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Alexus [3.1K]
2 years ago
10

Solve for y* w + py = e

Mathematics
1 answer:
Dmitry [639]2 years ago
3 0
69429 loser haha your going to fail
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A sample of blood pressure measurements is taken for a group of​ adults, and those values​ (mm Hg) are listed below. The values
nika2105 [10]

Answer:

For systolic pressure data:

\bar X =127.2\\\\S=17.91668\\\\CV=\frac{17.91668}{127.2}=0.14085

For diastolic pressure data:

\bar X =73.5\\\\S=12.54\\\\CV=\frac{73.5}{12.54}=0.17061

Systolic pressure is slightly less variable, among individuals in the sample, than diastolic pressure.

Step-by-step explanation:

The coefficient of variation is defined as the percentage relative variation of a set of data with respect to its average. And it is calculated like this:

CV=\frac{\bar X}{S}

S =\sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i-\bar{x})^2

\bar X={\frac{1}{n} \sum_{i=1}^n x_i

For systolic pressure data:

\bar X =127.2\\\\S=17.91668\\\\CV=\frac{17.91668}{127.2}=0.14085

For diastolic pressure data:

\bar X =73.5\\\\S=12.54\\\\CV=\frac{12.54}{73.5}=0.17061

It is observed that the systolic pressure shows greater standard deviation but less coefficient of variation. This is due to the greater magnitude of its measurement scale.

Systolic pressure is slightly less variable, among individuals in the sample, than diastolic pressure.

7 0
3 years ago
Between which two consecutive integers does √ 350 lie​
Annette [7]

Answer:

18 and 19

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
If 5 bulbs are $24.95 how much money is 1 bulb?
Elden [556K]

Answer:

4.99

Step-by-step explanation:

If you divide:

$24.95/5bulbs

you get

4.99/1 bulb  

5 0
3 years ago
By switching service providers, a family’s monthly bill decreased from $57 in May to
gavmur [86]

Answer:

38.6%

Step-by-step explanation:

8 0
3 years ago
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

The general formula for the margin of error is e = (zs)/√n, when the sample size is 30 or more (otherwise, we'd need to to a t-interval).

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.

4 0
3 years ago
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