All categories

2 years ago

When constructing inscribed polygons and parallel lines, how are the steps similar?

Mathematics

1 answer:

BlackZzzverrR [31]2 years ago

3 0

Don’t click the link they can get your address

You might be interested in

Chris borrows $9,000 at 8% annual simple interest. How much is owed after 1/2 of a year?

sergeinik [125]

Answer:

$360

Step-by-step explanation:

I am pretty sure this is the answer i am not that sure. But $9,000 at 8% is 760

and 360 is half of 760

3 0

2 years ago

Helppppppp me?????????????

egoroff_w [7]

Answer:

last option bottom right

x can be equal or greater than -2

4 0

2 years ago

Plz help will Mark brainlest!!

Verdich [7]

Answer:

Step-by-step explanation:

Well considering how the graph looks, from the x-axis(0,0), the line(x) moves upwards on the left before itf(x) moves downwards on the right

3 0

2 years ago

What should the following equation be multiplied by in order to eliminate the fractions? x/2 + x/3 = 25/3

tatuchka [14]

Answer:

Step-by-step explanation:

4 0

3 years ago

Use the information provided to determine a 95% confidence interval for the population variance. A researcher was interested in

Leno4ka [110]

Answer:

The 95% confidence interval for the population variance is (8.80, 32.45).

Step-by-step explanation:

The (1 - α)% confidence interval for the population variance is given as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

It is provided that:

n = 20

s = 3.9

Confidence level = 95%

⇒ α = 0.05

Compute the critical values of Chi-square:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2, (20-1)}=\chi^{2}_{0.025,19}=32.852\\\\\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.05/2, (20-1)}=\chi^{2}_{0.975,19}=8.907$

*Use a Chi-square table.

Compute the 95% confidence interval for the population variance as follows:

$\frac{(n-1)\cdot s^{2}}{\chi^{2}_{\alpha/2}}\leq \sigma^{2}\leq \frac{(n-1)\cdot s^{2}}{\chi^{2}_{1-\alpha/2}}$

$\frac{(20-1)\cdot (3.9)^{2}}{32.852}\leq \sigma^{2}\leq \frac{(20-1)\cdot (3.9)^{2}}{8.907}\\\\8.7967\leq \sigma^{2}\leq 32.4453\\\\8.80\leq \sigma^{2}\leq 32.45$

Thus, the 95% confidence interval for the population variance is (8.80, 32.45).

4 0

2 years ago