Discribe two different strategies to add 16+35+24+14

In circle F with m \angle EFG= 122m∠EFG=122 and EF=16EF=16 units, find the length of arc EG. Round to the nearest hundredth.

kow [346]

Given:

In circle F, $m\angle EFG=122^\circ,\ EF=16$ units.

To find:

The length of arc EG.

Solution:

It is given that, $m\angle EFG=122^\circ,\ EF=16$ units. It means the central angle of arc EG is 122 degrees and the radius of the circle F is 16 units.

Formula for arc length is:

$s=2\pi r\times \dfrac{\theta}{360^\circ }$

Where, r is the radius and $\theta$ is the central angle in degrees.

Putting $r=16,\ \theta=122^\circ,\ \pi=3.14$ , we get

$s=2(3.14)(16)\times \dfrac{122^\circ}{360^\circ }$

$s=100.48\times \dfrac{61}{180}$

$s=34.0515556$

$s\approx 34.05$

Therefore, the length of arc EG is 34.05 units.

6 0

3 years ago

The episodes of a TV program are each 50 minutes long.

Nimfa-mama [501]

Answer:

3 hours 20 minutes

Step-by-step explanation:

first multiply 50 by how many episodes were watched

50 * 4 = 200

then devide it by 60 (60 minutes in an hr)

200 / 60 = 3 with a remainder of 1/3

meaning that it is 3 1/3 hours, to convert this into minutes, just devide a full hour (60 minutes) by 3

60 / 3 = 20

therefore the answer is 3 hours and 20 minutes

6 0

3 years ago

Which transformation will always create congruent figures

Inga [223]

Rotations, reflections, and translations are isometric. That means that these transformations do not change the size of the figure. If the size and shape of the figureis not changed, then the figures are congruent.

5 0

3 years ago

I need help with this question

pickupchik [31]

The slope is 30... each week he gains 30 dollars

8 0

3 years ago

Read 2 more answers

SAT scores are normally distributed with a mean of 1,500 and a standard deviation of 300. An administrator at a college is inter

Rama09 [41]

Answer:

The administrator should sample 968 students.

Step-by-step explanation:

We have to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.88}{2} = 0.06$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a p-value of $1 - 0.06 = 0.94$ , so Z = 1.555.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Standard deviation of 300.

This means that $n = 300$

If the administrator would like to limit the margin of error of the 88% confidence interval to 15 points, how many students should the administrator sample?

This is n for which M = 15. So

$M = z\frac{\sigma}{\sqrt{n}}$

$15 = 1.555\frac{300}{\sqrt{n}}$