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Naddik [55]
3 years ago
5

A square is divided into smaller squares and portions are shaded. What is the are of the shaded portion?​

Mathematics
2 answers:
mariarad [96]3 years ago
8 0
The shaded are is 6/16
jasenka [17]3 years ago
8 0

Check the picture below.

and surely you know how much that is.

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A simple random sample from a population with a normal distribution of 99 body temperatures has xbar = 99.10°F and s = 0.64°F. C
motikmotik

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Since we're needing a 99% confidence interval, we need to know what the positive z-score associated with this two-tailed area under the normal curve is.  This can be a little tricky if you're using a standard normal table.  What you want is a two-tailed interval that has an area of .99.  What this means is that the remaining 0.01 is divided into 0.005 on each end.  This then means that, to the right of the mean (which is 0), the area is 0.005 less than the total right-half area of 0.5, or 0.495.  The total cumulative area, then, includes this plus the left half of .5, or 0.5 + 0.495 = .0995.

Then, if all we have to go on is a cumulative standard normal table, then we need to find the area closest to 0.995 and find the corresponding z-score.  This z-score is 2.58.

Now that we have the z-score, we can now plug in all the values into the formula.  

e = (2.58 · 0.66)/√102 = 0.1686.

So the 99% confidence interval will be that far above and below the mean (which is 98.8).  So the lower limit is 98.8 - 0.1686 = 98.631, and the upper limit is 98.8 + 0.1686 = 98.967.

The TI-84 method:

This is much easier, if you're allowed to use this technology.  You begin with the STAT button, and then move over to the TESTS menu.  Then select ZInterval.  We have summary stats rather than data in this problem, so we select Stats as the input.  For the standard deviation, we put 0.66, for x-bar 98.8, for n 102, and for C-level 0.99.  Then select Calculate, and it will provide the interval at the top of the screen (98.632, 98.968).  Note that the lower limit is slightly different by this method.  This is because, when doing it by hand, we had to approximate the z-score.  This created a rounding error, albeit  a small one.

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