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exis [7]
3 years ago
7

Which of the following does extrema represent

Mathematics
1 answer:
Viktor [21]3 years ago
3 0

I am sorry but you don't have posted the options

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Write the linear equation in standard form.<br> 12x = -9y+ 7
scZoUnD [109]

Answer:

Step-by-step explanation:

The linear equation in standard form is written in the form :

ax + by = c

Therefore : 12x = -9y + 7 in standard form will be :

12x + 9y = 7

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3 years ago
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What’s the answer for this one??
KIM [24]

Answer:

It is the second one.

Step-by-step explanation:

a and e are equal because they are mirrored. d and b are equal because they are on the same  line and at the same point where the parallel lines go through. b and f are also mirrored. That leaves c and e which don't match.

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I need help on letter b!
Likurg_2 [28]

40 of the toys weren't dolls. The reason why is 5/8 times 64 is equal to 40.

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Tính đạo hàm của y=2x-1-căn3x-0
adoni [48]

Answer:

The derivative is

\frac{dy}{dx} = 2 -  \frac{2\sqrt3}{\sqrt x}\\

Step-by-step explanation:

The function is given by

y = 2x - 1 - \sqrt {3x}

Differentiate with respect to x, we get

\frac{dy}{dx} = 2 - 0 - \frac{2\sqrt3}{\sqrt x}\\\\\frac{dy}{dx} = 2 -  \frac{2\sqrt3}{\sqrt x}\\

7 0
3 years ago
In a random sample of 80 teenagers, the average number of texts handled in a day is 50. The 96% confidence interval for the mean
Nastasia [14]

Answer:

a) \bar X =\frac{46+54}{2}=50

And the margin of error is given by:

ME= \frac{54-46}{2}= 4

The confidence level is 0.96 and the significance level is \alpha=1-0.96=0.04 and the value of \alpha/2 =0.02 and the margin of error is given by:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

We can calculate the critical value and we got:

z_{\alpha/2} = 2.05

And if we solve for the deviation like this:

\sigma = ME * \frac{\sqrt{n}}{z_{\alpha/2}}

And replacing we got:

\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45

b) ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor \sqrt{2}

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  

\mu population mean (variable of interest)  

\sigma represent the population standard deviation  

n=80 represent the sample size  

Solution to the problem

Part a

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

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)  

For this case we can calculate the mean like this:

\bar X =\frac{46+54}{2}=50

And the margin of error is given by:

ME= \frac{54-46}{2}= 4

The confidence level is 0.96 and the significance level is \alpha=1-0.96=0.04 and the value of \alpha/2 =0.02 and the margin of error is given by:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

We can calculate the critical value and we got:

z_{\alpha/2} = 2.05

And if we solve for the deviation like this:

\sigma = ME * \frac{\sqrt{n}}{z_{\alpha/2}}

And replacing we got:

\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45

Part b

For this case is the sample size is doubled the margin of error would be:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor \sqrt{2}

5 0
2 years ago
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