Carly has 15 stuffed animals. How many ways are there to organize her top 3 favorites?

Which of the following are exterior angles? Check all that apply.

3241004551 [841]

Angles 4, 6, and 2. Pls mark me brainliest. And have a great life ahead! If you mark me brainliest, it will really make me happy. Thank you so much!

6 0

2 years ago

Would love some help

natali 33 [55]

<h3>Answer: 90 cubic meters</h3>

Each triangle has a base of 10 and height of 3.6

area of triangle = 0.5*base*height = 0.5*10*3.6 = 18 square meters

multiply this with the height of the prism (5 meters) to get 18*5 = 90 cubic meters. I'm using the idea that,

volume of prism = (area of base face)*(height of prism)

The notation "m^3" is shorthand for "cubic meters"

We do not use the other dimensions of 9 m and 7 m.

7 0

3 years ago

Assume that the amount of beverage in a randomly selected 16-ounce beverage can has a normal distribution. Compute a 99% confide

ioda

Answer:

The question is incomplete, but the step-by-step procedures are given to solve the question.

Step-by-step explanation:

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

$\alpha = \frac{1 - 0.99}{2} = 0.005$

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

That is z with a pvalue of $1 - 0.005 = 0.995$ , so Z = 2.575.

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.

$M = 2.575\frac{\sigma}{\sqrt{n}}$

The lower end of the interval is the sample mean subtracted by M.

The upper end of the interval is the sample mean added to M.

The 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (lower end, upper end).

7 0

3 years ago

Put in least to highest order 6.86,6.8,7,6.9,6.827 and 12.03,1.2,12.3,1.203,12.301

Anon25 [30]

6.8,6.827,6.86,6.9

1.2,1.203,12.03, 12.301,12.31

6 0

3 years ago

An Article a Florida newspaper reported on the topics that teenagers most want to discuss with their parents. The findings, the

Sergio039 [100]

Answer:

The estimate is $P__{hat}} \pm E = 0.37 \pm 0.0348$

Step-by-step explanation:

From the question we are told that

The sample size is n = 522

The sample proportion of students would like to talk about school is $\r p__{hat}} = 0.37$

Given that the confidence level is 90 % then the level of significance can be mathematically evaluated as

$\alpha = 100 - 90$

$\alpha = 10\%$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{\alpha }{2} } =Z_{\frac{0.10}{2} } = 1.645$

Generally the margin of error can be mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\r P_{hat}(1- \r P_{hat} )}{n } }$

=> $E = 1.645 * \sqrt{\frac{0.37 (1- 0.37 )}{522 } }$

=> $E = 0.0348$

Generally the estimate the proportion of all teenagers who want more family discussions about school at 90% confidence level is

$P__{hat}} \pm E$

substituting values

$0.37 \pm 0.0348$

5 0

3 years ago