Ask question

Ask question

All categories

Mekhanik [1.2K]

2 years ago

11

An inscribed angle in a circle intercepts an arc whose measure is greater than 180

1 answer:

never [62]2 years ago

4 0

Answer:

The correct option is;

Never acute

Step-by-step explanation:

When an angle intercepts an ark, we have;

The legs of the angle meets with the endpoints of the arc

Therefore, given that the arc is greater than 180°, and that the angle of the arc is the angle defined at the center of the circle, while the angle defined by the inscribed angle is defined at the circumference of the circle

By circle theorem, the angle at the center is twice that at the circumference, whereby the angle of the arc > 180°, the angle at the subtended by the inscribed angle will > 180°/2, which is > 90° or an obtuse angle

You might be interested in

I need help please!!!

Alenkinab [10]

Answer:D

Step-by-step explanation:

5 0

2 years ago

PLZZZZZZZZZZZZ HELP THANKS :D

svetoff [14.1K]

Sdfff bhbhhb. Dddffff

3 0

3 years ago

Read 2 more answers

One method of indirect measurement involves setting up a right triangle and measuring one of the

Alekssandra [29.7K]

Obligue I think it should be

8 0

2 years ago

Read 2 more answers

Write in point-slope form, slope-intercept form, and standard form an equation that passes through (-1,2) with slope 4.

N76 [4]

point slope form:y-2=4(x+1)

then distribute 4 and simplify y-2=4x+4 , y=4x+6 is the slope -y intercept form

then, move 4x on the left side, you get -4x+y=6 , multiply every term by -1, you get 4x-y=-6 which is standard form.

4 0

2 years ago

A chemist examines 12 sedimentary samples for bromide concentraction. The mean bromide concentration for the sample date is 0.43

Umnica [9.8K]

Answer:

a) $z = 1.645$

b) Lower endpoint: 0.422cc/m³

Upper endpoint: 0.452 cc/m³

Step-by-step explanation:

Population is approximately normal, so we can find the normal confidence interval.

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

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$ . This is the critical value, the answer for a).

Now, find 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.

$M = 1.645*\frac{0.0325}{\sqrt{12}} = 0.015$

The lower end of the interval is the sample mean subtracted by M. So it is 0.437 - 0.015 = 0.422cc/m³.

The upper end of the interval is the sample mean added to M. So it is 0.437 + 0.015 = 0.452 cc/m³.

b)

Lower endpoint: 0.422cc/m³

Upper endpoint: 0.452 cc/m³

4 0

2 years ago