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Vanyuwa [196]
2 years ago
13

A stack of 500 pieces of paper is 1.875 in tall. Diego guess is that each piece of paper is 0.015 inches thick. Explain how you

know that Diego answer is incorrect. Compute the thickness of each piece of paper. Show you reasoning.
Mathematics
2 answers:
amid [387]2 years ago
5 0
You divide 1.875 by 500 and get 0.00375 in. per paper
Andrej [43]2 years ago
5 0

We can solve this problem with a proportion. On one side of the equals sign is one sheet paper of paper, on another side is for 500 sheets of paper. The top will be the number of sheets, the bottom the thickness.

\frac{1}{T} =\frac{500}{1.875}

We let T equal the thickness of one sheet of paper. We solve this proportion by cross multiplying.

500T = 1.875

T = 1.875 / 500 = 0.00375

Before, Diego guessed the paper was 0.0015 inches thick, and our proportion tells it is exactly 0.00375 inches thick. Diego overestimated the thickness.


Thus, one piece of paper has 0.0375 inches of thickness.

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Given: X is the midpoint of WY, WX = XZ<br> Prove: XY = XZ
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Answer:

proved XY=XZ

Step-by-step explanation:

As it is given that X is the mid point of WY, it means,

WX=XY-----(1)

Also it is given in the statement that

WX=XZ-----(2)

On equating equation 1 and 2 we have,

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A random sample of 160 car purchases are selected and categorized by age. The results are listed below. The age distribution of
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Answer:

The claim that all ages have purchase rates proportional to their driving rates is false.

Step-by-step explanation:

The complete question is:

A random sample of 160 car accidents are selected and categorized by the age of the driver determined to be at fault. The results are listed below. The age distribution of drivers for the given categories is 18% for the under 26 group, 39% for the 26-45 group, 31% for the 45-65 group, and 12% for the group over 65. Calculate the chi-square test statistic used to test the claim that all ages have crash rates proportional to their driving rates.

Age      >26     26-45       46-65      45<

Drivers 66    39            25          30

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      C)85.123

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Solution:

In this case we need to test whether there is sufficient evidence to warrant rejection of the claim that all ages have crash rates proportional to their driving rates.

A Chi-square test for goodness of fit will be used in this case.

The hypothesis can be defined as:

<em>H₀</em>: The observed frequencies are same as the expected frequencies.

<em>Hₐ</em>: The observed frequencies are not same as the expected frequencies.

The test statistic is given as follows:

\chi^{2}=\sum{\frac{(O-E)^{2}}{E}}

The values are computed in the table.

The test statistic value is 75.10.

The degrees of freedom of the test is:

n - 1 = 4 - 1 = 3

Compute the p-value of the test as follows:

p-value < 0.00001

*Use a Chi-square table.

p-value < 0.00001 < α = 0.05.

So, the null hypothesis will be rejected at any significance level.

Thus, there is sufficient evidence to warrant rejection of the claim that ages have crash rates proportional to their driving rates.

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Answer:

Step-by-step explanation:

150/x=sin 15

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