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3 years ago

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In a study of 250 adults, the mean heart rate was 70 beats per minute. Assume the population of heart rates is known to be appro

ximately normal with a standard deviation of 12 beats per minute. What is the 99% confidence interval for the mean beats per minute?

1 answer:

PSYCHO15rus [73]3 years ago

7 0

68 - 72!! hope this helps

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In a newspaper, it was reported that the number of yearly robberies in Springfield in

ryzh [129]

Answer:

The number of robberies in 2014 is 108

Step-by-step explanation:

Here, we want to calculate the number of robberies in 2014

from the question, the robberies went down by 40%

Let the number of robberies in 2014 be x

Thus;

40/100 * 180

= 72

This means that the value it went down was by 72

Thus;

x = 180-72

x = 108

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2 years ago

If y varies inversely as x and y=23 when x=8 find y when x=4

valina [46]

<h2>INVERSE VARIATION</h2>

<em>E</em><em>Q</em><em>U</em><em>A</em><em>T</em><em>I</em><em>O</em><em>N</em><em>:</em><em> </em><em> </em><em> </em><em> </em>

$y = \frac{k}{x}$

<em>–</em><em>–</em><em>–</em><em>–</em><em>–</em><em>–</em><em>–</em><em>–</em>

First, find the constant of variation <em>k</em> using the formula k=xy.

$23 = \frac{k}{8}$

$k = (8)(23)$

$k = 184$

<em>–</em><em>–––––––</em>

Then, find <em>y</em> if <em>x</em> is equal to four using the formula y=184/x.

$y = \frac{184}{x}$

$y = \frac{184}{4}$

$y = 46$

<em>F</em><em>i</em><em>n</em><em>a</em><em>l</em><em> </em><em>A</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>:</em>

<h2>y = 46</h2>

5 0

2 years ago

9x+14=2x+35 what is the answer to this I do not get it

son4ous [18]

$9x+14=2x+35\\
7x=21\\
x=3$

8 0

3 years ago

Read 2 more answers

The wholesale price for a dozen pairs of socks is $4.30. The retail price for the same dozen pairs of socks is $9.25. What is th

Mandarinka [93]

Answer:

the mark up used by the retailer is 115.12% of the wholesale price

Step-by-step explanation:

Mark-up is the percentage increase of the retail price over the wholesale price

Mark-up = (difference between retail and whole sale price / retail price ) x 100

difference between retail and whole sale price = $9.25 - $4.30 = $4.95

(4.95 / 4.3) x 100 = 115.12%.

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2 years ago

A sample of 66 obese adults was put on a low-carb diet for a year. The average weight loss was 11 pounds and the standard deviat

Illusion [34]

Answer:

The 99% lower confidence bound for the true average weight loss is 3.98 pounds.

This means that we can be 99% sure that the mean weight loss for all obese adults on a low-carb diet is positive.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.005 = 0.995$ , so $z = 2.575$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.575*\frac{19}{\sqrt{66}} = 6.02$

The lower end of the interval is the sample mean subtracted by M. So it is 11 - 6.02 = 3.98 pounds

The 99% lower confidence bound for the true average weight loss is 3.98 pounds.

This means that we can be 99% sure that the mean weight loss for all obese adults on a low-carb diet is positive.

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2 years ago

Read 2 more answers