Which of the following does extrema represent

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

3 0

3 years ago

Read 2 more answers

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.

6 0

3 years ago

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.

7 0

3 years ago

Read 2 more answers

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