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Aleksandr [31]
2 years ago
12

celina measured the height of a window frame as 5.55 feet, but the actual height was 6 feet. What is the percentage of error in

Celina's measurement?
Mathematics
1 answer:
skad [1K]2 years ago
6 0
(6/5.55) * 100 = 108.11 %

The percent error is 108.11 %
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The pretzels are 20% of the total weight of the snack.

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b) 0.0478 (or 4.78%)

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We need to find the area under the Normal curve with a mean of 4.5 mm and a standard deviation of 1.5 mm to the left of 3 mm

<h3>(See picture 1 attached) </h3>

We can do that with the help of a calculator or a spreadsheet.

<em>In Excel use </em>

<em>NORMDIST(3,4.5,1.5,1) </em>

<em>In OpenOffice Calc use  </em>

<em>NORMDIST(3;4.5;1.5;1) </em>

and we get the value 0.1587 (or 15.87%)

(b) the thickness is more than 7.0 mm

Now we need the area to the right of 7 (1 - the area to the left of 7)

<h3>(See picture 2 attached) </h3>

<em>In Excel use </em>

<em>1-NORMDIST(7,4.5,1.5,1)  </em>

<em>In OpenOffice Calc use  </em>

<em>1-NORMDIST(7;4.5;1.5;1) </em>

and we get the value 0.0478 (or 4.78%)

(c) the thickness is between 3.0 mm and 7.0 mm

We are looking for the area between 3 and 7

<h3>(See Picture 3) </h3>

Since the area under the Normal equals 1, we have

Area to the left of 3 + Area between 3 and 7 + Area to the right of 7 = 1

Hence,  

0.1587 +  Area between 3 and 7 + 0.0478 = 1

and

Area between 3 and 7 = 1 - 0.1587 - 0.0478 = 0.7935 (or 79.35%)

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3 years ago
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