4K]

Answer:

The point slope formula is

(y−y1)=m(x−x1)

Where m = the slope calculated as y2−y1x2−x1

(x1,y1)(x2,y2)

(-4 , 1) (-2 , 3)

Solve for the slope

m = 3−1−2−(−3) = 21 = m = 2

(y−3)=2(x−(−2)) Plug in known values.

y−3=2(x+2) Simplify signs

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7 0
2 years ago
An article includes the accompanying data on compression strength (lb) for a sample of 12-oz aluminum cans filled with strawberr
Veseljchak [2.6K]

Answer:

<em>The calculated value |t| = 2.375 > 2.1009  at 0.05 level of significance</em>

<em>Null hypothesis is rejected </em>

<em>Alternative hypothesis is accepted</em>

<em>The extra carbonation of cola results in a higher average compression strength</em>

<u><em /></u>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given data</em>

<em>First sample size (n₁) = 10 </em>

<em>mean of the first sample(x₁⁻) = 537</em>

<em>standard deviation of the first sample (S₁) = 22</em>

<em>second sample size (n₂) = 10 </em>

<em>mean of the second sample (x₂⁻) = 559</em>

<em>standard deviation of the second sample (S₂) = 17</em>

<u><em>Step(ii)</em></u><em>:-</em>

<u><em>Null hypothesis : H₀:</em></u><em>- The extra carbonation of cola results in a lower average compression strength</em>

<u>Alternative Hypothesis :H₁</u>

<em>The extra carbonation of cola results in a higher average compression strength</em>

<u><em>Step(iii)</em></u><em>:-</em>

<em>By using student's t -test for difference of means</em>

<u><em>Test statistic</em></u>

<em>       </em>t = \frac{x^{-} _{1}-x^{-} _{2}  }{\sqrt{S^{2} (\frac{1}{n_{1} }+\frac{1}{n_{2} }  } )}<em />

<em>  where </em>

<em>     </em>S^{2}  = \frac{n_{1}S^{2} _{1} + n_{2} S_{2} ^{2}  }{n_{1}+n_{2} -2 }<em />

<em>    </em>S^{2}  = \frac{10(22)^{2}  + 10 (17) ^{2}  }{10+10 -2 } = \frac{ 7730}{18} = 429.4<em />

<em>    </em>t = \frac{537-559 }{\sqrt{429.4 (\frac{1}{10 }+\frac{1}{10 }  } )}<em />

<em>   t =  -2.375</em>

<em>|t| = |-2.375| = 2.375</em>

<em>Degrees of freedom</em>

<em>γ = n₁+n₂ -2 = 10+10-2 =18</em>

<em />t_{\frac{\alpha }{2} ,n-1}=t_{(\frac{0.05}{2} ,18)} = t_{(0.025,18)}}=2.1009<em />

<em>The calculated value |t| = 2.375 > 2.1009  at 0.05 level of significance</em>

<em>Null hypothesis is rejected </em>

<em>Alternative hypothesis is accepted</em>

<u><em>Final answer:-</em></u>

<em>The extra carbonation of cola results in a higher average compression strength</em>

<u><em /></u>

<em />

<em> </em>

<em />

5 0
2 years ago
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