Use 3 for at Volume of a Cylinder

Carlos and Diya perform the same transformations on

KengaRu [80]

Answer: Carlos is right

Step-by-step explanation:

why because in dilation, the image and the original are similar, in that they are the same shape but not necessarily the same size. They are not congruent because that requires them to be the same shape and the same size, which they are not (unless the scale factor happens to be 1.0).

Diya is wrong because the picture is the not the same size && the image is going to stay the same

8 0

3 years ago

What shape would you get if you cut off one of the corners of a rectangular prism

Svetlanka [38]

Answer:

A Pyramid

Step-by-step explanation:

3 0

3 years ago

A digital camcorder repair service has set a goal not to exceed an average of 5 working days from the time the unit is brought i

garri49 [273]

Answer:

The degrees of freedom first given by:

$df=n-1=12-1=11$

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

$t_{\alpha}= 1.796$

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

Step-by-step explanation:

Information given

5, 7, 4, 6, 7, 5, 5, 6, 4, 4, 7, 5.

System of hypothesis

We want to test if the true mean is higher than 5, the system of hypothesis are :

Null hypothesis: $\mu \leq 5$

Alternative hypothesis: $\mu > 5$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

The degrees of freedom first given by:

$df=n-1=12-1=11$

Then we can find the critical value taking in count the degrees of freedom and the alternative hypothesis and then we need to find a critical value who accumulates 0.05 of the area in the right tail and we got:

$t_{\alpha}= 1.796$

And for this case the rejection region would be:

b) Reject H0 if tcalc >1.7960

6 0

3 years ago

Math:

Dahasolnce [82]

Answer:

a) $x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$ , $x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$ , b) $x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$ , $x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$