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Sliva [168]
3 years ago
12

What is the algebraic description for a transformation that translates the

Mathematics
1 answer:
MrRissso [65]3 years ago
7 0

Answer:

nhgtf

Step-by-step explanation:

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The answer should be less than
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I found this answer on the Internet. *If* it were an original answer, below is how I would grade it and why.
Dennis_Churaev [7]

Answer:

1) It is given that line AB is tangent to the circle at A.

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Please help on 19 and 20
Svetradugi [14.3K]

Answer:

c and d

Step-by-step explanation:

6 0
3 years ago
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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

6 0
3 years ago
Use inverse operations to solve each equation. Explain each step and identify the property used to reach each step.
maks197457 [2]
Hello,


The inverse operation's are easy.

is would be h/3-8=19


Hope this helps
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3 years ago
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