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Flauer [41]
3 years ago
9

The deepest part of a swimming pool is 12 feet deep the shallowest part of the pool is 3 feet deep. what is the ratio of the dep

th of the deepest part of the pool to the depth of the shallowest part of the pool?
Mathematics
2 answers:
Alexeev081 [22]3 years ago
8 0

Answer:

the answer to the question is 12:3

lora16 [44]3 years ago
7 0

Answer:

12 : 3

Step-by-step explanation:

You put the depth of the deep before the depth of the shallowness to get the ratio.  The ratio could be written like,

12 : 3

12 / 3

12 to 3

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it is true; just work them out, you should get what they got :))

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How to find the gradient of line segment joining each pair of points?
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Answer:

2

1/3

Step-by-step explanation:

A(xA, yA) = A (3,1)

B(xB,yB)=B(5,5)

the gradient=(yB-yA)/ (xB-xA)=(5-1)/(5-3)=4/2=2

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6 0
2 years ago
A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural freq
oksano4ka [1.4K]

Answer:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n=5 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

And for this case we know that the 95% confidence interval is given by:

\bar X=\frac{233.002 +229.266}{2}= 231.134

And the margin of error is given by:

ME = \frac{233.002 -229.266}{2}= 1.868

And the margin of error is given by:

ME= t_{\alpha/2} \frac{s}{\sqrt{n}}

The degrees of freedom are given by:

df = n-1 = 5-1=4

And the critical value for 95% of confidence is t_{\alpha/2}= 2.776

So then we can find the deviation like this:

s = \frac{ME \sqrt{n}}{t_{\alpha/2}}

s = \frac{1.868* \sqrt{5}}{2.776}= 1.506

And for the 99% confidence the critical value is: t_{\alpha/2}= 4.604

And the margin of error would be:

ME = 4.604 *\frac{1.506}{\sqrt{5}}= 3.098

And the interval is given by:

Lower = 231.134- 3.098=228.036

Upper = 231.134+ 3.098=234.232

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2 years ago
Carrie collected shells from the seashore in a box he takes out a handful of shells from the box is this a random sample of shel
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This would be random. This is because Carrie has no clue or has a limited knowledge of what type of shells he would grab. 
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