All categories

2 years ago

Which pair or pairs of polygons are congruent?

Mathematics

1 answer:

riadik2000 [5.3K]2 years ago

3 0

Answer:

Step-by-step explanation:

Congruent shapes are 2 shapes that are ewual in size angles and sides, but they can be rotated or moved around in any way

You might be interested in

PTC is a substance that has a strong bitter taste for some people and is tasteless for others. The ability to taste PTC is inher

lana66690 [7]

Answer:

a) $n=\frac{0.76(1-0.76)}{(\frac{0.01}{1.64})^2}=4905.83$

And rounded up we have that n=4906

b) For this case since we don't have prior info we need to use as estimatro for the proportion $\hat p =0.5$

$n=\frac{0.5(1-0.5)}{(\frac{0.01}{1.64})^2}=6724$

And rounded up we have that n=6724

Step-by-step explanation:

We need to remember that the confidence interval for the true proportion is given by :

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

Part a

The estimated proportion for this case is $\hat p =0.76$

Our interval is at 90% of confidence, and the significance level is given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$ . The critical values for this case are:

$z_{\alpha/2}=-1.64, t_{1-\alpha/2}=1.64$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

The margin of error desired is given $ME =\pm 0.01$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Replacing we got:

$n=\frac{0.76(1-0.76)}{(\frac{0.01}{1.64})^2}=4905.83$

And rounded up we have that n=4906

Part b

For this case since we don't have prior info we need to use as estimatro for the proportion $\hat p =0.5$

$n=\frac{0.5(1-0.5)}{(\frac{0.01}{1.64})^2}=6724$

And rounded up we have that n=6724

3 0

3 years ago

What happens to ice as it melts

Ludmilka [50]

It becomes a liquid
hope this helped

3 0

3 years ago

Read 2 more answers

How is multiplication using partial products different from multiplication using regrouping?

klio [65]

Partial products is breaking down every number in multiplication and adding them. Regrouping is grouping number them adding them.They are alike because they both involve breaking down the numbers and adding them. They are different because one is taking all the numbers broken down, while the other is just a couple of the number.

5 0

3 years ago

Read 2 more answers

Graph the linear function.

nexus9112 [7]

I dont know the awnser sorry

5 0

3 years ago

What is the equation of a line, in point-slope form, that passes through (−2, −6) and has a slope of 1/3?

Alexus [3.1K]

The equation of a line in point-slope form is y-6=1/3(x-2) remember point-slope is y1-y2=m(x1-x2)

8 0

3 years ago