What is the equivalent expression to y+ 2/5y

ZE is the angle bisector of AngleYEX and the perpendicular bisector of Line segment G F. Line segment G X is the angle bisector

frozen [14]

Answer:

Point A is the center of the circle that passes through points E, F, and G and the center of the circle that passes through points X, Y, and Z.

Step-by-step explanation:

A is the intersection of angle bisectors, so is the incenter of triangle EFG. It is also the intersection of the perpendicular bisectors of the sides of triangle EFG, so is the circumcenter.

The altitudes at X, Y, and Z are perpendicular to sides EF, EG, and FG, and pass through the incenter, so X, Y, Z are points on the incircle.

A is the center of circles through E, F, and G, and through X, Y, and Z.

6 0

3 years ago

Read 2 more answers

Find sum or difference.<br><br>-1/2 + 1/3

Free_Kalibri [48]

2/5 is the answer unless your multiplying

6 0

3 years ago

Which equation is equivalent to the formula below

dusya [7]

Answer:

C

Step-by-step explanation:

6 0

3 years ago

An accounting firm is planning for the next tax preparation season. From last years returns, the firm collects a systematic rand

Elena L [17]

Answer:

a)From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error for the mean would be:

$\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14$

b) We want this probability:

$P(\bar X >120)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143$

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

$P( z>2.143) = 1-P(Z$

Step-by-step explanation:

Previous concepts

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".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Part a

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error for the mean would be:

$\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14$

Part b

We want this probability:

$P(\bar X >120)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143$

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

$P( z>2.143) = 1-P(Z$

4 0

3 years ago

WILL MARK BRAINLIEST

slamgirl [31]

Answer:

down 6,right 4, rotate 180 degrees assuming we are starting from the blue heart.

Step-by-step explanation:

4 0

2 years ago